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Instructor’s Solution Manual
for
Fundamentals of Physics, 6/E
by Halliday, Resnick, and Walker
James B. Whitenton
Southern Polytechnic State University
ii
Preface
This booklet includes the solutions relevant to the EXERCISES & PROBLEMS sections of the 6th edition
of Fundamentals of Physics, by Halliday, Resnick, and Walker. We also include solutions to problems in
the Problems Supplement. We have not included solutions or discussions which pertain to the QUESTIONS
sections.
I am very grateful for helpful input from J. Richard Christman, Meighan Dillon, Barbara Moore, and Jearl
Walker regarding the development of this document.
iii
iv PREFACE
Chapter 1
1. The metric prefixes (micro, pico, nano, . . .) are given for ready reference on the inside front cover of the
textbook (see also Table 1-2).
(a) Since 1 km = 1× 103m and 1m = 1× 106 µm,
1 km = 103m = (103m)(106 µm/m) = 109 µm .
The given measurement is 1.0 km (two significant figures), which implies our result should be
written as 1.0× 109 µm.
(b) We calculate the number of microns in 1 centimeter. Since 1 cm = 10−2m,
1 cm = 10−2m = (10−2m)(106 µm/m) = 104 µm .
We conclude that the fraction of one centimeter equal to 1.0µm is 1.0× 10−4.
(c) Since 1 yd = (3 ft)(0.3048m/ft) = 0.9144m,
1.0 yd = (0.91m)(106 µm/m) = 9.1× 105 µm .
2. The customer expects 20× 7056 in3 and receives 20× 5826 in3, the difference being 24600 cubic inches,
or (
24600 in3
)(2.54 cm
1 inch
)3(
1 L
1000 cm3
)
= 403 L
where Appendix D has been used (see also Sample Problem 1-2).
3. Using the given conversion factors, we find
(a) the distance d in rods to be
d = 4.0 furlongs =
(4.0 furlongs)(201.168m/furlong)
5.0292m/rod
= 160 rods ,
(b) and that distance in chains to be
d =
(4.0 furlongs)(201.168m/furlong)
20.117m/chain
= 40 chains .
4. (a) Recalling that 2.54 cm equals 1 inch (exactly), we obtain
(0.80 cm)
(
1 inch
2.54 cm
)(
6 picas
1 inch
)(
12 points
1 pica
)
≈ 23 points ,
(b) and
(0.80 cm)
(
1 inch
2.54 cm
)(
6 picas
1 inch
)
≈ 1.9 picas .
1
2 CHAPTER 1.
5. Various geometric formulas are given in Appendix E.
(a) Substituting
R =
(
6.37× 106m) (10−3 km/m) = 6.37× 103 km
into circumference= 2πR, we obtain 4.00× 104 km.
(b) The surface area of Earth is
4πR2 = 4π
(
6.37× 103 km)2 = 5.10× 108 km2 .
(c) The volume of Earth is
4π
3
R3 =
4π
3
(
6.37× 103 km)3 = 1.08× 1012 km3 .
6. (a) Using the fact that the area A of a rectangle is width×length, we find
Atotal = (3.00 acre) + (25.0 perch)(4.00 perch)
= (3.00 acre)
(
(40 perch)(4 perch)
1 acre
)
+ 100 perch2
= 580 perch2 .
We multiply this by the perch2 → rood conversion factor (1 rood/40 perch2) to obtain the answer:
Atotal = 14.5 roods.
(b) We convert our intermediate result in part (a):
Atotal = (580 perch
2)
(
16.5 ft
1 perch
)2
= 1.58× 105 ft2 .
Now, we use the feet→ meters conversion given in Appendix D to obtain
Atotal =
(
1.58× 105 ft2)( 1 m
3.281 ft
)2
= 1.47× 104 m2 .
7. The volume of ice is given by the product of the semicircular surface area and the thickness. The
semicircle area is A = πr2/2, where r is the radius. Therefore, the volume is
V =
π
2
r2 z
where z is the ice thickness. Since there are 103 m in 1 km and 102 cm in 1 m, we have
r = (2000 km)
(
103m
1km
)(
102 cm
1m
)
= 2000× 105 cm .
In these units, the thickness becomes
z = (3000m)
(
102 cm
1m
)
= 3000× 102 cm .
Therefore,
V =
π
2
(
2000× 105 cm)2 (3000× 102 cm) = 1.9× 1022 cm3 .
8. The total volume V of the real house is that of a triangular prism (of height h = 3.0 m and base area
A = 20× 12 = 240 m2) in addition to a rectangular box (height h′ = 6.0 m and same base). Therefore,
V =
1
2
hA+ h′A =
(
h
2
+ h′
)
A = 1800 m3 .
3
(a) Each dimension is reduced by a factor of 1/12, and we find
Vdoll =
(
1800 m3
)( 1
12
)3
≈ 1.0 m3 .
(b) In this case, each dimension (relative to the real house) is reduced by a factor of 1/144. Therefore,
Vminiature =
(
1800 m3
)( 1
144
)3
≈ 6.0× 10−4 m3 .
9. We use the conversion factors found in Appendix D.
1 acre · ft = (43, 560 ft2) · ft = 43, 560 ft3 .
Since 2 in. = (1/6) ft, the volume of water that fell during the storm is
V = (26 km2)(1/6 ft) = (26 km2)(3281 ft/km)2(1/6 ft) = 4.66× 107 ft3 .
Thus,
V =
4.66× 107 ft3
4.3560× 104 ft3/acre · ft = 1.1× 10
3 acre · ft .
10. The metric prefixes (micro (µ), pico, nano, . . .) are given for ready reference on the inside front cover of
the textbook (also, Table 1-2).
1µcentury =
(
10−6 century
)( 100 y
1 century
)(
365 day
1 y
)(
24 h
1 day
)(
60min
1 h
)
= 52.6min .
The percent difference is therefore
52.6min− 50min
50min
= 5.2% .
11. We use the conversion factors given in Appendix D and the definitions of the SI prefixes given in Table 1-
2 (also listed on the inside front cover of the textbook). Here, “ns” represents the nanosecond unit, “ps”
represents the picosecond unit, and so on.
(a) 1m = 3.281 ft and 1 s = 109 ns. Thus,
3.0× 108m/s =
(
3.0× 108m
s
)(
3.281 ft
m
)( s
109 ns
)
= 0.98 ft/ns .
(b) Using 1m = 103mm and 1 s = 1012 ps, we find
3.0× 108m/s =
(
3.0× 108m
s
)(
103mm
m
)(
s
1012 ps
)
= 0.30 mm/ps .
12. The number of seconds in a year is 3.156×107. This is listed in Appendix D and results from the product
(365.25 day/y)(24 h/day)(60min/h)(60 s/min) .
(a) The number of shakes in a second is 108; therefore, there are indeed more shakes per second than
there are seconds per year.
4 CHAPTER 1.
(b) Denoting the age of the universe as 1 u-day (or 86400 u-sec), then the time during which humans
have existed is given by
106
1010
= 10−4 u-day ,
which we may also express as
(
10−4 u-day
)(86400 u-sec
1 u-day
)
= 8.6 u-sec .
13. None of the clocks advance by exactly 24 h in a 24-h period but this is not the most important criterion
for judging their quality for measuring time intervals. What is important is that the clock advance by
the same amount in each 24-h period. The clock reading can then easily be adjusted to give the correct
interval. If the clock reading jumps around from one 24-h period to another, it cannot be corrected since
it would impossible to tell what the correction should be. The following gives the corrections (in seconds)
that must be applied to the reading on each clock for each 24-h period. The entries were determined by
subtracting the clock reading at the end of the interval from the clock reading at the beginning.
CLOCK Sun. Mon. Tues. Wed. Thurs. Fri.
-Mon. -Tues. -Wed. -Thurs. -Fri. -Sat
A −16 −16 −15 −17 −15 −15
B −3 +5 −10 +5 +6 −7
C −58 −58 −58 −58 −58 −58
D +67 +67 +67 +67 +67 +67
E +70 +55 +2 +20 +10 +10
Clocks C and D are both good timekeepers in the sense that each is consistent in its daily drift (relative
to WWF time); thus, C and D are easily made “perfect” with simple and predictable corrections. The
correction for clock C is less than the correction for clock D, so we judge clock C to be the best and
clock D to be the next best. The correction that must be applied to clock A is in the range from 15 s
to 17s. For clock B it is the range from −5 s to +10 s, for clock E it is in the range from −70 s to −2 s.
After C and D, A has the smallest range of correction, B has the next smallest range, and E has the
greatest range. From best the worst, the ranking of the clocks is C, D, A, B, E.
14. The time on any of these clocks is a straight-line function of that on another, with slopes 6= 1 and
y-intercepts 6= 0. From the data in the figure we deduce
tC =
2
7
tB +
594
7
tB =
33
40
tA − 662
5
.
These are
used in obtaining the following results.
(a) We find
t′B − tB =
33
40
(t′A − tA) = 495 s
when t′A − tA = 600 s.
(b) We obtain
t′C − tC =
2
7
(t′B − tB) =
2
7
(495) = 141 s .
(c) Clock B reads tB = (33/40)(400)− (662/5) ≈ 198 s when clock A reads tA = 400 s.
(d) From tC = 15 = (2/7)tB + (594/7), we get tB ≈ −245 s.
5
15. We convert meters to astronomical units, and seconds to minutes, using
1000m = 1 km
1AU = 1.50× 108 km
60 s = 1min .
Thus, 3.0× 108m/s becomes(
3.0× 108m
s
)(
1 km
1000m
)(
AU
1.50× 108 km
)(
60 s
min
)
= 0.12AU/min .
16. Since a change of longitude equal to 360◦ corresponds to a 24 hour change, then one expects to change
longitude by 360◦/24 = 15◦ before resetting one’s watch by 1.0 h.
17. The last day of the 20 centuries is longer than the first day by
(20 century)(0.001 s/century) = 0.02 s .
The average day during the 20 centuries is (0 + 0.02)/2 = 0.01 s longer than the first day. Since the
increase occurs uniformly, the cumulative effect T is
T = (average increase in length of a day)(number of days)
=
(
0.01 s
day
)(
365.25 day
y
)
(2000 y)
= 7305 s
or roughly two hours.
18. We denote the pulsar rotation rate f (for frequency).
f =
1 rotation
1.55780644887275× 10−3 s
(a) Multiplying f by the time-interval t = 7.00 days (which is equivalent to 604800 s, if we ignore
significant figure considerations for a moment), we obtain the number of rotations:
N =
(
1 rotation
1.55780644887275× 10−3 s
)
(604800 s) = 388238218.4
which should now be rounded to 3.88 × 108 rotations since the time-interval was specified in the
problem to three significant figures.
(b) We note that the problem specifies the exact number of pulsar revolutions (one million). In this
case, our unknown is t, and an equation similar to the one we set up in part (a) takes the form
N = ft
1× 106 =
(
1 rotation
1.55780644887275× 10−3 s
)
t
which yields the result t = 1557.80644887275 s (though students who do this calculation on their
calculator might not obtain those last several digits).
(c) Careful reading of the problem shows that the time-uncertainty per revolution is ±3 × 10−17 s.
We therefore expect that as a result of one million revolutions, the uncertainty should be (±3 ×
10−17)(1 × 106) = ±3× 10−11 s.
6 CHAPTER 1.
19. IfME is the mass of Earth,m is the averagemass of an atom in Earth, andN is the number of atoms, then
ME = Nm or N =ME/m. We convert mass m to kilograms using Appendix D (1 u = 1.661×10−27 kg).
Thus,
N =
ME
m
=
5.98× 1024 kg
(40 u)(1.661× 10−27 kg/u) = 9.0× 10
49 .
20. To organize the calculation, we introduce the notion of density (which the students have probably seen
in other courses):
ρ =
m
V
.
(a) We take the volume of the leaf to be its area A multiplied by its thickness z. With density
ρ = 19.32 g/cm3 and mass m = 27.63 g, the volume of the leaf is found to be
V =
m
ρ
= 1.430 cm3 .
We convert the volume to SI units:
(
1.430 cm3
)( 1 m
100 cm
)3
= 1.430× 10−6 m3 .
And since V = Az where z = 1× 10−6 m (metric prefixes can be found in Table 1-2), we obtain
A =
1.430× 10−6m3
1× 10−6m = 1.430 m
2 .
(b) The volume of a cylinder of length ℓ is V = Aℓ where the cross-section area is that of a circle:
A = πr2. Therefore, with r = 2.500× 10−6 m and V = 1.430× 10−6m3, we obtain
ℓ =
V
πr2
= 7.284× 104 m .
21. We introduce the notion of density (which the students have probably seen in other courses):
ρ =
m
V
and convert to SI units: 1 g = 1× 10−3 kg.
(a) For volume conversion, we find 1 cm3 = (1 × 10−2m)3 = 1 × 10−6m3. Thus, the density in kg/m3
is
1 g/cm3 =
(
1 g
cm3
)(
10−3 kg
g
)(
cm3
10−6m3
)
= 1× 103 kg/m3 .
Thus, the mass of a cubic meter of water is 1000 kg.
(b) We divide the mass of the water by the time taken to drain it. The mass is found from M = ρV
(the product of the volume of water and its density):
M = (5700m3)(1× 103 kg/m3) = 5.70× 106 kg .
The time is t = (10 h)(3600 s/h) = 3.6× 104 s, so the mass flow rate R is
R =
M
t
=
5.70× 106 kg
3.6× 104 s = 158 kg/s .
7
22. The volume of the water that fell is
V = (26 km2)(2.0 in.)
= (26 km2)
(
1000m
1 km
)2
(2.0 in.)
(
0.0254m
1 in.
)
= (26× 106m2)(0.0508m)
= 1.3× 106 m3 .
We write the mass-per-unit-volume (density) of the water as:
ρ =
m
V
= 1× 103 kg/m3 .
The mass of the water that fell is therefore given by m = ρV :
m =
(
1× 103 kg/m3
) (
1.3× 106m3)
= 1.3× 109 kg .
23. We introduce the notion of density (which the students have probably seen in other courses):
ρ =
m
V
and convert to SI units: 1000 g = 1 kg, and 100 cm = 1 m.
(a) The density ρ of a sample of iron is therefore
ρ =
(
7.87 g/cm
3
)( 1 kg
1000 g
)(
100 cm
1m
)3
which yields ρ = 7870 kg/m3. If we ignore the empty spaces between the close-packed spheres, then
the density of an individual iron atom will be the same as the density of any iron sample. That is,
if M is the mass and V is the volume of an atom, then
V =
M
ρ
=
9.27× 10−26 kg
7.87× 103 kg/m3 = 1.18× 10
−29 m3 .
(b) We set V = 4πR3/3, where R is the radius of an atom (Appendix E contains several geometry
formulas). Solving for R, we find
R =
(
3V
4π
)1/3
=
(
3(1.18× 10−29m3)
4π
)1/3
= 1.41× 10−10m .
The center-to-center distance between atoms is twice the radius, or 2.82× 10−10 m.
24. The metric prefixes (micro (µ), pico, nano, . . .) are given for ready reference on the inside front cover of the
textbook (see also Table 1-2). The surface area A of each grain of sand of radius r = 50µm = 50×10−6 m
is given by A = 4π(50×10−6)2 = 3.14×10−8 m2 (Appendix E contains a variety of geometry formulas).
We introduce the notion of density (which the students have probably seen in other courses):
ρ =
m
V
so that the mass can be found from m = ρV , where ρ = 2600 kg/m3. Thus, using V = 4πr3/3, the mass
of each grain is
m =
(
4π
(
50× 10−6m)3
3
)(
2600
kg
m3
)
= 1.36× 10−9 kg .
8 CHAPTER 1.
We observe that (because a cube has six equal faces) the indicated surface area is 6 m2. The number of
spheres (the grains of sand) N which have a total surface area of 6 m2 is given by
N =
6m2
3.14× 10−8m2 = 1.91× 10
8 .
Therefore, the total mass M is given by
M = Nm =
(
1.91× 108) (1.36× 10−9 kg) = 0.260 kg .
25. From the Figure we see that, regarding differences in positions ∆x, 212 S is equivalent to 258 W and
180 S is equivalent to 156 Z. Whether or not the origin of the Zelda path coincides with the origins of
the other paths is immaterial to consideration of ∆x.
(a)
∆x = (50.0 S)
(
258 W
212 S
)
= 60.8 W
(b)
∆x = (50.0 S)
(
156 Z
180 S
)
= 43.3 Z
26. The first two conversions are easy enough that a formal conversion is not especially called for, but in
the interest of practice makes perfect we go ahead and proceed formally:
(a)
(11 tuffet)
(
2 peck
1 tuffet
)
= 22 peck
(b)
(11 tuffet)
(
0.50 bushel
1 tuffet
)
= 5.5 bushel
(c)
(5.5 bushel)
(
36.3687 L
1 bushel
)
≈ 200 L
27. We make the assumption that the clouds are directly overhead, so that Figure 1-3 (and the calculations
that accompany it) apply. Following the steps in Sample Problem 1-4, we have
θ
360◦
=
t
24 h
which, for t = 38 min = 38/60 h yields θ = 9.5◦. We obtain the altitude h from the relation
d2 = r2 tan2 θ = 2rh
which is discussed in that Sample Problem, where r = 6.37×106 m is the radius of the earth. Therefore,
h = 8.9× 104 m.
9
28. In the simplest approach, we set up a ratio for the total increase in horizontal depth x (where ∆x = 0.05 m
is the increase in horizontal depth per step)
x = Nsteps∆x
=
(
4.57
0.19
)
(0.05) = 1.2 m .
However, we can approach this more carefully by noting that if there are N = 4.57/.19 ≈ 24 rises then
under normal circumstances we would expect N −1 = 23 runs (horizontal pieces) in that staircase. This
would yield (23)(0.05) = 1.15 m, which – to two significant figures – agrees with our first result.
29. Abbreviating wapentake as “wp” and assuming a hide to be 110 acres, we set up the ratio 25wp/11 barn
along with appropriate conversion factors:
(25wp)
(
100 hide
1 wp
) (
110 acre
1 hide
) (
4047m2
1 acre
)
(11 barn)
(
1×10−28 m2
1 barn
) ≈ 1× 1036 .
30. It is straightforward to compute how many seconds in a year (about 3 × 107). Now, if we estimate
roughly one breath per second (or every two seconds, or three seconds – it won’t affect the result) then
to within an order of magnitude, a person takes 107 breaths in a year.
31. A day is equivalent to 86400 seconds and a meter is equivalent to a million micrometers, so
(3.7m)(106 µm/m)
(14 day)(86400 s/day)
= 3.1 µm/s .
32. The mass in kilograms is
(28.9 piculs)
(
100 gin
1 picul
)(
16 tahil
1 gin
)(
10 chee
1 tahil
)(
10 hoon
1 chee
)(
0.3779 g
1 hoon
)
which yields 1.747× 106 g or roughly 1750 kg.
33. (a) In atomic mass units, the mass of one molecule is 16 + 1 + 1 = 18 u. Using Eq. 1-9, we find
(18 u)
(
1.6605402× 10−27 kg
1 u
)
= 3.0× 10−26 kg .
(b) We divide the total mass by the mass of each molecule and obtain the (approximate) number of
water molecules:
1.4× 1021
3.0× 10−26 ≈ 5× 10
46 .
34. (a) We find the volume in cubic centimeters
(193 gal)
(
231 in3
1 gal
)(
2.54 cm
1 in
)3
= 7.31× 105 cm3
and subtract this from 1× 106 cm3 to obtain 2.69× 105 cm3. The conversion gal→ in3 is given in
Appendix D (immediately below the table of Volume conversions).
(b) The volume found in part (a) is converted (by dividing by (100 cm/m)3) to 0.731 m3, which corre-
sponds to a mass of (
1000 kg/m3
) (
0.731m2
)
= 731 kg
using the density given in the problem statement. At a rate of 0.0018 kg/min, this can be filled in
731 kg
0.0018 kg/min
= 4.06× 105 min
which we convert to 0.77 y, by dividing by the number of minutes in a year (365 days)(24 h/day)(60min/h).
10 CHAPTER 1.
35. (a) When θ is measured in radians, it is equal to the arclength divided by the radius. For very large
radius circles and small values of θ, such as we deal with in this problem,
the arcs may be
approximated as
straight lines –
which for our
purposes corre-
spond to the di-
ameters d and
D of the Moon
and Sun, respec-
tively. Thus,
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..
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.......................................................................
θ =
d
RMoon
=
D
RSun
=⇒ RSun
RMoon
=
D
d
which yields D/d = 400.
(b) Various geometric formulas are given in Appendix E. Using rs and rm for the radius of the Sun and
Moon, respectively (noting that their ratio is the same as D/d), then the Sun’s volume divided by
that of the Moon is
4
3πr
3
s
4
3πr
3
m
=
(
rs
rm
)3
= 4003 = 6.4× 107 .
(c) The angle should turn out to be roughly 0.009 rad (or about half a degree). Putting this into the
equation above, we get
d = θRMoon = (0.009)
(
3.8× 105) ≈ 3.4× 103 km .
36. (a) For the minimum (43 cm) case, 9 cubit converts as follows:
(9 cubit)
(
0.43m
1 cubit
)
= 3.9 m .
And for the maximum (43 cm) case we obtain
(9 cubit)
(
0.53 m
1 cubit
)
= 4.8 m .
(b) Similarly, with 0.43m → 430mm and 0.53m → 530mm, we find 3.9×103 mm and 4.8×103 mm,
respectively.
(c) We can convert length and diameter first and then compute the volume, or first compute the volume
and then convert. We proceed using the latter approach (where d is diameter and ℓ is length).
Vcylinder,min =
π
4
ℓ d2 = 28 cubit3
=
(
28 cubit3
)( 0.43m
1 cubit
)3
= 2.2 m3 .
Similarly, with 0.43m replaced by 0.53m, we obtain Vcylinder,max = 4.2m
3.
37. (a) Squaring the relation 1 ken = 1.97m, and setting up the ratio, we obtain
1 ken2
1m2
=
1.972m2
1m2
= 3.88 .
11
(b) Similarly, we find
1 ken3
1m3
=
1.973m3
1m3
= 7.65 .
(c) The volume of a cylinder is the circular area of its base multiplied by its height. Thus,
πr2h = π(3.00)2(5.50) = 155.5 ken3 .
(d) If we multiply this by the result of part (b), we determine the volume in cubic meters: (155.5)(7.65) =
1.19× 103 m3.
38. Although we can look up the distance from Cleveland to Los Angeles, we can just as well (for an order of
magnitude calculation) assume it’s some relatively small fraction of the circumference of Earth – which
suggests that (again, for an order of magnitude calculation) we can estimate the distance to be roughly
r, where r ≈ 6 × 106 m is the radius of Earth. If we take each toilet paper sheet to be roughly 10 cm
(0.1 m) then the number of sheets needed is roughly 6× 106/0.1 = 6× 107 ≈ 108.
39. Using the (exact) conversion 2.54 cm = 1 in. we find that 1 ft = (12)(2.54)/100 = 0.3048 m (which also
can be found in Appendix D). The volume of a cord of wood is 8 × 4 × 4 = 128 ft3, which we convert
(multiplying by 0.30483 ) to
3.6 m3. Therefore, one cubic meter of wood corresponds to 1/3.6 ≈ 0.3 cord.
40. (a) When θ is measured in radians, it is equal to the arclength s divided by the radius R. For a
very large radius circle and small value of θ, such as we deal with in Fig. 1-9, the arc may be
approximated as the straight line-segment of length 1 AU. First, we convert θ = 1 arcsecond to
radians:
(1 arcsecond)
(
1 arcminute
60 arcsecond
)(
1◦
60 arcminute
)(
2π radian
360◦
)
which yields θ = 4.85× 10−6 rad. Therefore, one parsec is
Ro =
s
θ
=
1AU
4.85× 10−6 = 2.06 × 10
5AU .
Now we use this to convert R = 1 AU to parsecs:
R = (1AU)
(
1 pc
2.06 × 105AU
)
= 4.9× 10−6 pc .
(b) Also, since it is straightforward to figure the number of seconds in a year (about 3.16× 107 s), and
(for constant speeds) distance= speed×time, we have
1 ly = (186, 000mi/s)
(
3.16× 107 s) 5.9× 1012 mi
which we convert to AU by dividing by 92.6 × 106 (given in the problem statement), obtaining
6.3× 104 AU. Inverting, the result is 1AU = 1/6.3× 104 = 1.6× 10−5 ly.
(c) As found in the previous part, 1 ly = 5.9× 1012mi.
(d) We now know what one parsec is in AU (denoted above as Ro ), and we also know how many miles
are in an AU. Thus, one parsec is equivalent to(
92.9× 106mi/AU) (2.06 × 105AU) = 1.9× 1013 mi .
41. We reduce the stock amount to British teaspoons:
1 breakfastcup = 2× 8× 2× 2 = 64 teaspoons
1 teacup = 8× 2× 2 = 32 teaspoons
6 tablespoons = 6× 2× 2 = 24 teaspoons
1 dessertspoon = 2 teaspoons
12 CHAPTER 1.
which totals to 122 teaspoons – which corresponds (since liquid measure is being used) to 122 U.S.
teaspoons. Since each U.S cup is 48 teaspoons, then upon dividing 122/48 ≈ 2.54, we find this amount
corresponds to two-and-a-half U.S. cups plus a remainder of precisely 2 teaspoons. For the nettle tops,
one-half quart is still one-half quart. For the rice, one British tablespoon is 4 British teaspoons which
(since dry-goods measure is being used) corresponds to 2 U.S. teaspoons. Finally, a British saltspoon is
1
2 British teaspoon which corresponds (since dry-goods measure is again being used) to 1 U.S. teaspoon.
42. (a) Megaphone.
(b) microphone.
(c) dekacard (“deck of cards”).
(d) Gigalow (“gigalo”).
(e) terabull (“terrible”).
(f) decimate.
(g) centipede.
(h) nanonannette. (“No No Nanette”).
(i) picoboo (“peek-a-boo”).
(j) attoboy (“at-a-boy”).
(k) Two hectowithit (“to heck with it”).
(l) Two kilomockingbird (“to kill a mockingbird”).
43. The volume removed in one year is
V =
(
75× 104m2) (26m) ≈ 2× 107 m3
which we convert to cubic kilometers:
V =
(
2× 107m3)( 1 km
1000m
)3
= 0.020 km3 .
44. (a) Using Appendix D, we have 1 ft = 0.3048m, 1 gal = 231 in.3, and 1 in.3 = 1.639× 10−2 L. From the
latter two items, we find that 1 gal = 3.79 L. Thus, the quantity 460ft2/gal becomes
(
460 ft2
gal
)(
1m
3.28 ft
)2(
1 gal
3.79 L
)
= 11.3 m2/L .
(b) Also, since 1 m3 is equivalent to 1000 L, our result from part (a) becomes(
11.3m2
L
)(
1000 L
1m3
)
= 1.13× 104 m−1 .
(c) The inverse of the original quantity is (460 ft2/gal)−1 = 2.17 × 10−3 gal/ft2, which is the volume
of the paint (in gallons) needed to cover a square foot of area. From this, we could also figure the
paint thickness (it turns out to be about a tenth of a millimeter, as one sees by taking the reciprocal
of the answer in part (b)).
Chapter 2
1. Assuming the horizontal velocity of the ball is constant, the horizontal displacement is
∆x = v∆t
where ∆x is the horizontal distance traveled, ∆t is the time, and v is the (horizontal) velocity. Converting
v to meters per second, we have 160 km/h = 44.4m/s. Thus
∆t =
∆x
v
=
18.4m
44.4m/s
= 0.414 s .
The velocity-unit conversion implemented above can be figured “from basics” (1000 m = 1 km, 3600 s
= 1 h) or found in Appendix D.
2. Converting to SI units, we use Eq. 2-3 with d for distance.
savg =
d
t
(110.6 km/h)
(
1000m/km
3600 s/h
)
=
200.0m
t
which yields t = 6.510 s. We converted the speed km/h→m/s by converting each unit (km→m, h→ s)
individually. But we mention that the “one-step” conversion can be found in Appendix D (1 km/h =
0.2778m/s).
3. We use Eq. 2-2 and Eq. 2-3. During a time tc when the velocity remains a positive constant, speed is
equivalent to velocity, and distance is equivalent to displacement, with ∆x = v tc .
(a) During the first part of the motion, the displacement is ∆x1 = 40 km and the time interval is
t1 =
(40 km)
(30 km/h)
= 1.33 h .
During the second part the displacement is ∆x2 = 40 km and the time interval is
t2 =
(40 km)
(60 km/h)
= 0.67 h .
Both displacements are in the same direction, so the total displacement is ∆x = ∆x1 + ∆x2 =
40 km + 40 km = 80 km. The total time for the trip is t = t1 + t2 = 2.00 h. Consequently, the
average velocity is
vavg =
(80 km)
(2.0 h)
= 40 km/h .
(b) In this example, the numerical result for the average speed is the same as the average velocity
40 km/h.
13
14 CHAPTER 2.
(c) In the interest of saving space, we briefly describe the graph (with kilometers and hours understood):
two contiguous line segments, the first having a slope of 30 and connecting the origin to (t1, x1) =
(1.33, 40) and the second having a slope of 60 and connecting (t1, x1) to (t, x) = (2.00, 80). The
average velocity, from the graphical point of view, is the slope of a line drawn from the origin to
(t, x).
4. If the plane (with velocity v) maintains its present course, and if the terrain continues its upward slope
of 4.3◦, then the plane will strike the ground after traveling
∆x =
h
tan θ
=
35m
tan 4.3◦
= 465.5 m ≈ 0.465 km .
This corresponds to a time of flight found from Eq. 2-2 (with v = vavg since it is constant)
t =
∆x
v
=
0.465 km
1300 km/h
= 0.000358 h ≈ 1.3 s .
This, then, estimates the time available to the pilot to make his correction.
5. (a) Denoting the travel time and distance from San Antonio to Houston as T and D, respectively, the
average speed is
savg 1 =
D
T
=
(55 km/h)T2 + (90 km/h)
T
2
T
= 72.5 km/h
which should be rounded to 73 km/h.
(b) Using the fact that time=distance/speed while the speed is constant, we find
savg 2 =
D
T
=
D
D/2
55 km/h +
D/2
90 km/h
= 68.3 km/h
which should be rounded to 68 km/h.
(c) The total distance traveled (2D) must not be confused with the net displacement (zero). We obtain
for the two-way trip
savg =
2D
D
72.5 km/h +
D
68.3 km/h
= 70 km/h .
(d) Since the net displacement vanishes, the average velocity for the trip in its entirety is zero.
(e) In asking for a sketch, the problem is allowing the student to arbitrarily set the distanceD (the intent
is not to make the student go to an Atlas to look it up); the student can just as easily arbitrarily
set T instead of D, as will be clear in the following discussion. In the interest of saving space, we
briefly describe the graph (with kilometers-per-hour understood for the slopes): two contiguous line
segments, the first having a slope of 55 and connecting the origin to (t1, x1) = (T/2, 55T/2) and
the second having a slope of 90 and connecting (t1, x1) to (T,D) where D = (55 + 90)T/2. The
average velocity, from the graphical point of view, is the slope of a line drawn from the origin to
(T,D).
6. (a) Using the fact that time=distance/velocity while the velocity is constant, we find
vavg =
73.2m+ 73.2m
73.2m
1.22m/s +
73.2m
3.05m
= 1.74m/s .
(b) Using the fact that distance = vt while the velocity v is constant, we find
vavg =
(1.22m/s)(60 s) + (3.05m/s)(60 s)
120 s
= 2.14m/s .
15
(c) The graphs are shown below (with meters and seconds understood). The first consists of two (solid)
line segments, the first having
a slope of 1.22 and the second having a slope of 3.05. The slope of
the dashed line represents the average velocity (in both graphs). The second graph also consists of
two (solid) line segments, having the same slopes as before – the main difference (compared to the
first graph) being that the stage involving higher-speed motion lasts much longer.
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t60 84
73
146
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x
t60 120
73
256
7. Using x = 3t− 4t2 + t3 with SI units understood is efficient (and is the approach we will use), but if we
wished to make the units explicit we would write x = (3m/s)t− (4m/s2)t2 + (1m/s3)t3. We will quote
our answers to one or two significant figures, and not try to follow the significant figure rules rigorously.
(a) Plugging in t = 1 s yields x = 0. With t = 2 s we get x = −2 m. Similarly, t = 3 s yields x = 0 and
t = 4 s yields x = 12 m. For later reference, we also note that the position at t = 0 is x = 0.
(b) The position at t = 0 is subtracted from the position at t = 4 s to find the displacement ∆x = 12 m.
(c) The position at t = 2 s is subtracted from the position at t = 4 s to give the displacement ∆x = 14 m.
Eq. 2-2, then, leads to
vavg =
∆x
∆t
=
14
2
= 7 m/s .
(d) The horizontal axis is 0 ≤ t ≤ 4 with SI units understood.
Not shown is
a straight line
drawn from
the point at
(t, x) = (2,−2)
to the highest
point shown (at
t = 4 s) which
would represent
the answer for
part (c).
t
0
10
x
8. Recognizing that the gap between the trains is closing at a constant rate of 60 km/h, the total time
which elapses before they crash is t = (60 km)/(60 km/h) = 1.0 h. During this time, the bird travels a
distance of x = vt = (60 km/h)(1.0 h) = 60 km.
16 CHAPTER 2.
9. Converting to seconds, the running times are t1 = 147.95 s and t2 = 148.15 s, respectively. If the runners
were equally fast, then
savg 1 = savg 2 =⇒
L1
t1
=
L2
t2
.
From this we obtain
L2 − L1 =
(
148.15
147.95
− 1
)
L1 ≈ 1.35 m
where we set L1 ≈ 1000 m in the last step. Thus, if L1 and L2 are no different than about 1.35 m, then
runner 1 is indeed faster than runner 2. However, if L1 is shorter than L2 than 1.4 m then runner 2 is
actually the faster.
10. The velocity (both magnitude and sign) is determined by the slope of the x versus t curve, in accordance
with Eq. 2-4.
(a) The armadillo is to the left of the coordinate origin on the axis between t = 2.0 s and t = 4.0 s.
(b) The velocity is negative between t = 0 and t = 3.0 s.
(c) The velocity is positive between t = 3.0 s and t = 7.0 s.
(d) The velocity is zero at the graph minimum (at t = 3.0 s).
11. We use Eq. 2-4.
(a) The velocity of the particle is
v =
dx
dt
=
d
dt
(
4− 12t+ 3t2) = −12 + 6t .
Thus, at t = 1 s, the velocity is v = (−12 + (6)(1)) = −6 m/s.
(b) Since v < 0, it is moving in the negative x direction at t = 1 s.
(c) At t = 1 s, the speed is |v| = 6 m/s.
(d) For 0 < t < 2 s, |v| decreases until it vanishes. For 2 < t < 3 s, |v| increases from zero to the value
it had in part (c). Then, |v| is larger than that value for t > 3 s.
(e) Yes, since v smoothly changes from negative values (consider the t = 1 result) to positive (note
that as t→ +∞, we have v → +∞). One can check that v = 0 when t = 2 s.
(f) No. In fact, from v = −12 + 6t, we know that v > 0 for t > 2 s.
12. We use Eq. 2-2 for average velocity and Eq. 2-4 for instantaneous velocity, and work with distances in
centimeters and times in seconds.
(a) We plug into the given equation for x for t = 2.00 s and t = 3.00 s and obtain x2 = 21.75 cm and
x3 = 50.25 cm, respectively. The average velocity during the time interval 2.00 ≤ t ≤ 3.00 s is
vavg =
∆x
∆t
=
50.25 cm− 21.75 cm
3.00 s− 2.00 s
which yields vavg = 28.5 cm/s.
(b) The instantaneous velocity is v = dxdt = 4.5t
2, which yields v = (4.5)(2.00)2 = 18.0 cm/s at time
t = 2.00 s.
(c) At t = 3.00 s, the instantaneous velocity is v = (4.5)(3.00)2 = 40.5 cm/s.
(d) At t = 2.50 s, the instantaneous velocity is v = (4.5)(2.50)2 = 28.1 cm/s.
(e) Let tm stand for the moment when the particle is midway between x2 and x3 (that is, when the
particle is at xm = (x2 + x3)/2 = 36 cm). Therefore,
xm = 9.75 + 1.5t
3
m =⇒ tm = 2.596
in seconds. Thus, the instantaneous speed at this time is v = 4.5(2.596)2 = 30.3 cm/s.
17
(f) The answer to part (a) is given by the slope of the straight line
between t = 2 and
t = 3 in this x-
vs-t plot. The an-
swers to parts (b),
(c), (d) and (e)
correspond to the
slopes of tangent
lines (not shown
but easily imag-
ined) to the curve
at the appropriate
points.
(cm)x
(a)
20
40
60
2 3t
13. Since v = dxdt (Eq. 2-4), then ∆x =
∫
v dt, which corresponds to the area under the v vs t graph. Dividing
the total area A into rectangular (base×height) and triangular (12base×height) areas, we have
A = A0<t<2 +A2<t<10 +A10<t<12 +A12<t<16
=
1
2
(2)(8) + (8)(8) +
(
(2)(4) +
1
2
(2)(4)
)
+ (4)(4)
with SI units understood. In this way, we obtain ∆x = 100 m.
14. From Eq. 2-4 and Eq. 2-9, we note that the sign of the velocity is the sign of the slope in an x-vs-t plot,
and the sign of the acceleration corresponds to whether such a curve is concave up or concave down. In
the interest of saving space, we indicate sample points for parts (a)-(d) in a single figure; this means
that all points are not at t = 1 s (which we feel is an acceptable modification of the problem – since the
datum t = 1 s is not used).
(c) (a)
(b)
(d)
Any change from zero
to non-zero values
of ~v represents in-
creasing |~v| (speed).
Also, ~v ‖ ~a implies
that the particle is
going faster. Thus,
points (a), (b) and
(d) involve increasing
speed.
15. We appeal to Eq. 2-4 and Eq. 2-9.
(a) This is v2 – that is, the velocity squared.
(b) This is the acceleration a.
(c) The SI units for these quantities are (m/s)
2
= m2/s2 and m/s2, respectively.
16. Eq. 2-9 indicates that acceleration is the slope of the v-vs-t graph.
18 CHAPTER 2.
Based on this, we
show here a sketch
of the acceleration (in
m/s2) as a function
of time. The values
along the acceleration
axis should not be
taken too seriously. –20
–10
0
10
a
5t
17. We represent its initial direction of motion as the +x direction, so that v0 = +18 m/s and v = −30 m/s
(when t = 2.4 s). Using Eq. 2-7 (or Eq. 2-11, suitably interpreted) we find
aavg =
(−30)− (+18)
2.4
= −20 m/s2
which indicates that the average acceleration has magnitude 20 m/s2 and is in the opposite direction to
the particle’s initial velocity.
18. We use Eq. 2-2 (average velocity) and Eq. 2-7 (average acceleration). Regarding our coordinate choices,
the initial position of the man is taken as the origin and his direction of motion during 5min ≤ t ≤ 10min
is taken to be the positive x direction. We also use the fact that ∆x = v∆t′ when the velocity is constant
during a time interval ∆t′.
(a) Here, the entire interval considered is ∆t = 8− 2 = 6 min which is equivalent to 360 s, whereas the
sub-interval in which he is moving is only ∆t′ = 8 − 5 = 3 min= 180 s. His position at t = 2 min
is x = 0 and his position at t = 8 min is x = v∆t′ = (2.2)(180) = 396 m. Therefore,
vavg =
396m− 0
360 s
= 1.10 m/s .
(b) The man is at rest at t = 2 min and has velocity v = +2.2 m/s at t = 8 min. Thus, keeping the
answer to 3 significant figures,
aavg =
2.2m/s− 0
360 s
= 0.00611 m/s2 .
(c) Now, the entire interval considered is ∆t = 9 − 3 = 6 min (360 s again), whereas the sub-interval
in which he is moving is ∆t′ = 9 − 5 = 4 min= 240 s). His position at t = 3 min is x = 0 and his
position at t = 9 min is x = v∆t′ = (2.2)(240) = 528 m. Therefore,
vavg =
528m− 0
360 s
= 1.47 m/s .
(d) The man is at rest at t = 3 min and has velocity v = +2.2 m/s at t = 9 min. Consequently,
aavg = 2.2/360 = 0.00611 m/s
2 just as in part (b).
(e) The horizontal line near the bottom of this x-vs-t graph represents
19
the man stand-
ing at x = 0 for
0 ≤ t < 300 s
and the linearly
rising line for
300 ≤ t ≤ 600 s
represents his
constant-velocity
motion. The dotted
lines represent the
answers to part (a)
and (c) in the sense
that their slopes
yield those results.
(c)
(a)
0
500
x
0 500t
The graph of v-vs-t is not shown here, but would consist of two horizontal “steps” (one at v = 0
for 0 ≤ t < 300 s and the next at v = 2.2 m/s for 300 ≤ t ≤ 600 s). The indications of the
average accelerations found in parts (b) and (d) would be dotted lines connected the “steps” at the
appropriate t values (the slopes of the dotted lines representing the values of aavg).
19. In this solution, we make use of the notation x(t) for the value of x at a particular t. Thus, x(t) =
50t+ 10t2 with SI units (meters and seconds) understood.
(a) The average velocity during the first 3 s is given by
vavg =
x(3)− x(0)
∆t
=
(50)(3) + (10)(3)2 − 0
3
= 80 m/s .
(b) The instantaneous velocity at time t is given by v = dx/dt = 50 + 20t, in SI units. At t = 3.0 s,
v = 50 + (20)(3.0) = 110 m/s.
(c) The instantaneous acceleration at time t is given by a = dv/dt = 20 m/s2. It is constant, so the
acceleration at any time is 20m/s
2
.
(d) and (e) The graphs below show the coordinate x and velocity v as functions of time, with SI units
understood. The dotted line marked (a) in the first graph runs from t = 0, x = 0 to t = 3.0 s,
x = 240m. Its slope is the average velocity during the first 3 s of motion. The dotted line marked
(b) is tangent to the x curve at t = 3.0 s. Its slope is the instantaneous velocity at t = 3.0 s.
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1.0 2.0 3.0 4.0
0
100
200
300
400
(a)
(b)
t
x
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1.0 2.0 3.0 4.0
50
0
100
150
200
t
v
20. Using the general property ddx exp(bx) = b exp(bx), we write
v =
dx
dt
=
(
d (19t)
dt
)
· e−t + (19t) ·
(
d e−t
dt
)
.
20 CHAPTER 2.
If a concern develops about the appearance of an argument of the exponential (−t) apparently having
units, then an explicit factor of 1/T where T = 1 second can be inserted and carried through
the
computation (which does not change our answer). The result of this differentiation is
v = 16(1− t)e−t
with t and v in SI units (s and m/s, respectively). We see that this function is zero when t = 1 s.
Now that we know when it stops, we find out where it stops by plugging our result t = 1 into the given
function x = 16te−t with x in meters. Therefore, we find x = 5.9 m.
21. In this solution, we make use of the notation x(t) for the value of x at a particular t. The notations v(t)
and a(t) have similar meanings.
(a) Since the unit of ct2 is that of length, the unit of c must be that of length/time2, or m/s2 in the SI
system. Since bt3 has a unit of length, b must have a unit of length/time3, or m/s3.
(b) When the particle reaches its maximum (or its minimum) coordinate its velocity is zero. Since the
velocity is given by v = dx/dt = 2ct− 3bt2, v = 0 occurs for t = 0 and for
t =
2c
3b
=
2(3.0m/s
2
)
3(2.0m/s
3
)
= 1.0 s .
For t = 0, x = x0 = 0 and for t = 1.0 s, x = 1.0m > x0. Since we seek the maximum, we reject the
first root (t = 0) and accept the second (t = 1 s).
(c) In the first 4 s the particle moves from the origin to x = 1.0m, turns around, and goes back to
x(4 s) = (3.0m/s
2
)(4.0 s)2 − (2.0m/s3)(4.0 s)3 = −80m .
The total path length it travels is 1.0m+ 1.0m+ 80m = 82m.
(d) Its displacement is given by ∆x = x2 − x1, where x1 = 0 and x2 = −80m. Thus, ∆x = −80m.
(e) The velocity is given by v = 2ct− 3bt2 = (6.0m/s2)t− (6.0m/s3)t2. Thus
v(1 s) = (6.0m/s
2
)(1.0 s)− (6.0m/s3)(1.0 s)2 = 0
v(2 s) = (6.0m/s
2
)(2.0 s)− (6.0m/s3)(2.0 s)2 = −12m/s
v(3 s) = (6.0m/s
2
)(3.0 s)− (6.0m/s3)(3.0 s)2 = −36.0m/s
v(4 s) = (6.0m/s2)(4.0 s)− (6.0m/s3)(4.0 s)2 = −72m/s .
(f) The acceleration is given by a = dv/dt = 2c− 6b = 6.0m/s2 − (12.0m/s3)t. Thus
a(1 s) = 6.0m/s
2 − (12.0m/s3)(1.0 s) = −6.0m/s2
a(2 s) = 6.0m/s
2 − (12.0m/s3)(2.0 s) = −18m/s2
a(3 s) = 6.0m/s2 − (12.0m/s3)(3.0 s) = −30m/s2
a(4 s) = 6.0m/s
2 − (12.0m/s3)(4.0 s) = −42m/s2 .
22. For the automobile ∆v = 55− 25 = 30 km/h, which we convert to SI units:
a =
∆v
∆t
=
(30 km/h)
(
1000m/km
3600 s/h
)
(0.50min)(60 s/min)
= 0.28 m/s
2
.
The change of velocity for the bicycle, for the same time, is identical to that of the car, so its acceleration
is also 0.28 m/s2.
23. The constant-acceleration condition permits the use of Table 2-1.
21
(a) Setting v = 0 and x0 = 0 in v
2 = v20 + 2a(x− x0), we find
x = −1
2
v20
a
= −1
2
(
5.00× 106
−1.25× 1014
)
= 0.100 m .
Since the muon is slowing, the initial velocity and the acceleration must have opposite signs.
(b) Below are the time-plots of the position x and velocity v of the muon from the moment it enters
the field to the time it stops. The computation in part (a) made no reference to t, so that other
equations from Table 2-1 (such as v = v0 + at and x = v0t+
1
2at
2) are used in making these plots.
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........
10 20 30 40
t (ns)
0
2.5
5.0
7.5
10
x (cm)
................................................................................................................................................................................................................................................................................................................................................................................................
10 20 30 40
t (ns)
0
2.0
4.0
6.0
8.0
v (Mm/s)
24. The time required is found from Eq. 2-11 (or, suitably interpreted, Eq. 2-7). First, we convert the
velocity change to SI units:
∆v = (100 km/h)
(
1000m/km
3600 s/h
)
= 27.8 m/s .
Thus, ∆t = ∆v/a = 27.8/50 = 0.556 s.
25. We use v = v0 + at, with t = 0 as the instant when the velocity equals +9.6 m/s.
(a) Since we wish to calculate the velocity for a time before t = 0, we set t = −2.5 s. Thus, Eq. 2-11
gives
v = (9.6m/s) +
(
3.2m/s
2
)
(−2.5 s) = 1.6 m/s .
(b) Now, t = +2.5 s and we find
v = (9.6m/s) +
(
3.2m/s2
)
(2.5 s) = 18 m/s .
26. The bullet starts at rest (v0 = 0) and after traveling the length of the barrel (∆x = 1.2 m) emerges with
the given velocity (v = 640 m/s), where the direction of motion is the positive direction. Turning to the
constant acceleration equations in Table 2-1, we use
∆x =
1
2
(v0 + v) t .
Thus, we find t = 0.00375 s (about 3.8 ms).
27. The constant acceleration stated in the problem permits the use of the equations in Table 2-1.
(a) We solve v = v0 + at for the time:
t =
v − v0
a
=
1
10
(
3.0× 108m/s)
9.8m/s2
= 3.1× 106 s
which is equivalent to 1.2 months.
22 CHAPTER 2.
(b) We evaluate x = x0 + v0t+
1
2at
2, with x0 = 0. The result is
x =
1
2
(
9.8m/s
2
) (
3.1× 106 s)2 = 4.7× 1013m .
28. From Table 2-1, v2 − v20 = 2a∆x is used to solve for a. Its minimum value is
amin =
v2 − v20
2∆xmax
=
(360 km/h)2
2(1.80 km)
= 36000 km/h2
which converts to 2.78 m/s2.
29. Assuming constant acceleration permits the use of the equations in Table 2-1. We solve v2 = v20+2a(x−
x0) with x0 = 0 and x = 0.010 m. Thus,
a =
v2 − v20
2x
=
(
5.7× 105)2 − (1.5× 105)2
2(0.01)
= 1.62× 1015 m/s2 .
30. The acceleration is found from Eq. 2-11 (or, suitably interpreted, Eq. 2-7).
a =
∆v
∆t
=
(1020 km/h)
(
1000m/km
3600 s/h
)
1.4 s
= 202.4 m/s
2
.
In terms of the gravitational acceleration g, this is expressed as a multiple of 9.8 m/s2 as follows:
a =
202.4
9.8
g = 21g .
31. We choose the positive direction to be that of the initial velocity of the car (implying that a < 0 since
it is slowing down). We assume the acceleration is constant and use Table 2-1.
(a) Substituting v0 = 137 km/h = 38.1m/s, v = 90 km/h = 25m/s, and a = −5.2m/s2 into v = v0+at,
we obtain
t =
25m/s− 38m/s
−5.2m/s2 = 2.5 s .
(b) We take the car to be at x = 0 when the brakes are applied
(at time t = 0). Thus, the
coordinate of the car as a
function of time is given by
x = (38)t+
1
2
(−5.2)t2
in SI units. This function
is plotted from t = 0 to t =
2.5 s on the graph to the
right. We have not shown
the v-vs-t graph here; it is
a descending straight line
from v0 to v.
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0
20
40
60
80
0.5 1.0 1.5 2.0 2.5
x (m)
t (s)
32. From the figure, we see that x0 = −2.0 m. From Table 2-1, we can apply x − x0 = v0t + 12at2 with
t = 1.0 s, and then again with t = 2.0 s. This yields two equations for the two unknowns, v0 and a. SI
units are understood.
0.0− (−2.0) = v0 (1.0) + 1
2
a(1.0)2
6.0− (−2.0) = v0 (2.0) + 1
2
a(2.0)2 .
23
Solving these simultaneous equations yields the results v0 = 0.0 and a = 4.0 m/s
2. The fact that the
answer is positive tells us that the acceleration vector points in the +x direction.
33. The problem statement (see part (a)) indicates that a =constant, which allows us to use Table 2-1.
(a) We take x0 = 0, and solve x = v0t +
1
2at
2 (Eq. 2-15) for the acceleration: a = 2(x − v0t)/t2.
Substituting x = 24.0m, v0 = 56.0 km/h = 15.55m/s and t = 2.00 s, we find
a =
2 (24.0m− (15.55m/s)(2.00 s))
(2.00 s)2
= −3.56m/s2 .
The negative sign indicates that the acceleration is opposite to the direction of motion of the car.
The car is slowing down.
(b) We evaluate v = v0 + at as follows:
v = 15.55m/s−
(
3.56m/s
2
)
(2.00 s) = 8.43 m/s
which is equivalent to 30.3 km/h.
34. We take the moment of applying brakes to be t = 0. The deceleration is constant so that Table 2-1 can
be used. Our primed variables (such as v′o = 72 km/h = 20 m/s) refer to one train (moving in the +x
direction and located at the origin when t = 0) and unprimed variables refer to the other (moving in
the −x direction and located at x0 = +950 m when t = 0). We note that the acceleration vector of the
unprimed train points in the positive direction, even though the train is slowing down; its initial velocity
is vo = −144 km/h = −40 m/s. Since the primed train has the lower initial speed, it should stop sooner
than the other train would (were it not for the collision). Using Eq 2-16, it should stop (meaning v′ = 0)
at
x′ =
(v′)2 − (v′o)2
2a′
=
0− 202
−2 = 200 m .
The speed of the other train, when it reaches that location, is
v =
√
v2o + 2a∆x =
√
(−40)2 + 2(1.0)(200− 950) =
√
100 = 10 m/s
using Eq 2-16 again. Specifically, its velocity at that moment would be −10 m/s since it is still traveling
in the −x direction when it crashes. If the computation of v had failed (meaning that a negative number
would have been inside the square root) then we would have looked at the possibility that there was no
collision and examined how far apart they finally were. A concern that can be brought up is whether the
primed train collides before it comes to rest; this can be studied by computing the time it stops (Eq. 2-11
yields t = 20 s) and seeing where the unprimed train is at that moment (Eq. 2-18 yields x = 350 m, still
a good distance away from contact).
35. The acceleration is constant and we may use the equations in Table 2-1.
(a) Taking the first point as coordinate origin and time to be zero when the car is there, we apply
Eq. 2-17 (with SI units understood):
x =
1
2
(v + v0) t =
1
2
(15 + v0) (6) .
With x = 60.0 m (which takes the direction of motion as the +x direction) we solve for the initial
velocity: v0 = 5.00 m/s.
(b) Substituting v = 15 m/s, v0 = 5 m/s and t = 6 s into a = (v − v0)/t (Eq. 2-11), we find
a = 1.67 m/s2.
(c) Substituting v = 0 in v2 = v20 + 2ax and solving for x, we obtain
x = − v
2
0
2a
= − 5
2
2(1.67)
= −7.50 m .
24 CHAPTER 2.
(d) The graphs require computing the time when v = 0, in which case, we use v = v0 + at
′ = 0. Thus,
t′ =
−v0
a
=
−5
1.67
= −3.0 s
indicates the moment the car was at rest. SI units are assumed.
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−5 5 10
t
−20
20
40
60 x
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.
−5 0 5 10
t
5
10
15
v
36. We denote the required time as t, assuming the light turns green when the clock reads zero. By this
time, the distances traveled by the two vehicles must be the same.
(a) Denoting the acceleration of the automobile as a and the (constant) speed of the truck as v then
∆x =
(
1
2
at2
)
car
= (vt)truck
which leads to
t =
2v
a
=
2(9.5)
2.2
= 8.6 s .
Therefore,
∆x = vt = (9.5)(8.6) = 82 m .
(b) The speed of the car at that moment is
vcar = at = (2.2)(8.6) = 19 m/s .
37. We denote tr as the reaction time and tb as the braking time. The motion during tr is of the constant-
velocity (call it v0) type. Then the position of the car is given by
x = v0tr + v0tb +
1
2
at2b
where v0 is the initial velocity and a is the acceleration (which we expect to be negative-valued since we
are taking the velocity in the positive direction and we know the car is decelerating). After the brakes
are applied the velocity of the car is given by v
= v0+atb. Using this equation, with v = 0, we eliminate
tb from the first equation and obtain
x = v0tr − v
2
0
a
+
1
2
v20
a
= v0tr − 1
2
v20
a
.
We write this equation for each of the initial velocities:
x1 = v01tr − 1
2
v201
a
25
and
x2 = v02tr − 1
2
v202
a
.
Solving these equations simultaneously for tr and a we get
tr =
v202x1 − v201x2
v01v02(v02 − v01)
and
a = −1
2
v02v
2
01 − v01v202
v02x1 − v01x2 .
Substituting x1 = 56.7m, v01 = 80.5 km/h = 22.4m/s, x2 = 24.4 m and v02 = 48.3 km/h = 13.4 m/s,
we find
tr =
13.42(56.7)− 22.42(24.4)
(22.4)(13.4)(13.4− 22.4) = 0.74 s
and
a = −1
2
(13.4)22.42 − (22.4)13.42
(13.4)(56.7)− (22.4)(24.4) = −6.2 m/s
2 .
The magnitude of the deceleration is therefore 6.2 m/s2. Although rounded off values are displayed
in the above substitutions, what we have input into our calculators are the “exact” values (such as
v02 =
161
12 m/s).
38. In this solution we elect to wait until the last step to convert to SI units. Constant acceleration is
indicated, so use of Table 2-1 is permitted. We start with Eq. 2-17 and denote the train’s initial velocity
as vt and the locomotive’s velocity as vℓ (which is also the final velocity of the train, if the rear-end
collision is barely avoided). We note that the distance ∆x consists of the original gap between them D
as well as the forward distance traveled during this time by the locomotive vℓt. Therefore,
vt + vℓ
2
=
∆x
t
=
D + vℓt
t
=
D
t
+ vℓ .
We now use Eq. 2-11 to eliminate time from the equation. Thus,
vt + vℓ
2
=
D
(vℓ − vt) /a + vℓ
leads to
a =
(
vt + vℓ
2
− vℓ
)(
vℓ − vt
D
)
= − 1
2D
(vℓ − vt)2 .
Hence,
a = − 1
2(0.676 km)
(
29
km
h
− 161 km
h
)2
= −12888 km/h2
which we convert as follows:
a =
(
−12888 km/h2
)(1000m
1 km
)(
1 h
3600 s
)2
= −0.994 m/s2
so that its magnitude is 0.994 m/s2. A graph is shown below for the case where a collision is just avoided
(x along the vertical axis is in meters and t along the horizontal axis is in seconds). The top (straight)
line shows the motion of the locomotive and the bottom curve shows the motion of the passenger train.
26 CHAPTER 2.
The other case
(where the colli-
sion is not quite
avoided) would
be similar except
that the slope of
the bottom curve
would be greater
than that of the
top line at the
point where they
meet.
0
200
400
600
800
x
10 20 30
t
39. We assume the periods of acceleration (duration t1) and deceleration (duration t2) are periods of constant
a so that Table 2-1 can be used. Taking the direction of motion to be +x then a1 = +1.22 m/s
2 and
a2 = −1.22 m/s2. We use SI units so the velocity at t = t1 is v = 305/60 = 5.08 m/s.
(a) We denote ∆x as the distance moved during t1, and use Eq. 2-16:
v2 = v20 + 2a1∆x =⇒ ∆x =
5.082
2(1.22)
which yields ∆x = 10.59 ≈ 10.6 m.
(b) Using Eq. 2-11, we have
t1 =
v − v0
a1
=
5.08
1.22
= 4.17 s .
The deceleration time t2 turns out to be the same so that t1 + t2 = 8.33 s. The distances traveled
during t1 and t2 are the same so that they total to 2(10.59) = 21.18 m. This implies that for a
distance of 190 − 21.18 = 168.82 m, the elevator is traveling at constant velocity. This time of
constant velocity motion is
t3 =
168.82m
5.08m/s
= 33.21 s .
Therefore, the total time is 8.33 + 33.21 ≈ 41.5 s.
40. Neglect of air resistance justifies setting a = −g = −9.8 m/s2 (where down is our −y direction) for the
duration of the fall. This is constant acceleration motion, and we may use Table 2-1 (with ∆y replacing
∆x).
(a) Using Eq. 2-16 and taking the negative root (since the final velocity is downward), we have
v = −
√
v20 − 2g∆y = −
√
0− 2(9.8)(−1700) = −183
in SI units. Its magnitude is therefore 183 m/s.
(b) No, but it is hard to make a convincing case without more analysis. We estimate the mass of a
raindrop to be about a gram or less, so that its mass and speed (from part (a)) would be less than
that of a typical bullet, which is good news. But the fact that one is dealing with many raindrops
leads us to suspect that this scenario poses an unhealthy situation. If we factor in air resistance,
the final speed is smaller, of course, and we return to the relatively healthy situation with which
we are familiar.
41. We neglect air resistance, which justifies setting a = −g = −9.8 m/s2 (taking down as the −y direction)
for the duration of the fall. This is constant acceleration motion, which justifies the use of Table 2-1
(with ∆y replacing ∆x).
27
(a) Starting the clock at the moment the wrench is dropped (vo = 0), then v
2 = v2o − 2g∆y leads to
∆y = − (−24)
2
2(9.8)
= −29.4 m
so that it fell through a height of 29.4 m.
(b) Solving v = v0 − gt for time, we find:
t =
v0 − v
g
=
0− (−24)
9.8
= 2.45 s .
(c) SI units are used in the graphs, and the initial position is taken as the coordinate origin. In
the interest of saving space, we do not show the acceleration graph, which is a horizontal line at
−9.8m/s2.
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1 2 3
t
0
−10
−20
−30
y
.......................................................................................................................................................................................................................................................................................
1 2 3
t
0
−10
−20
−30
v
42. We neglect air resistance, which justifies setting a = −g = −9.8 m/s2 (taking down as the −y direction)
for the duration of the fall. This is constant acceleration motion, which justifies the use of Table 2-1
(with ∆y replacing ∆x).
(a) Noting that ∆y = y − y0 = −30 m, we apply Eq. 2-15 and the quadratic formula (Appendix E) to
compute t:
∆y = v0t− 1
2
gt2 =⇒ t = v0 ±
√
v20 − 2g∆y
g
which (with v0 = −12 m/s since it is downward) leads, upon choosing the positive root (so that
t > 0), to the result:
t =
−12 +√(−12)2 − 2(9.8)(−30)
9.8
= 1.54 s .
(b) Enough information is now known that any of the equations in Table 2-1 can be used to obtain v;
however, the one equation that does not use our result from part (a) is Eq. 2-16:
v =
√
v20 − 2g∆y = 27.1 m/s
where the positive root has been chosen in order to give speed (which is the magnitude of the
velocity vector).
43. We neglect air resistance for the duration of the motion (between “launching” and “landing”), so a =
−g = −9.8 m/s2 (we take downward to be the −y direction). We use the equations in Table 2-1 (with
∆y replacing ∆x) because this is a =constant motion.
(a) At the highest point the velocity of the ball vanishes. Taking y0 = 0, we set v = 0 in v
2 = v20 − 2gy,
and solve for the initial velocity: v0 =
√
2gy. Since y = 50m we find v0 = 31 m/s.
28 CHAPTER 2.
(b) It will be in the air from the time it leaves the ground until the time it returns to the ground
(y = 0). Applying Eq. 2-15 to the entire motion (the rise and the fall, of total time t > 0) we have
y = v0t− 1
2
gt2 =⇒ t = 2v0
g
which (using our result from part (a))
produces t = 6.4 s. It is possible to obtain this without
using part (a)’s result; one can find the time just for the rise (from ground to highest point) from
Eq. 2-16 and then double it.
(c) SI units are understood in the x and v graphs shown. In the interest of saving space, we do not
show the graph of a, which is a horizontal line at −9.8m/s2.
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2 4 6 8 t
0
20
40
60
y
............................................................................................................................................................................................................................................................................................................................................................................
2 4 6 8
t
0
−20
20
40
v
44. There is no air resistance, which makes it quite accurate to set a = −g = −9.8 m/s2 (where downward
is the −y direction) for the duration of the fall. We are allowed to use Table 2-1 (with ∆y replacing
∆x) because this is constant acceleration motion; in fact, when the acceleration changes (during the
process of catching the ball) we will again assume constant acceleration conditions; in this case, we have
a2 = +25g = 245 m/s
2.
(a) The time of fall is given by Eq. 2-15 with v0 = 0 and y = 0. Thus,
t =
√
2y0
g
=
√
2(145)
9.8
= 5.44 s .
(b) The final velocity for its free-fall (which becomes the initial velocity during the catching process)
is found from Eq. 2-16 (other equations can be used but they would use the result from part (a)).
v = −
√
v20 − 2g (y − y0) = −
√
2gy0 = −53.3 m/s
where the negative root is chosen since this is a downward velocity.
(c) For the catching process, the answer to part (b) plays the role of an initial velocity (v0 = −53.3 m/s)
and the final velocity must become zero. Using Eq. 2-16, we find
∆y2 =
v2 − v20
2a2
=
−(−53.3)2
2(245)
= −5.80 m
where the negative value of ∆y2 signifies that the distance traveled while arresting its motion is
downward.
45. Taking the +y direction downward and y0 = 0, we have y = v0t +
1
2gt
2 which (with v0 = 0) yields
t =
√
2y/g.
29
(a) For this part of the motion, y = 50 m so that
t =
√
2(50)
9.8
= 3.2 s .
(b) For this next part of the motion, we note that the total displacement is y = 100 m. Therefore, the
total time is
t =
√
2(100)
9.8
= 4.5 s .
The difference between this and the answer to part (a) is the time required to fall through that
second 50 m distance: 4.5− 3.2 = 1.3 s.
46. We neglect air resistance, which justifies setting a = −g = −9.8 m/s2 (taking down as the −y direction)
for the duration of the motion. We are allowed to use Table 2-1 (with ∆y replacing ∆x) because this is
constant acceleration motion. The ground level is taken to correspond to y = 0.
(a) With y0 = h and v0 replaced with −v0, Eq. 2-16 leads to
v =
√
(−v0)2 − 2g (y − y0) =
√
v20 + 2gh .
The positive root is taken because the problem asks for the speed (the magnitude of the velocity).
(b) We use the quadratic formula to solve Eq. 2-15 for t, with v0 replaced with −v0,
∆y = −v0t− 1
2
gt2 =⇒ t =
−v0 +
√
(−v0)2 − 2g∆y
g
where the positive root is chosen to yield t > 0. With y = 0 and y0 = h, this becomes
t =
√
v20 + 2gh− v0
g
.
(c) If it were thrown upward with that speed from height h then (in the absence of air friction) it would
return to height h with that same downward speed and would therefore yield the same final speed
(before hitting the ground) as in part (a). An important perspective related to this is treated later
in the book (in the context of energy conservation) .
(d) Having to travel up before it starts its descent certainly requires more time than in part (b). The
calculation is quite similar, however, except for now having +v0 in the equation where we had put
in −v0 in part (b). The details follow:
∆y = v0t− 1
2
gt2 =⇒ t = v0 +
√
v20 − 2g∆y
g
with the positive root again chosen to yield t > 0. With y = 0 and y0 = h, we obtain
t =
√
v20 + 2gh+ v0
g
.
47. We neglect air resistance, which justifies setting a = −g = −9.8 m/s2 (taking down as the −y direction)
for the duration of the motion. We are allowed to use Table 2-1 (with ∆y replacing ∆x) because this is
constant acceleration motion. The ground level is taken to correspond to the origin of the y axis.
(a) Using y = v0t− 12gt2, with y = 0.544 m and t = 0.200 s, we find
v0 =
y + 12gt
2
t
=
0.544 + 12 (9.8)(0.200)
2
0.200
= 3.70 m/s .
30 CHAPTER 2.
(b) The velocity at y = 0.544 m is
v = v0 − gt = 3.70− (9.8)(0.200) = 1.74 m/s .
(c) Using v2 = v20 − 2gy (with different values for y and v than before), we solve for the value of y
corresponding to maximum height (where v = 0).
y =
v20
2g
=
3.72
2(9.8)
= 0.698 m .
Thus, the armadillo goes 0.698− 0.544 = 0.154 m higher.
48. We neglect air resistance, which justifies setting a = −g = −9.8 m/s2 (taking down as the −y direction)
for the duration of the motion. We are allowed to use Table 2-1 (with ∆y replacing ∆x) because this is
constant acceleration motion. The ground level is taken to correspond to the origin of the y axis. The
total time of fall can be computed from Eq. 2-15 (using the quadratic formula).
∆y = v0t− 1
2
gt2 =⇒ t = v0 +
√
v20 − 2g∆y
g
with the positive root chosen. With y = 0, v0 = 0 and y0 = h = 60 m, we obtain
t =
√
2gh
g
=
√
2h
g
= 3.5 s .
Thus, “1.2 s earlier” means we are examining where the rock is at t = 2.3 s:
y − h = v0(2.3)− 1
2
g(2.3)2 =⇒ y = 34 m
where we again use the fact that h = 60 m and v0 = 0.
49. The speed of the boat is constant, given by vb = d/t. Here, d is the distance of the boat from the bridge
when the key is dropped (12m) and t is the time the key takes in falling. To calculate t, we put the
origin of the coordinate system at the point where the key is dropped and take the y axis to be positive
in the downward direction. Taking the time to be zero at the instant the key is dropped, we compute
the time t when y = 45m. Since the initial velocity of the key is zero, the coordinate of the key is given
by y = 12gt
2. Thus
t =
√
2y
g
=
√
2(45m)
9.8m/s
2 = 3.03 s .
Therefore, the speed of the boat is
vb =
12m
3.03 s
= 4.0 m/s .
50. With +y upward, we have y0 = 36.6 m and y = 12.2 m. Therefore, using Eq. 2-18 (the last equation in
Table 2-1), we find
y − y0 = vt+ 1
2
gt2 =⇒ v = −22 m/s
at t = 2.00 s. The term speed refers to the magnitude of the velocity vector, so the answer is |v| =
22.0 m/s.
51. We first find the velocity of the ball just before it hits the ground. During contact with the ground its
average acceleration is given by
aavg =
∆v
∆t
31
where ∆v is the change in its velocity during contact with the ground and ∆t = 20.0 × 10−3 s is the
duration of contact. Now, to find the velocity just before contact, we put the origin at the point where
the ball is dropped (and take +y upward) and take t = 0 to be when it is dropped. The ball strikes the
ground at y = −15.0 m. Its velocity there is found from Eq. 2-16: v2 = −2gy. Therefore,
v = −
√
−2gy = −
√
−2(9.8)(−15.0) = −17.1 m/s
where the negative sign is chosen since the ball is traveling downward at the moment of contact. Con-
sequently, the average acceleration during contact with the ground is
aavg =
0− (−17.1)
20.0× 10−3 = 857 m/s
2
.
The fact that the result is positive indicates that this acceleration vector points upward. In a later
chapter, this will be directly related to the magnitude and direction of the force exerted by the ground
on the ball during the collision.
52. The y axis is arranged so that ground level is y = 0 and +y is upward.
(a) At the point where its fuel gets exhausted, the rocket has reached a height of
y′ =
1
2
at2 =
(4.00)(6.00)2
2
= 72.0 m .
From Eq. 2-11, the speed of the rocket (which had started at rest) at this instant is
v′ = at = (4.00)(6.00) = 24.0 m/s .
The additional height ∆y1 the rocket can attain (beyond y
′) is given by Eq. 2-16 with vanishing
final speed: 0 = v′ 2 − 2g∆y1. This gives
∆y1 =
v′ 2
2g
=
(24.0)2
2(9.8)
= 29.4 m .
Recalling our value for y′, the total height the rocket attains is seen to be 72.0 + 29.4 = 101 m.
(b) The time of free-fall flight (from y′ until it returns to y = 0) after the fuel gets exhausted is found
from Eq. 2-15:
−y′ = v′t− 1
2
gt2 =⇒ −72.0 = (24.0)t− 9.80
2
t2 .
Solving for t (using the quadratic formula) we obtain t = 7.00 s. Recalling the upward acceleration
time used in part (a), we see the total time of flight is 7.00 + 6.00 = 13.0 s.
53. The average acceleration during contact with the floor is given by aavg = (v2 − v1)/∆t, where v1 is its
velocity just before striking the floor, v2 is its velocity just as it leaves the floor, and ∆t is the duration
of contact with the floor (12× 10−3 s). Taking the y axis to be positively upward and placing the origin
at the point where the ball is dropped, we first find the velocity just before striking the floor, using
v21 = v
2
0 − 2gy. With v0 = 0 and y = −4.00 m, the result is
v1 = −
√
−2gy = −
√
−2(9.8)(−4.00) = −8.85 m/s
where the negative root is chosen because the ball is traveling downward. To find the velocity just after
hitting the floor (as it ascends without air friction to a height of 2.00 m), we use v2 = v22−2g(y−y0) with
v = 0, y = −2.00 m (it ends up two meters below its initial drop height), and y0 = −4.00 m. Therefore,
v2 =
√
2g(y − y0) =
√
2(9.8)(−2.00 + 4.00) = 6.26 m/s .
32 CHAPTER 2.
Consequently, the average acceleration is
aavg =
v2 − v1
∆t
=
6.26 + 8.85
12.0× 10−3 = 1.26× 10
3 m/s
2
.
The positive nature of the result indicates that the acceleration vector points upward. In a later chapter,
this will be directly related to the magnitude and direction of the force exerted by the ground on the
ball during the collision.
54. The height reached by the player is y = 0.76 m (where we have taken the origin of the y axis at the floor
and +y to be upward).
(a) The initial velocity v0 of the player is
v0 =
√
2gy =
√
2(9.8)(0.76) = 3.86 m/s .
This is a consequence of Eq. 2-16 where velocity v vanishes. As the player reaches y1 = 0.76−0.15 =
0.61 m, his speed v1 satisfies v
2
0 − v21 = 2gy1, which yields
v1 =
√
v20 − 2gy1 =
√
(3.86)2 − 2(9.80)(0.61) = 1.71 m/s .
The time t1 that the player spends ascending in the top ∆y1 = 0.15 m of the jump can now be
found from Eq. 2-17:
∆y1 =
1
2
(v1 + v) t1 =⇒ t1 = 2(0.15)
1.71 + 0
= 0.175 s
which means that the total time spend in that top 15 cm (both ascending and descending) is
2(0.17) = 0.35 s = 350 ms.
(b) The time t2 when the player reaches a height of 0.15 m is found from Eq. 2-15:
0.15 = v0t2 − 1
2
gt22 = (3.86)t2 −
9.8
2
t22 ,
which yields (using the quadratic formula, taking the smaller of the two positive roots) t2 = 0.041 s
= 41 ms, which implies that the total time spend in that bottom 15 cm (both ascending and
descending) is 2(41) = 82 ms.
55. We neglect air resistance, which justifies setting a = −g = −9.8 m/s2 (taking down as the −y direction)
for the duration of the motion. We are allowed to use Table 2-1 (with ∆y replacing ∆x) because this is
constant acceleration motion. The ground level is taken to correspond to the origin of the y axis. The
time drop 1 leaves the nozzle is taken as t = 0 and its time of landing on the floor t1 can be computed
from Eq. 2-15, with v0 = 0 and y1 = −2.00 m.
y1 = −1
2
gt21 =⇒ t1 =
√−2y
g
=
√
−2(−2.00)
9.8
= 0.639 s .
At that moment,the fourth drop begins to fall, and from the regularity of the dripping we conclude that
drop 2 leaves the nozzle at t = 0.639/3 = 0.213 s and drop 3 leaves the nozzle at t = 2(0.213) = 0.426 s.
Therefore, the time in free fall (up to the moment drop 1 lands) for drop 2 is t2 = t1 − 0.213 = 0.426 s
and the time in free fall (up to the moment drop 1 lands) for drop 3 is t3 = t1 − 0.426 = 0.213 s. Their
positions at that moment are
y2 = −1
2
gt22 = −
1
2
(9.8)(0.426)2 = −0.889 m
y3 = −1
2
gt23 = −
1
2
(9.8)(0.213)2 = −0.222 m ,
respectively. Thus, drop 2 is 89 cm below the nozzle and drop 3 is 22 cm below the nozzle when drop 1
strikes the floor.
33
56. The graph shows y = 25 m to be the highest point (where the speed momentarily vanishes). The neglect
of “air friction” (or whatever passes for that on the distant planet) is certainly reasonable due to the
symmetry of the graph.
(a) To find the acceleration due to gravity gp on that planet, we use Eq. 2-15 (with +y up)
y − y0 = vt+ 1
2
gpt
2 =⇒ 25− 0 = (0)(2.5) + 1
2
gp(2.5)
2
so that gp = 8.0 m/s
2.
(b) That same (max) point on the graph can be used to find the initial velocity.
y − y0 = 1
2
(v0 + v) t =⇒ 25− 0 = 1
2
(v0 + 0) (2.5)
Therefore, v0 = 20 m/s.
57. Taking +y to be upward and placing the origin at the point from which the objects are dropped, then the
location of diamond 1 is given by y1 = − 12gt2 and the location of diamond 2 is given by y2 = − 12g(t−1)2.
We are starting the clock when the first object is dropped. We want the time for which y2− y1 = 10 m.
Therefore,
−1
2
g(t− 1)2 + 1
2
gt2 = 10 =⇒ t = (10/g) + 0.5 = 1.5 s .
58. We neglect air resistance, which justifies setting a = −g = −9.8 m/s2 (taking down as the −y direction)
for the duration of the motion. We are allowed to use Table 2-1 (with ∆y replacing ∆x) because this is
constant acceleration motion. When something is thrown straight up and is caught at the level it was
thrown from (with a trajectory similar to that shown in Fig. 2-25), the time of flight t is half of its time
of ascent ta, which is given by Eq. 2-18 with ∆y = H and v = 0 (indicating the maximum point).
H = vta +
1
2
gt2a =⇒ ta =
√
2H
g
Writing these in terms of the total time in the air t = 2ta we have
H =
1
8
gt2 =⇒ t = 2
√
2H
g
.
We consider two
throws, one to height H1 for total time t1 and another to height H2 for total time t2,
and we set up a ratio:
H2
H1
=
1
8gt
2
2
1
8gt
2
1
=
(
t2
t1
)2
from which we conclude that if t2 = 2t1 (as is required by the problem) then H2 = 2
2H1 = 4H1.
59. We neglect air resistance, which justifies setting a = −g = −9.8 m/s2 (taking down as the −y direction)
for the duration of the motion. We are allowed to use Table 2-1 (with ∆y replacing ∆x) because this is
constant acceleration motion. We placing the coordinate origin on the ground. We note that the initial
velocity of the package is the same as the velocity of the balloon, v0 = +12 m/s and that its initial
coordinate is y0 = +80 m.
(a) We solve y = y0+v0t− 12gt2 for time, with y = 0, using the quadratic formula (choosing the positive
root to yield a positive value for t).
t =
v0 +
√
v20 + 2gy0
g
=
12 +
√
122 + 2(9.8)(80)
9.8
= 5.4 s
34 CHAPTER 2.
(b) If we wish to avoid using the result from part (a), we could use Eq. 2-16, but if that is not a
concern, then a variety of formulas from Table 2-1 can be used. For instance, Eq. 2-11 leads to
v = v0 − gt = 12− (9.8)(5.4) = −41 m/s. Its final speed is 41 m/s.
60. We neglect air resistance, which justifies setting a = −g = −9.8 m/s2 (taking down as the −y direction)
for the duration of the motion. We are allowed to use Eq. 2-15 (with ∆y replacing ∆x) because this is
constant acceleration motion. We use primed variables (except t) with the first stone, which has zero
initial velocity, and unprimed variables with the second stone (with initial downward velocity −v0, so
that v0 is being used for the initial speed). SI units are used throughout.
∆y′ = 0(t)− 1
2
gt2
∆y = (−v0) (t− 1)− 1
2
g(t− 1)2
Since the problem
indicates ∆y′ =
∆y = −43.9 m,
we solve the
first equation
for t (finding
t = 2.99 s) and
use this result to
solve the second
equation for the
initial speed of
the second stone:
21
0
20
40
y
2t
−43.9 = (−v0) (1.99)− 1
2
(9.8)(1.99)2
which leads to v0 = 12.3 m/s.
61. We neglect air resistance, which justifies setting a = −g = −9.8 m/s2 (taking down as the −y direction)
for the duration of the motion of the shot ball. We are allowed to use Table 2-1 (with ∆y replacing
∆x) because the ball has constant acceleration motion. We use primed variables (except t) with the
constant-velocity elevator (so v′ = 20 m/s), and unprimed variables with the ball (with initial velocity
v0 = v
′ + 10 = 30 m/s, relative to the ground). SI units are used throughout.
(a) Taking the time to be zero at the instant the ball is shot, we compute its maximum height y (relative
to the ground) with v2 = v20 − 2g(y− yo), where the highest point is characterized by v = 0. Thus,
y = yo +
v20
2g
= 76 m
where yo = y
′
o + 2 = 30 m (where y
′
o = 28 m is given in the problem) and v0 = 30 m/s relative to
the ground as noted above.
(b) There are a variety of approaches to this question. One is to continue working in the frame of
reference adopted in part (a) (which treats the ground as motionless and “fixes” the coordinate
origin to it); in this case, one describes the elevator motion with y′ = y′o + v
′t and the ball motion
with Eq. 2-15, and solves them for the case where they reach the same point at the same time.
Another is to work in the frame of reference of the elevator (the boy in the elevator might be
oblivious to the fact the elevator is moving since it isn’t accelerating), which is what we show here
in detail:
∆ye = v0et− 12gt
2 =⇒ t = v0e +
√
v02e − 2g∆ye
g
35
where v0e = 20 m/s is the initial velocity of the ball relative to the elevator and ∆ye = −2.0 m is
the ball’s displacement relative to the floor of the elevator. The positive root is chosen to yield a
positive value for t ; the result is t = 4.2 s.
62. We neglect air resistance, which justifies setting a = −g = −9.8 m/s2 (taking down as the −y direction)
for the duration of the stone’s motion. We are allowed to use Table 2-1 (with ∆x replaced by y) because
the ball has constant acceleration motion (and we choose yo = 0).
(a) We apply Eq. 2-16 to both measurements, with SI units understood.
v2B = v
2
0 − 2gyB =⇒
(
1
2
v
)2
+ 2g(yA + 3) = v
2
0
v2A = v
2
0 − 2gyA =⇒ v2 + 2gyA = v20
We equate the two expressions that each equal v20 and obtain
1
4
v2 + 2gyA + 2g(3) = v
2 + 2gyA =⇒ 2g(3) = 3
4
v2
which yields v =
√
2g(4) = 8.85 m/s.
(b) An object moving upward at A with speed v = 8.85 m/s will reach a maximum height y − yA =
v2/2g = 4.00 m above point A (this is again a consequence of Eq. 2-16, now with the “final” velocity
set to zero to indicate the highest point). Thus, the top of its motion is 1.00 m above point B.
63. The object, once it is dropped (v0 = 0) is in free-fall (a = −g = −9.8 m/s2 if we take down as the −y
direction), and we use Eq. 2-15 repeatedly.
(a) The (positive) distance D from the lower dot to the mark corresponding to a certain reaction time
t is given by ∆y = −D = − 12gt2, or D = gt2/2. Thus for t1 = 50.0ms
D1 =
(9.8m/s
2
)(50.0× 10−3 s)2
2
= 0.0123m = 1.23 cm .
(b) For t2 = 100ms
D2 =
(9.8m/s
2
)(100× 10−3 s)2
2
= 0.049m = 4D1 ;
for t3 = 150ms
D3 =
(9.8m/s2)(150× 10−3 s)2
2
= 0.11m = 9D1 ;
for t4 = 200ms
D4 =
(9.8m/s
2
)(200× 10−3 s)2
2
= 0.196m = 16D1 ;
and for t4 = 250ms
D5 =
(9.8m/s
2
)(250× 10−3 s)2
2
= 0.306m = 25D1 .
64. During free fall, we ignore the air resistance and set a = −g = −9.8 m/s2 where we are choosing down
to be the −y direction. The initial velocity is zero so that Eq. 2-15 becomes ∆y = − 12gt2 where ∆y
represents the negative of the distance d she has fallen. Thus, we can write the equation as d = 12gt
2 for
simplicity.
36 CHAPTER 2.
(a) The time t1 during which the parachutist is in free fall is (using Eq. 2-15) given by
d1 = 50m =
1
2
gt21 =
1
2
(
9.80m/s
2
)
t21
which yields t1 = 3.2 s. The speed of the parachutist just before he opens the parachute is given by
the positive root v21 = 2gd1, or
v1 =
√
2gh1 =
√
(2)(9.80m/s
2
)(50m) = 31m/s .
If the final speed is v2, then the time interval t2 between the opening of the parachute and the
arrival of the parachutist at the ground level is
t2 =
v1 − v2
a
=
31m/s− 3.0m/s
2m/s
2 = 14 s .
This is a result of Eq. 2-11 where speeds are used instead of the (negative-valued) velocities (so that
final-velocity minus initial-velocity turns out to equal initial-speed minus final-speed); we also note
that the acceleration vector for this part of the motion is positive since it points upward (opposite
to the direction of motion – which makes it a deceleration). The total time of flight is therefore
t1 + t2 = 17 s.
(b) The distance through which the parachutist falls after the parachute is opened is given by
d =
v21 − v22
2a
=
(31m/s)2 − (3.0m/s)2
(2)(2.0m/s
2
)
≈ 240m .
In the computation, we have used Eq. 2-16 with both sides multiplied by −1 (which changes the
negative-valued ∆y into the positive d on the left-hand side, and switches the order of v1 and v2
on the right-hand side). Thus the fall begins at a height of h = 50 + d ≈ 290m.
65. The time t the pot spends passing in front of the window of length L = 2.0 m is 0.25 s each way. We use
v for its velocity as it passes the top of the window (going up). Then, with a = −g = −9.8 m/s2 (taking
down to be the −y direction), Eq. 2-18 yields
L = vt− 1
2
gt2 =⇒ v = L
t
− 1
2
gt .
The distance H the pot goes above the top of the window is therefore (using Eq. 2-16 with the final
velocity being zero to indicate the highest point)
H =
v2
2g
=
(L/t− gt/2)2
2g
=
(2.00/0.25− (9.80)(0.25)/2)2
(2) (9.80)
= 2.34m .
66. The time being considered is 6 years and roughly 235
days, which is approximately ∆t = 2.1 × 107 s.
Using Eq. 2-3, we find the average speed to be
30600× 103m
2.1× 107 s = 0.15 m/s .
67. We assume constant velocity motion and use Eq. 2-2 (with vavg = v > 0). Therefore,
∆x = v∆t =
(
303
km
h
(
1000m/km
3600 s/h
))(
100× 10−3 s) = 8.4m .
68. For each rate, we use distance d = vt and convert to SI using 0.0254 cm = 1 inch (from which we derive
the factors appearing in the computations below).
37
(a) The total distance d comes from summing
d1 =
(
120
steps
min
)(
0.762m/step
60 s/min
)
(5 s) = 7.62m
d2 =
(
120
steps
min
)(
0.381m/step
60 s/min
)
(5 s) = 3.81m
d3 =
(
180
steps
min
)(
0.914m/step
60 s/min
)
(5 s) = 13.72m
d4 =
(
180
steps
min
)(
0.457m/step
60 s/min
)
(5 s) = 6.86m
so that d = d1 + d2 + d3 + d4 = 32 m.
(b) Average velocity is computed using Eq. 2-2: vavg = 32/20 = 1.6 m/s, where we have used the fact
that the total time is 20 s.
(c) The total time t comes from summing
t1 =
8m(
120 stepsmin
) ( 0.762m/step
60 s/min
) = 5.25 s
t2 =
8m(
120 stepsmin
) ( 0.381m/step
60 s/min
) = 10.5 s
t3 =
8m(
180 stepsmin
) ( 0.914m/step
60 s/min
) = 2.92 s
t4 =
8m(
180 stepsmin
) ( 0.457m/step
60 s/min
) = 5.83 s
so that t = t1 + t2 + t3 + t4 = 24.5 s.
(d) Average velocity is computed using Eq. 2-2: vavg = 32/24.5 = 1.3 m/s, where we have used the
fact that the total distance is 4(8) = 32 m.
69. The statement that the stoneflies have “constant speed along a straight path” means we are dealing with
constant velocity motion (Eq. 2-2 with vavg replaced with vs or vns, as the case may be).
(a) We set up the ratio and simplify (using d for the common distance).
vs
vns
=
d/ts
d/tns
=
tns
ts
=
25.0
7.1
= 3.52
(b) We examine ∆t and simplify until we are left with an expression having numbers and no variables
other than vs. Distances are understood to be in meters.
tns − ts = 2
vns
− 2
vs
=
2
(vs/3.52)
− 2
vs
=
2
vs
(3.52− 1)
≈ 5
vs
38 CHAPTER 2.
70. We orient +x along the direction of motion (so a will be negative-valued, since it is a deceleration), and
we use Eq. 2-7 with aavg = −3400g = −3400(9.8) = −3.33 × 104 m/s2 and v = 0 (since the recorder
finally comes to a stop).
aavg =
v − v0
∆t
=⇒ v0 =
(
3.33× 104m/s2
) (
6.5× 10−3 s)
which leads to v0 = 217 m/s.
71. (a) It is the intent of this problem to treat the v0 = 0 condition rigidly. In other words, we are not
fitting the distance to just any second-degree polynomial in t; rather, we requiring d = At2 (which
meets the condition that d and its derivative is zero when t = 0). If we perform a leastsquares fit
with this expression, we obtain A = 3.587 (SI units understood). We return to this discussion in
part (c). Our expectation based on Eq. 2-15, assuming no error in starting the clock at the moment
the acceleration begins, is d = 12at
2 (since he started at the coordinate origin, the location of which
presumably is something we can be fairly certain about).
(b) The graph (d on the vertical axis, SI units understood) is shown.
The horizontal
axis is t2 (as
indicated by
the problem
statement) so
that we have
a straight line
instead of a
parabola.
0
20
40
d
10
t_squared
(c) Comparing our two expressions for d, we see the parameter A in our fit should correspond to 12a,
so a = 2(3.587) ≈ 7.2m/s2. Now, other approaches might be considered (trying to fit the data with
d = Ct2 +B for instance, which leads to a = 2C = 7.0m/s2 and B 6= 0), and it might be useful to
have the class discuss the assumptions made in each approach.
72. (a) We estimate x ≈ 2 m at t = 0.5 s, and x ≈ 12 m at t = 4.5 s. Hence, using the definition of average
velocity Eq. 2-2, we find
vavg =
12− 2
4.5− 0.5 = 2.5 m/s .
(b) In the region 4.0 ≤ t ≤ 5.0, the graph depicts a straight line, so its slope represents the instantaneous
velocity for any point in that interval. Its slope is the average velocity between t = 4.0 s and t = 5.0
s:
vavg =
16.0− 8.0
5.0− 4.0 = 8.0 m/s .
Thus, the instantaneous velocity at t = 4.5 s is 8.0 m/s. (Note: similar reasoning leads to a value
needed in the next part: the slope of the 0 ≤ t ≤ 1 region indicates that the instantaneous velocity
at t = 0.5 s is 4.0 m/s.)
(c) The average acceleration is defined by Eq. 2-7:
aavg =
v2 − v1
t2 − t1 =
8.0− 4.0
4.5− 0.5 = 1.0 m/s
2
.
(d) The instantaneous acceleration is the instantaneous rate-of-change of the velocity, and the constant
x vs. t slope in the interval 4.0 ≤ t ≤ 5.0 indicates that the velocity is constant during that interval.
Therefore, a = 0 at t = 4.5 s.
39
73. We use the functional notation x(t), v(t) and a(t) and find the latter two quantities by differentiating:
v(t) =
dx(t)
t
= 6.0t2 and a(t) =
dv(t)
dt
= 12t
with SI units understood. This expressions are used in the parts that follow.
(a) Using the definition of average velocity, Eq. 2-2, we find
vavg =
x(2)− x(1)
2.0− 1.0 =
2(2)3 − 2(1)3
1.0
= 14 m/s .
(b) The average acceleration is defined by Eq. 2-7:
aavg =
v(2)− v(1)
2.0− 1.0 =
6(2)2 − 6(1)2
1.0
= 18 m/s2 .
(c) The value of v(t) when t = 1.0 s is v(1) = 6(1)2 = 6.0 m/s.
(d) The value of a(t) when t = 1.0 s is a(1) = 12(1) = 12 m/s2.
(e) The value of v(t) when t = 2.0 s is v(2) = 6(2)2 = 24 m/s.
(f) The value of a(t) when t = 2.0 s is a(2) = 12(2) = 24 m/s2.
(g) We don’t expect average values of a quantity, say, heights of trees, to equal any particular height
for a specific tree, but we are sometimes surprised at the different kinds of averaging that can be
performed. Now, the acceleration is a linear function (of time) so its average as defined by Eq. 2-7
is, not surprisingly, equal to the arithmetic average of its a(1) and a(2) values. The velocity is not
a linear function so the result of part (a) is not equal to the arithmetic average of parts (c) and
(e) (although it is fairly close). This reminds us that the calculus-based definition of the average a
function (equivalent to Eq. 2-2 for vavg ) is not the same as the simple idea of an arithmetic average
of two numbers; in other words,
1
t′ − t
∫ t′
t
f(τ) dτ 6= f(t
′) + f(t)
2
except in very special cases (like with linear functions).
(h) The graphs are shown below, x(t) on the left and v(t) on the right. SI units are understood. We
do not show the tangent lines (representing instantaneous slope values) at t = 1 and t = 2, but we
do show line segments representing the average quantities computed in parts (a) and (b).
0
10
20
30
x
1 2t
v
0
10
20
30
1 2t
74. We choose down as the +y direction and set the coordinate origin at the point where it was dropped
(which is when we start the clock). We denote the 1.00 s duration mentioned in the problem as t − t′
where t is the value of time when it lands and t′ is one second prior to that. The corresponding distance
is y − y′ = 0.50h, where y denotes the location of the ground. In these terms, y is the same as h, so we
have h− y′ = 0.50h or 0.50h = y′.
40 CHAPTER 2.
(a) We find t′ and t from Eq. 2-15 (with v0 = 0):
y′ =
1
2
gt′ 2 =⇒ t′ =
√
2y′
g
y =
1
2
gt2 =⇒ t =
√
2y
g
.
Plugging in y = h and y′ = 0.50h, and dividing these two equations, we obtain
t′
t
=
√
2(0.50h)/g
2h/g
=
√
0.50 .
Letting t′ = t− 1.00 (SI units understood) and cross-multiplying, we find
t− 1.00 = t
√
0.50 =⇒ t = 1.00
1−√0.50
which yields t = 3.41 s.
(b) Plugging this result into y = 12gt
2 we find h = 57 m.
(c) In our approach, we did not use the quadratic formula, but we did “choose a root” when we assumed
(in the last calculation in part (a)) that
√
0.50 = +2.236 instead of −2.236.
If we had instead let√
0.50 = −2.236 then our answer for t would have been roughly 0.6 s which would imply that
t′ = t− 1 would equal a negative number (indicating a time before it was dropped) which certainly
does not fit with the physical situation described in the problem.
75. (a) Let the height of the diving board be h. We choose down as the +y direction and set the coordinate
origin at the point where it was dropped (which is when we start the clock). Thus, y = h designates
the location where the ball strikes the water. Let the depth of the lake be D, and the total time
for the ball to descend be T . The speed of the ball as it reaches the surface of the lake is then
v =
√
2gh (from Eq. 2-16), and the time for the ball to fall from the board to the lake surface is
t1 =
√
2h/g (from Eq. 2-15). Now, the time it spends descending in the lake (at constant velocity
v) is
t2 =
D
v
=
D√
2gh
.
Thus, T = t1 + t2 =
√
2h
g +
D√
2gh
, which gives
D = T
√
2gh− 2h = (4.80)
√
(2)(9.80)(5.20)− (2)(5.20) = 38.1 m .
(b) Using Eq. 2-2, the average velocity is
vavg =
D + h
T
=
38.1 + 5.20
4.80
= 9.02 m/s
where (recalling our coordinate choices) the positive sign means that the ball is going downward
(if, however, upwards had been chosen as the positive direction, then this answer would turn out
negative-valued).
(c) We find v0 from ∆y = v0t+
1
2gt
2 with t = T and ∆y = h+D. Thus,
v0 =
h+D
T
− gT
2
=
5.20 + 38.1
4.80
− (9.8)(4.80)
2
= −14.5 m/s
where (recalling our coordinate choices) the negative sign means that the ball is being thrown
upward.
41
76. The time ∆t is 2(60) + 41 = 161 min and the displacement ∆x = 370 cm. Thus, Eq. 2-2 gives
vavg =
∆x
∆t
=
370
161
= 2.3 cm/min .
77. We use the functional notation x(t), v(t) and a(t) and find the latter two quantities by differentiating:
v(t) =
dx(t)
t
= −15t2 + 20 and a(t) = dv(t)
dt
= −30t
with SI units understood. This expressions are used in the parts that follow.
(a) From 0 = −15t2+20, we see that the only positive value of t for which the particle is (momentarily)
stopped is t =
√
20/15 = 1.2 s.
(b) From 0 = −30t, we find a(0) = 0 (that is, it vanishes at t = 0).
(c) It is clear that a(t) = −30t is negative for t > 0 and positive for t < 0.
(d) We show the two of the graphs below (the third graph, a(t), which is a straight line through the
origin with slope= −30 is omitted in the interest of saving space). SI units are understood.
–20
0
x
t
–50
0
v
t
78. (a) It follows from Eq. 2-8 that v − v0 =
∫
a dt, which has the geometric interpretation of being the
area under the graph. Thus, with v0 = 2.0 m/s and that area amounting to 3.0 m/s (adding that
of a triangle to that of a square, over the interval 0 ≤ t ≤ 2 s), we find v = 2.0 + 3.0 = 5.0 m/s
(which we will denote as v2 in the next part). The information given that x0 = 4.0 m is not used
in this solution.
(b) During 2 < t ≤ 4 s, the graph of a is a straight line with slope 1.0 m/s3. Extrapolating, we see that
the intercept of this line with the a axis is zero. Thus, with SI units understood,
v = v2 +
∫ t
2.0
a dτ = 5.0 +
∫ t
2.0
(1.0)τ dτ = 5.0 +
(1.0)t2 − (1.0)(2.0)2
2
which yields v = 3.0 + 0.50t2 in m/s.
79. We assume the train accelerates from rest (v0 = 0 and x0 = 0) at a1 = +1.34 m/s
2 until the midway
point and then decelerates at a2 = −1.34 m/s2 until it comes to a stop (v2 = 0) at the next station. The
velocity at the midpoint is v1 which occurs at x1 = 806/2 = 403 m.
(a) Eq. 2-16 leads to
v21 = v
2
0 + 2a1x1 =⇒ v1 =
√
2(1.34)(403)
which yields v1 = 32.9 m/s.
(b) The time t1 for the accelerating stage is (using Eq. 2-15)
x1 = v0t1 +
1
2
a1t
2
1 =⇒ t1 =
√
2(403)
1.34
which yields t1 = 24.53 s. Since the time interval for the decelerating stage turns out to be the
same, we double this result and obtain t = 49.1 s for the travel time between stations.
42 CHAPTER 2.
(c) With a “dead time” of 20 s, we have T = t+20 = 69.1 s for the total time between start-ups. Thus,
Eq. 2-2 gives
vavg =
806m
69.1 s
= 11.7 m/s .
(d) We show the two of the graphs below. The third graph, a(t), is not shown to save space; it consists
of three horizontal “steps” – one at 1.34 during 0 < t < 24.53 and the next at −1.34 during
24.53 < t < 49.1 and the last at zero during the “dead time” 49.1 < t < 69.1). SI units are
understood.
x
0
200
400
600
800
10 20 30 40 50 60t
0
10
20
30
v
0 10 20 30 40 50 60t
80. Average speed, as opposed to average velocity, relates to the total distance, as opposed to the net
displacement. The distance D up the hill is, of course, the same as the distance down the hill, and since
the speed is constant (during each stage of the motion) we have speed= D/t. Thus, the average speed
is
Dup +Ddown
tup + tdown
=
2D
D
vup
+ Dvdown
which, after canceling D and plugging in vup = 40 km/h and vdown = 60 km/h, yields 48 km/h for the
average speed.
81. During Tr the velocity v0 is constant (in the direction we choose as +x) and obeys v0 = Dr/Tr where
we note that in SI units the velocity is v0 = 200(1000/3600) = 55.6 m/s. During Tb the acceleration is
opposite to the direction of v0 (hence, for us, a < 0) until the car is stopped (v = 0).
(a) Using Eq. 2-16 (with ∆xb = 170 m) we find
v2 = v20 + 2a∆xb =⇒ a = −
v20
2∆xb
which yields |a| = 9.08 m/s2.
(b) We express this as a multiple of g by setting up a ratio:
a =
(
9.08
9.8
)
9 = 0.926g .
(c) We use Eq. 2-17 to obtain the braking time:
∆xb =
1
2
(v0 + v)Tb =⇒ Tb = 2(170)
55.6
= 6.12 s .
(d) We express our result for Tb as a multiple of the reaction time Tr by setting up a ratio:
Tb =
(
6.12
400× 10−3
)
Tr = 15.3Tr .
43
(e) We are only asked what the increase in distance D is, due to ∆Tr = 0.100 s, so we simply have
∆D = v0∆Tr = (55.6)(0.100) = 5.56 m .
82. We take +x in the direction of motion. We use subscripts 1 and 2 for the data. Thus, v1 = +30 m/s,
v2 = +50 m/s and x2 − x1 = +160 m.
(a) Using these subscripts, Eq. 2-16 leads to
a =
v22 − v21
2 (x2 − x1) =
502 − 302
2(160)
= 5.0 m/s2 .
(b) We find the time interval corresponding to the displacement x2 − x1 using Eq. 2-17:
t2 − t1 = 2 (x2 − x1)
v1 + v2
=
2(160)
30 + 50
= 4.0 s .
(c) Since the train is at rest (v0 = 0) when the clock starts, we find the value of t1 from Eq. 2-11:
v1 = v0 + at1 =⇒ t1 = 30
5.0
= 6.0 s .
(d) The coordinate origin is taken to be the location at which the train was initially at rest (so x0 = 0).
Thus, we are asked to find the value of x1 . Although any of several equations could be used, we
choose Eq. 2-17:
x1 =
1
2
(v0 + v1) t1 =
1
2
(30)(6.0) = 90 m .
(e) The graphs are shown below, with SI units assumed.
0
200
x
10t 0
50
v
10t
83. Direction of +x is implicit in the problem statement. The initial position (when the clock starts) is
x0 = 0 (where v0 = 0), the end of the speeding-up motion occurs at x1 = 1100/2 = 550 m, and the
subway comes to a halt (v2 = 0) at x2 = 1100 m.
(a) Using Eq. 2-15, the subway reaches x1 at
t1 =
√
2x1
a1
=
√
2(550)
1.2
= 30.3 s .
The time interval t2 − t1 turns out to be the same value (most easily seen using Eq. 2-18 so the
total time is t2 = 2(30.3) = 60.6 s.
(b) Its maximum speed occurs at t1 and equals
v1 = v0 + a1t1 = 36.3 m/s .
(c) The graphs are not shown here, in the interest of saving space. They are very similar to those
shown in the solution for problem 79, above.
44 CHAPTER 2.
84. We note that the running time for Bill Rodgers is ∆tR = 2(3600) + 10(60) = 7800 s. We also note that
the magnitude of the average velocity (Eq. 2-2) and Eq. 2-3 (for average speed) agree
in this exercise
(which is not usually the case).
(a) Denoting the Lewis’ average velocity as vL (similarly for Rodgers), we find
vL =
100m
10 s
= 10 m/s vR =
42000m
7800 s
= 5.4 m/s .
(b) If Lewis continued at this rate, he would covered D = 42000 m in
∆tL =
D
vL
=
42000
10
= 4200 s
which is equivalent to 1 h and 10 min.
85. We choose down as the +y direction and use the equations of Table 2-1 (replacing x with y) with a = +g,
v0 = 0 and y0 = 0. We use subscript 2 for the elevator reaching the ground and 1 for the halfway point.
(a) Eq. 2-16, v22 = v
2
0 + 2a(y2 − y0), leads to
v2 =
√
2gy2 =
√
2(9.8)(120) = 48.5 m/s .
(b) The time at which it strikes the ground is (using Eq. 2-15)
t2 =
√
2y2
g
=
√
2(120)
9.8
= 4.95 s .
(c) Now Eq. 2-16, in the form v21 = v
2
0 + 2a(y1 − y0), leads to
v1 =
√
2gy1 =
√
2(9.8)(60) = 34.2 m/s .
(d) The time at which it reaches the halfway point is (using Eq. 2-15)
t1 =
√
2y1
g
=
√
2(60)
9.8
= 3.50 s .
86. To find the “launch” velocity of the rock, we apply Eq. 2-11 to the maximum height (where the speed
is momentarily zero)
v = v0 − gt =⇒ 0 = v0 − (9.8)(2.5)
so that v0 = 24.5 m/s (with +y up). Now we use Eq. 2-15 to find the height of the tower (taking y0 = 0
at the ground level)
y − y0 = v0t+ 1
2
at2 =⇒ y − 0 = (24.5)(1.5)− 1
2
(9.8)(1.5)2 .
Thus, we obtain y = 26 m.
87. We take the direction of motion as +x, so a = −5.18 m/s2, and we use SI units, so v0 = 55(1000/3600) =
15.28 m/s.
(a) The velocity is constant during the reaction time T , so the distance traveled during it is dr =
v0T − (15.28)(0.75) = 11.46 m. We use Eq. 2-16 (with v = 0) to find the distance db traveled during
braking:
v2 = v20 + 2adb =⇒ db = −
15.282
2(−5.18)
which yields db = 22.53 m. Thus, the total distance is dr+db = 34.0 m, which means that the driver
is able to stop in time. And if the driver were to continue at v0, the car would enter the intersection
in t = (40m)/(15.28m/s) = 2.6 s which is (barely) enough time to enter the intersection before the
light turns, which many people would consider an acceptable situation.
45
(b) In this case, the total distance to stop (found in part (a) to be 34 m) is greater than the distance to
the intersection, so the driver cannot stop without the front end of the car being a couple of meters
into the intersection. And the time to reach it at constant speed is 32/15.28 = 2.1 s, which is too
long (the light turns in 1.8 s). The driver is caught between a rock and a hard place.
88. We assume v0 = 0 and integrate the acceleration to find the velocity. In the graphs below (the first is
the acceleration, like Fig. 2-35 but with some numbers we adopted, and the second is the velocity) we
modeled the curve in the textbook with straight lines and circular arcs for the rounded corners, and
literally integrated it. The intent of the textbook was not, however, to go through such an involved
procedure, and one should be able to obtain a close approximation to the shape of the velocity graph
below (the one on the right) just by applying the idea that constant nonzero acceleration means a linearly
changing velocity.
–0.4
–0.2
0
0.2
0.4
0.6
0.8
1t
–0.2
0
0.2
v
t
89. We take the direction of motion as +x, take x0 = 0 and use SI units, so v = 1600(1000/3600) = 444 m/s.
(a) Eq. 2-11 gives 444 = a(1.8) or a = 247 m/s2. We express this as a multiple of g by setting up a
ratio:
a =
(
247
9.8
)
g = 25g .
(b) Eq. 2-17 readily yields
x =
1
2
(v0 + v) t =
1
2
(444)(1.8) = 400 m .
90. The graph is shown below. We assumed each interval described in the problem was one time unit long.
A marks where the curve is steepest and B is where it is least steep (where it, in fact, has zero slope).
x
B
A
–4
–2
0
t
91. We use the functional notation x(t), v(t) and a(t) in this solution, where the latter two quantities are
obtained by differentiation:
v(t) =
dx(t)
dt
= −12t and a(t) = dv(t)
dt
= −12
with SI units understood.
46 CHAPTER 2.
(a) From v(t) = 0 we find it is (momentarily) at rest at t = 0.
(b) We obtain x(0) = 4.0 m
(c) Requiring x(t) = 0 in the expression x(t) = 4.0− 6.0t2 leads to t = ±0.82 s for the times when the
particle can be found passing through the origin.
(d) We show both the asked-for graph (on the left) as well as the “shifted” graph which is relevant to
part (e). In both cases, the time axis is given by −3 ≤ t ≤ 3 (SI units understood).
–100
0
x
t
–100
0
x
t
(e) We arrived at the graph on the right (shown above) by adding 20t to the x(t) expression.
(f) Examining where the slopes of the graphs become zero, it is clear that the shift causes the v = 0
point to correspond to a larger value of x (the top of the second curve shown in part (d) is higher
than that of the first).
92. (a) The slope of the graph (at a point) represents the velocity there, and the up-or-down concavity
of the curve there indicates the ± sign of the acceleration. Thus, during AB we have v > 0 and
a = 0 (since it is a straight line). During BC, we still have v > 0 but there is some curvature
and a downward concavity is indicated (so a < 0). The segment CD is horizontal, implying the
particle remains at the same position for some time; thus, v = a = 0 during CD. Clearly, the slope
is negative during DE (so v < 0) but whether or not the graph is curved is less clear; we believe it
is, with an upward concavity (a > 0).
(b) The key word is “obviously.” Since it seems plausible to us that the curved portions can be “fit”
with parabolic arcs (indications of constant acceleration by Eq. 2-15), then our answer is “no.”
(c) Neither signs of slopes nor the sign of the concavity depends on a global shift in one axis or another
(or, for that matter, on rescalings of the axes themselves) so the answer again is “no.”
93. (a) The slope of the graph (at a point) represents the velocity there, and the up-or-down concavity of
the curve there indicates the ± sign of the acceleration. Thus, during AB we have positive slope
(v > 0) and a < 0 (since it is concave downward). The segment BC is horizontal, implying the
particle remains at the same position for some time; thus, v = a = 0 during BC. During CD we
have v > 0 and a > 0 (since it is concave upward). Clearly, the slope is positive during DE (so
v > 0) but whether or not the graph is curved is less clear; we believe it is not, so a = 0.
(b) The key word is “obviously.” Since it seems plausible to us that the curved portions can be “fit”
with parabolic arcs (indications of constant acceleration by Eq. 2-15), then our answer is “no.”
(c) Neither signs of slopes nor the sign of the concavity depends on a global shift in one axis or another
(or, for that matter, on rescalings of the axes themselves) so the answer again is “no.”
94. This problem consists of two parts: part 1 with constant acceleration (so that the equations in Table 2-1
apply), v0 = 0, v = 11.0 m/s, x = 12.0 m, and x0 = 0 (adopting the starting line as the coordinate
origin); and, part 2 with constant velocity (so that x − x0 = vt applies) with v = 11.0 m/s, x0 = 12.0,
and x = 100.0 m.
(a) We obtain the time for part 1 from Eq. 2-17
x− x0 = 1
2
(v0 + v) t1 =⇒ 12.0− 0 = 1
2
(0 + 11.0)t1
47
so that t1 = 2.2 s, and we find the time for part 2 simply from 88.0 = (11.0)t2 → t2 = 8.0 s.
Therefore, the total time is t1 + t2 = 10.2 s.
(b) Here, the total time is required to be 10.0 s, and we are to locate the point xp where the runner
switches from accelerating to proceeding at constant speed. The equations for parts 1 and 2, used
above, therefore become
xp − 0 = 1
2
(0 + 11.0)t1
100.0− xp = (11.0)(10.0− t1)
where in the latter equation, we use the fact that t2 = 10.0− t1. Solving the equations
for the two
unknowns, we find that t1 = 1.8 s and xp = 10.0 m.
95. We take +x in the direction of motion, so v0 = +24.6 m/s and a = −4.92 m/s2. We also take x0 = 0.
(a) The time to come to a halt is found using Eq. 2-11:
0 = v0 + at =⇒ t = − 24.6−4.92 = 5.00 s .
(b) Although several of the equations in Table 2-1 will yield the result, we choose Eq. 2-16 (since it
does not depend on our answer to part (a)).
0 = v20 + 2ax =⇒ x = −
24.62
2(−4.92) = 61.5 m .
(c) Using these results, we plot v0t+
1
2at
2 (the x graph, shown below, on the left) and v0 + at (the v
graph, below right) over 0 ≤ t ≤ 5 s, with SI units understood.
0
20
40
60
x
1 2 3 4 5t 0
10
20
v
1 2 3 4 5t
96. We take +x in the direction of motion, so
v = (60 km/h)
(
1000m/km
3600 s/h
)
= +16.7 m/s
and a > 0. The location where it starts from rest (v0 = 0) is taken to be x0 = 0.
(a) Eq. 2-7 gives aavg = (v − v0)/t where t = 5.4 s and the velocities are given above. Thus, aavg =
3.1 m/s2.
(b) The assumption that a =constant permits the use of Table 2-1. From that list, we choose Eq. 2-17:
x =
1
2
(v0 + v) t =
1
2
(16.7)(5.4) = 45 m .
(c) We use Eq. 2-15, now with x = 250 m:
x =
1
2
at2 =⇒ t =
√
2x
a
=
√
2(250)
3.1
which yields t = 13 s.
48 CHAPTER 2.
97. Converting to SI units, we have v = 3400(1000/3600) = 944 m/s (presumed constant) and ∆t = 0.10 s.
Thus, ∆x = v∆t = 94 m.
98. The (ideal) driving time before the change was t = ∆x/v, and after the change it is t′ = ∆x/v′. The
time saved by the change is therefore
t− t′ = ∆x
(
1
v
− 1
v′
)
= ∆x
(
1
55
− 1
65
)
= ∆x(0.0028 h/mi)
which becomes, converting ∆x = 700/1.61 = 435 mi (using a conversion found on the inside front cover
of the textbook), t− t′ = (435)(0.0028) = 1.2 h. This is equivalent to 1 h and 13 min.
99. (a) With the understanding that these are good to three significant figures, we write the function (in
SI units) as
x(t) = −32 + 24t2e−0.03t
and find the velocity and acceleration functions by differentiating (calculus is reviewed Appendix E).
We find
v(t) = 24t(2− 0.03t)e−0.03t and a(t) = 24 (2− 0.12t+ 0.0009t2) e−0.03t .
(b) The v(t) and a(t) graphs are shown below (SI units understood). The time axis in both cases runs
from t = 0 to t = 100 s. We include the x(t) graph in the next part, accompanying our discussion
of its root (which is, as suggested by the graph, a small positive value of t).
0
200v
20 40 60 80 100t 0
20
40
a
20 40 60 80 100t
(c) We seek to find a positive value of t for which 24t2e−0.03t = 32. We turn to the calculator or to
a computer for its (numerical) solution. In this case, we ignore the roots outside the 0 ≤ t ≤ 100
range (such as t = −1.14 s and
t = 387.77 s)
and choose
t = 1.175 s as
our answer.
All of these
are rounded-
off values.
We find
v = 53.5 m/s
and
a = 43.1 m/s2
at this time. 0
10000
x
20 40 60 80 100t
(d) It is much easier to find when 24t(2 − 0.03t)e−0.03t = 0 since the roots are clearly t1 = 0 and
t2 = 2/0.03 = 66.7 s. We find x(t1) = −32.0 m and a(t1) = 48.0 m/s2 at the first root, and we find
x(t2) = 1.44× 104 m and a(t2) = −6.50 m/s2 at the second root.
49
100. We take +x in the direction of motion, so v0 = +30 m/s, v1 = +15 m/s and a < 0. The acceleration is
found from Eq. 2-11: a = (v1 − v0)/t1 where t1 = 3.0 s. This gives a = −5.0 m/s2. The displacement
(which in this situation is the same as the distance traveled) to the point it stops (v2 = 0) is, using
Eq. 2-16,
v22 = v
2
0 + 2a∆x =⇒ ∆x = −
302
2(−5) = 90 m .
101. We choose the direction of motion as the positive direction. We work with the kilometer and hour units,
so we write ∆x = 0.088 km.
(a) Eq. 2-16 leads to
a =
v2 − v20
2∆x
=
652 − 852
2(0.088)
which yields a = −1.7× 104 km/h2.
(b) In this case, we obtain
a =
602 − 802
2(0.088)
= −1.6× 104 km/h2 .
(c) In this final situation, we find
a =
402 − 502
2(0.088)
= −5.1× 103 km/h2 .
102. Let the vertical distances between Jim’s and Clara’s feet and the jump-off level be HJ and HC , respec-
tively. At the instant this photo was taken, Clara has fallen for a time TC , while Jim has fallen for TJ .
Thus (using Eq. 2-15 with v0 = 0) we have
HJ =
1
2
gT 2J and HC =
1
2
gT 2C .
Measuring directly from the photo, we get HJ ≈ 3.6 m and HC ≈ 6.3 m, which yields TJ ≈ 0.86 s and
TC ≈ 1.13 s. Jim’s waiting time is therefore TC − TJ ≈ 0.3 s.
103. We choose down as the +y direction and place the coordinate origin at the top of the building (which
has height H). During its fall, the ball passes (with velocity v1 ) the top of the window (which is at
y1 ) at time t1, and passes the bottom (which is at y2 ) at time t2 . We are told y2 − y1 = 1.20 m and
t2 − t1 = 0.125 s. Using Eq. 2-15 we have
y2 − y1 = v1 (t2 − t1) + 1
2
g (t2 − t1)2
which immediately yields
v1 =
1.20− 12 (9.8)(0.125)2
0.125
= 8.99 m/s .
From this, Eq. 2-16 (with v0 = 0) reveals the value of y1 :
v21 = 2gy1 =⇒ y1 =
8.992
2(9.8)
= 4.12 m .
It reaches the ground (y3 = H) at t3 . Because of the symmetry expressed in the problem (“upward flight
is a reverse of the fall”) we know that t3− t2 = 2.00/2 = 1.00 s. And this means t3− t1 = 1.00+0.125 =
1.125 s. Now Eq. 2-15 produces
y3 − y1 = v1 (t3 − t1) + 1
2
g (t3 − t1)2
y3 − 4.12 = (8.99)(1.125) + 1
2
(9.8)(1.125)2
which yields y3 = H = 20.4 m.
50 CHAPTER 2.
104. (a) Using the fact that the area of a triangle is 12 (base)(height) (and the fact that the integral corre-
sponds to area under the curve) we find, from t = 0 through t = 5 s, the integral of v with respect
to t is 15 m. Since we are told that x0 = 0 then we conclude that x = 15 m when t = 5.0 s.
(b) We see directly from the graph that v = 2.0 m/s when t = 5.0 s.
(c) Since a = dvdt = slope of the graph, we find that the acceleration during the interval 4 < t < 6 is
uniformly equal to −2.0m/s2.
(d) Thinking of x(t) in terms of accumulated area (on the graph), we note that x(1) = 1 m; using this
and the value found in part (a), Eq. 2-2 produces
vavg =
x(5)− x(1)
5− 1 =
15− 1
4
= 3.5 m/s .
(e) From Eq. 2-7 and the values v(t) we read directly from the graph, we find
aavg =
v(5)− v(1)
5− 1 =
2− 2
4
= 0.
105. (First problem of Cluster 1)
The two parts of this problem are as follows. Part 1 (motion fromA to B) consists of constant acceleration
(so Table 2-1 applies) and involves the data v0 = 0, v = 10.0 m/s, x0 = 0 and x = 40.0 m (taking point
A as the coordinate origin and orienting the positive x axis towards B and C). Part 2 (from B to C)
consists of constant velocity motion (so the simple equation ∆x∆t = v applies) with v = 10.0 m/s and
∆t = 10.0 s.
(a) Eq. 2-16 is an efficient way of finding the part 1 acceleration:
v2 = v20 + 2a (x− x0) =⇒ (10.0)2 = 0 + 2a(40.0)
from which we obtain a = 1.25 m/s
2
.
(b) Using Eq. 2-17 avoids using the result from part (a) and finds the time readily.
x− x0 = 1
2
(v0 + v) t =⇒ 40.0− 0 = 1
2
(0 + 10.0)t
This leads to t = 8.00 s, for part 1.
(c) We find the distance traveled in part 2 with ∆x = v∆t = (10.0)(10.0) = 100 m.
(d) The average velocity is defined by Eq. 2-2
vavg =
xC − xA
tC − tA =
140− 0
18− 0 = 7.78 m/s .
106. (Second problem of Cluster 1)
The two parts of this problem are as follows. Part 1 (motion fromA to B) consists of constant acceleration
(so Table 2-1 applies) and involves the data v0 = 20.0 m/s, v = 30.0 m/s, x0 = 0 and t1 = 10.0 s (taking
point A as the coordinate origin, orienting the positive x axis towards B and C, and starting the clock
when it passes point A). Part 2 (from B to C) also involves uniformly accelerated motion but now with
the data v0 = 30.0 m/s, v = 15.0 m/s, and ∆x = x− x0 = 150 m.
(a) The distance for part 1 is given
by
x− x0 = 1
2
(v0 + v) t1 =
1
2
(20.0 + 30.0)(10.0)
which yields x = 250 m.
51
(b) The time t2 for part 2 is found from the same formula as in part (a).
x− x0 = 1
2
(v0 + v) t2 =⇒ 150 = 1
2
(30.0 + 15.0)t2 .
This results in t2 = 6.67 s.
(c) The definition of average velocity is given by Eq. 2-2:
vavg =
xC − xA
tC − tA =
400− 0
16.7
= 24.0 m/s .
(d) The definition of average acceleration is given by Eq. 2-7:
aavg =
vC − vA
tC − tA =
15.0− 20.0
16.7
= −0.30 m/s2 .
107. (Third problem of Cluster 1)
The problem consists of two parts (A to B at constant velocity, then B to C with constant acceleration).
The constant velocity in part 1 is 20 m/s (taking the positive direction in the direction of motion) and
t1 = 5.0 s. In part 2, we have v0 = 20 m/s, v = 0, and t2 = 10 s.
(a) We find the distance in part 1 from x − x0 = vt1, so we obtain x = 100 m (taking A to be at the
origin). And the position at the end of part 2 is then found using Eq. 2-17.
x = x0 +
1
2
(v0 + v) t2 = 100 +
1
2
(20 + 0)(10) = 200 m .
(b) The acceleration in part (a) can be found using Eq. 2-11.
v = v0 + at2 =⇒ 0 = 20 + a(10) .
Thus, we find a = −2.0 m/s2. The negative sign indicates that the acceleration vector points
opposite to the chosen positive direction (the direction of motion), which is what we expect of a
deceleration.
108. (Fourth problem of Cluster 1)
The part 1 motion in this problem is simply that of constant velocity, so xB − x0 = v1t1 applies with
t1 = 5.00 s and x0 = xA = 0 if we choose point A as the coordinate origin. Next, the part 2 motion
consists of constant acceleration (so the equations of Table 2-1, such as Eq. 2-17, apply) with x0 = xB
(an unknown), v0 = vB (also unknown, but equal to the v1 above), xC = 300 m, vC = 10.0 m/s, and
t2 = 20.0 s. The equations describing parts 1 and 2, respectively, are therefore
xB − xA = v1t1 =⇒ xB = v1(5.00)
xC − xB = 1
2
(vB + vC) t2 =⇒ 300− xB = 1
2
(vB + 10.0) (20.0)
(a) We use the fact that vA = v1 = vB in solving this set of simultaneous equations. Adding equations,
we obtain the result v1 = 13.3 m/s.
(b) In order to find the acceleration, we use our result from part (a) as the initial velocity in Eq. 2-14
(applied to the part 2 motion):
v = v0 + at2 =⇒ 10.0 = 13.3 + a(20.0)
Thus, a = −0.167 m/s2.
52 CHAPTER 2.
109. (Fifth problem of Cluster 1)
The problem consists of two parts, where part 1 (A to B) involves constant velocity motion for t1 = 5.00 s
and part 2 (B to C) involves uniformly accelerated motion. Assuming the coordinate origin is at point
A and the positive axis is directed towards B and C, then we have xC = 250 m, a2 = −0.500 m/s2, and
vC = 0.
(a) We set up the uniform velocity equation for part 1 (∆x = vt) and Eq. 2-16 for part 2 (v2 =
v20 + 2a∆x) as a simultaneous set of equations to be solved:
xB − 0 = v1(5.00)
02 = v2B + 2(−0.500) (250− xB) .
Bearing in mind that vA = v1 = vB, we can solve the equations by, for instance, substituting the
first into the second – eliminating xB and leading to a quadratic equation for v1 :
v21 + 5v1 − 150 = 0 .
The positive root gives us v1 = 13.5 m/s.
(b) We obtain the duration t2 of part 2 from Eq. 2-11:
v = v0 + at2 =⇒ 0 = 13.5 + (−0.500)t2
which yields the value t2 = 27.0 s. Therefore, the total time is t1 + t2 = 32.0 s.
110. (Sixth problem of Cluster 1)
Both part 1 and part 2 of this problem involve uniformly accelerated motion, but at different rates a1
and a2. We take the coordinate origin at point A and direct the positive axis towards B and C. In these
terms, we are given xA = 0, xC = 1300 m, vA = 0, and vC = 50 m/s. Further, the time-duration for
each part is given: t1 = 20 s and t2 = 40 s.
(a) We have enough information to apply Eq. 2-17 (∆x = 12 (v0 + v)t) to parts 1 and 2 and solve the
simultaneous set:
xB − xA = 1
2
(vA + vB) t1 =⇒ xB = 1
2
vB (20)
xC − xB = 1
2
(vB + vC) t2 =⇒ 1300− xB = 1
2
(vB + 50) (40)
Adding equations, we find vB = 10 m/s.
(b) The other unknown in the above set of equations is now easily found by plugging the result for vB
back in: xB = 100 m.
(c) We can find a1 a variety of ways, using the just-obtained results. We note that Eq. 2-11 is especially
easy to use.
v = v0 + a1t1 =⇒ 10 = 0 + a1(20)
This leads to a1 = 0.50 m/s
2
.
(d) To find a2 we proceed as just as we did in part (c), so that Eq. 2-11 for part 2 becomes 50 =
10 + a2(40). Therefore, the acceleration is a2 = 1.0m/s
2.
Chapter 3
1. The vectors should be parallel to achieve a resultant 7 m long (the unprimed case shown below), antipar-
allel (in opposite directions) to achieve a resultant 1 m long (primed case shown), and perpendicular to
achieve a resultant
√
32 + 42 = 5 m long (the double-primed case shown). In each sketch, the vectors
are shown in a “head-to-tail” sketch but the resultant is not shown. The resultant would be a straight
line drawn from beginning to end; the beginning is indicated by A (with or without primes, as the case
may be) and the end is indicated by B.
b
A
- b -
B
bA
′
- bff
B′
bA
′′
- b
6B′′
2. A sketch of the displace-
ments is shown. The re-
sultant (not shown) would
be a straight line from
start (Bank) to finish
(Walpole). With a careful
drawing, one should find
that the resultant vector
has length 29.5 km at 35◦
west of south.
West East
South
Bank
@
@
@
@
@
@
@
@
@@RH
HH
HH
HH
HH
HH
HH
HHYB
B
B
B
B
B
BN
Walpole
3. The x component of ~a is given by ax = 7.3 cos 250
◦ = −2.5 and the y component is given by ay =
7.3 sin250◦ = −6.9. In considering the variety of ways to compute these, we note that the vector is
70◦ below the −x axis, so the components could also have been found from ax = −7.3 cos 70◦ and
ay = −7.3 sin 70◦. In a similar vein, we note that the vector is 20◦ from the −y axis, so one could use
ax = −7.3 sin20◦ and ay = −7.3 cos20◦ to achieve the same results.
53
54 CHAPTER 3.
4. The angle described by a full circle is 360◦ = 2π rad, which is the basis of our conversion factor. Thus,
(20.0◦)
2π rad
360◦
= 0.349 rad
and (similarly) 50.0◦ = 0.873 rad and 100◦ = 1.75 rad. Also,
(0.330 rad)
360◦
2π rad
= 18.9◦
and (similarly) 2.10 rad = 120◦ and 7.70 rad = 441◦.
5. The textbook’s approach to this sort of problem is through the use of Eq. 3-6, and is illustrated in Sample
Problem 3-3. However, most modern graphical calculators can produce the results quite efficiently using
rectangular ↔ polar “shortcuts.”
(a)
√
(−25)2 + 402 = 47.2 m
(b) Recalling that tan (θ) = tan (θ + 180◦), we note that the two possibilities for tan−1 (40/− 25)
are −58◦ and 122◦. Noting that the vector is in the third quadrant (by the signs of its x and y
components) we see that 122◦ is the correct answer. The graphical calculator “shortcuts” mentioned
above are designed to correctly choose the right possibility.
6. The x component of ~r is given by 15 cos 30◦ = 13 m and the y component is given by 15 sin30◦ = 7.5 m.
7. The point P is displaced vertically by 2R, where R is the radius of the wheel. It is displaced horizontally
by half the circumference of the wheel, or πR. Since R = 0.450 m, the horizontal component of the
displacement is 1.414 m and the vertical component of the displacement is 0.900 m. If the x axis is
horizontal and the y axis is vertical, the vector displacement (in meters) is ~r = (1.414 ıˆ + 0.900 jˆ). The
displacement has a magnitude of
|~r| =
√
(πR)2 + (2R)2 = R
√
π2 + 4 = 1.68 m
and an angle of
tan−1
(
2R
πR
)
= tan−1
(
2
π
)
= 32.5◦
above the floor. In physics there are no “exact” measurements, yet that angle computation seemed to
yield something exact. However, there has to be some uncertainty in the observation that the wheel
rolled half of a revolution,
which introduces some indefiniteness in our result.
8. Although we think of this as a three-dimensional movement, it is rendered effectively two-dimensional
by referring measurements to its well-defined plane of the fault.
(a) The magnitude of the net displacement is
| ~AB| =
√
|AD|2 + |AC|2 =
√
172 + 222 = 27.8 m .
(b) The magnitude of the vertical component of ~AB is |AD| sin 52.0◦ = 13.4 m.
9. The length unit meter is understood throughout the calculation.
(a) We compute the distance from one corner to the diametrically opposite corner: d =
√
3.002 + 3.702 + 4.302 =
6.42.
55
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............................
~d
ℓ
w
h
L
(b) The displacement vector is along the straight line from the beginning to the end point of the trip.
Since a straight line is the shortest distance between two points, the length of the path cannot be
less than the magnitude of the displacement.
(c) It can be greater, however. The fly might, for example, crawl along the edges of the room. Its
displacement would be the same but the path length would be ℓ+ w + h.
(d) The path length is the same as the magnitude of the displacement if the fly flies along the displace-
ment vector.
(e) We take the x axis to be out of the page, the y axis to be to the right, and the z axis to be upward.
Then the x component of the displacement is w = 3.70, the y component of the displacement is
4.30, and the z component is 3.00. Thus ~d = 3.70 ıˆ + 4.30 jˆ + 3.00 kˆ. An equally correct answer is
gotten by interchanging the length, width, and height.
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ℓ
w
h
(f) Suppose the path of the fly is as shown by the dotted lines on the upper diagram. Pretend there
is a hinge where the front wall of the room joins the floor and lay the wall down as shown on the
lower diagram. The shortest walking distance between the lower left back of the room and the
upper right front corner is the dotted straight line shown on the diagram. Its length is
Lmin =
√
(w + h)2 + ℓ2 =
√
(3.70 + 3.00)2 + 4.302 = 7.96 m .
10. We label the displacement vectors ~A, ~B and ~C (and denote the result
56 CHAPTER 3.
of their vector sum as ~r). We
choose east as the ıˆ direction (+x
direction) and north as the jˆ di-
rection (+y direction). All dis-
tances are understood to be in
kilometers. We note that the an-
gle between ~C and the x axis is
60◦. Thus,
-ff 6
?
eastwest
north
south
e
~A - e
6~B
e
~C
�
�
�
�
�
���
~A = 50 ıˆ
~B = 30 jˆ
~C = 25 cos (60◦) ıˆ + 25 sin (60◦) jˆ
~r = ~A+ ~B + ~C = 62.50 ıˆ + 51.65 jˆ
which means
that its magnitude is
|~r| =
√
62.502 + 51.652 ≈ 81 km .
and its angle (counterclockwise from +x axis) is tan−1 (51.65/62.50) ≈ 40◦, which is to say that it points
40◦ north of east. Although the resultant ~r is shown in our sketch, it would be a direct line from the
“tail” of ~A to the “head” of ~C.
11. The diagram shows the displacement vectors for the two segments of her walk, labeled ~A and ~B, and
the total (“final”) displacement vector, labeled ~r. We take east to be the +x direction and north to be
the +y direction. We observe that the angle between ~A and the x axis is 60◦. Where the units are not
explicitly shown, the distances are understood
to be in meters. Thus, the components
of ~A are Ax = 250 cos60
◦ = 125 and
Ay = 250 sin30
◦ = 216.5. The compo-
nents of ~B are Bx = 175 and By = 0.
The components of the total displace-
ment are rx = Ax + Bx = 125 + 175 =
300 and ry = Ay + By = 216.5 + 0 =
216.5. .....
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......................
North
East
~A
~B
~r
(a) The magnitude of the resultant displacement is
|~r| =
√
r2x + r
2
y =
√
3002 + 216.52 = 370 m .
(b) The angle the resultant displacement makes with the +x axis is
tan−1
(
ry
rx
)
= tan−1
(
216.5
300
)
= 36◦ .
(c) The total distance walked is d = 250 + 175 = 425 m.
(d) The total distance walked is greater than the magnitude of the resultant displacement. The diagram
shows why: ~A and ~B are not collinear.
12. We label the displacement vectors ~A, ~B and ~C (and denote the result
57
of their vector sum as ~r). We
choose east as the ıˆ direction (+x
direction) and north as the jˆ di-
rection (+y direction). All dis-
tances are understood to be in
kilometers. Therefore,
-ff 6
?
eastwest
north
south
e
~A
6
eff
~B
e
~C
?
~A = 3.1 jˆ
~B = −2.4 ıˆ
~C = −5.2 jˆ
~r = ~A+ ~B + ~C = −2.1 ıˆ− 2.4 jˆ
which means
that its magnitude is
|~r| =
√
(−2.1)2 + (−2.4)2 ≈ 3.2 km .
and the two possibilities for its angle are
tan−1
(−2.4
−2.1
)
= 41◦, or 221◦ .
We choose the latter possibility since ~r is in the third quadrant. It should be noted that many graphical
calculators have polar↔ rectangular “shortcuts” that automatically produce the correct answer for angle
(measured counterclockwise from the +x axis). We may phrase the angle, then, as 221◦ counterclockwise
from East (a phrasing that sounds peculiar, at best) or as 41◦ south from west or 49◦ west from south.
The resultant ~r is not shown in our sketch; it would be an arrow directed from the “tail” of ~A to the
“head” of ~C.
13. We write ~r = ~a + ~b. When not explicitly displayed, the units here are assumed to be meters. Then
rx = ax + bx = 4.0− 13 = −9.0 and ry = ay + by = 3.0 + 7.0 = 10. Thus ~r = (−9.0m) ıˆ + (10m) jˆ . The
magnitude of the resultant is
r =
√
r2x + r
2
y =
√
(−9.0)2 + (10)2 = 13 m .
The angle between the resultant and the +x axis is given by tan−1 (ry/rx) = tan−1 10/(−9.0) which is
either −48◦ or 132◦. Since the x component of the resultant is negative and the y component is positive,
characteristic of the second quadrant, we find the angle is 132◦ (measured counterclockwise from +x
axis).
14. The x, y and z components (with meters understood) of ~r are
rx = cx + dx = 7.4 + 4.4 = 12
ry = cy + dy = −3.8− 2.0 = −5.8
rz = cz + dz = −6.1 + 3.3 = −2.8 .
15. The vectors are shown on the diagram. The x axis runs from west
to east and the y axis run
from south to north. Then
ax = 5.0m, ay = 0, bx =
−(4.0m) sin35◦ = −2.29m,
and by = (4.0m) cos 35
◦ =
3.28m.
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............................
~a
~b
~a+~b
E
N
50.5◦
35◦
58 CHAPTER 3.
(a) Let ~c = ~a+~b. Then cx = ax+ bx = 5.0m−2.29m = 2.71m and cy = ay+ by = 0+3.28m = 3.28m.
The magnitude of c is
c =
√
c2x + c
2
y =
√
(2.71m)2 + (3.28m)2 = 4.3 m .
(b) The angle θ that ~c = ~a+~b makes with the +x axis is
θ = tan−1
cy
cx
= tan−1
3.28m
2.71m
= 50.4◦ .
The second possibility (θ = 50.4◦ + 180◦ = 126◦) is rejected because it would point in a direction
opposite to ~c.
(c) The vector ~b− ~a is found by adding −~a to ~b. The result is shown
on the diagram to the
right. Let ~c = ~b −
~a. Then cx = bx −
ax = −2.29m − 5.0m =
−7.29m and cy = by −
ay = 3.28m. The magni-
tude of ~c is c =
√
c2x + c
2
y
= 8.0m .
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−~a
~b −~a+~b
W
N
(d) The tangent of the angle θ that ~c makes with the +x axis (east) is
tan θ =
cy
cx
=
3.28m
−7.29m = −4.50, .
There are two solutions: −24.2◦ and 155.8◦. As the diagram shows, the second solution is correct.
The vector ~c = −~a+~b is 24◦ north of west.
16. All distances in this solution are understood to be in meters.
(a) ~a+~b = (3.0 ıˆ + 4.0 jˆ) + (5.0 ıˆ− 2.0 jˆ) = 8.0 ıˆ + 2.0 jˆ .
(b) The magnitude of ~a+~b is
|~a+~b| =
√
8.02 + 2.02 = 8.2 m .
(c) The angle between this vector and the +x axis is tan−1(2.0/8.0) = 14◦.
(d) ~b− ~a = (5.0 ıˆ− 2.0 jˆ)− (3.0 ıˆ + 4.0 jˆ) = 2.0 ıˆ− 6.0 jˆ .
(e) The magnitude of the difference vector ~b− ~a is
|~b− ~a| =
√
2.02 + (−6.0)2 = 6.3 m .
(f) The angle between this vector and the +x axis is tan−1(−6.0/2.0) = −72◦. The vector is 72◦
clockwise from the axis defined by ıˆ.
17. All distances in this solution are understood to be in meters.
(a) ~a+~b = (4.0 + (−1.0)) ıˆ + ((−3.0) + 1.0) jˆ + (1.0 + 4.0) kˆ = 3.0 ıˆ− 2.0 jˆ + 5.0 kˆ .
(b) ~a−~b = (4.0− (−1.0)) ıˆ + ((−3.0)− 1.0) jˆ + (1.0− 4.0) kˆ = 5.0 ıˆ− 4.0 jˆ− 3.0 kˆ .
(c) The requirement ~a−~b+ ~c = 0 leads to ~c = ~b − ~a, which we note is the opposite of what we found
in part (b). Thus, ~c = −5.0 ıˆ + 4.0 jˆ + 3.0 kˆ .
59
18. Many of the operations are done efficiently on most modern graphical calculators using their built-in
vector manipulation and rectangular↔ polar “shortcuts.” In this solution, we employ the “traditional”
methods (such as Eq. 3-6).
(a) The magnitude of ~a is
√
42 + (−3)2 = 5.0 m.
(b) The angle between ~a and the +x axis is tan−1(−3/4) = −37◦. The vector is 37◦ clockwise from the
axis defined by ıˆ .
(c) The magnitude of ~b is
√
62 + 82 = 10 m.
(d) The angle between ~b and the +x axis is tan−1(8/6) = 53◦.
(e) ~a+~b = (4 + 6) ıˆ + ((−3) + 8) jˆ = 10 ıˆ + 5 jˆ , with the unit meter understood. The magnitude of this
vector is
√
102 + 52 = 11 m; we rounding to two significant figures in our results.
(f) The angle between the vector described in part (e) and the +x axis is tan−1(5/10) = 27◦.
(g) ~b−~a = (6− 4) ıˆ + (8− (−3)) jˆ = 2 ıˆ + 11 jˆ , with the unit meter understood. The magnitude of this
vector is
√
22 + 112 = 11 m, which is, interestingly, the same result as in part (e) (exactly, not just
to 2 significant figures) (this curious coincidence is made possible by the fact that ~a ⊥ ~b).
(h) The angle between the vector described in part (g) and the +x axis is tan−1(11/2) = 80◦.
(i) ~a −~b = (4 − 6) ıˆ + ((−3) − 8) jˆ = −2 ıˆ− 11 jˆ , with the unit meter understood. The magnitude of
this vector is
√
(−2)2 + (−11)2 = 11 m.
(j) The two possibilities presented by a simple calculation for the angle between the vector described in
part (i) and the +x direction are tan−1(11/2) = 80◦, and 180◦+80◦ = 260◦. The latter possibility
is the correct answer (see part (k) for a further observation related to this result).
(k) Since ~a−~b = (−1)(~b− ~a), they point in opposite (antiparallel) directions; the angle between them
is 180◦.
19. Many of the operations are done efficiently on most modern graphical calculators using their built-in
vector manipulation and rectangular↔ polar “shortcuts.” In this solution, we employ the “traditional”
methods (such as Eq. 3-6). Where the length unit is not displayed, the unit meter should be understood.
(a) Using unit-vector notation,
~a = 50 cos (30◦) ıˆ + 50 sin (30◦) jˆ
~b = 50 cos (195◦) ıˆ + 50 sin (195◦) jˆ
~c = 50 cos (315◦) ıˆ + 50 sin (315◦) jˆ
~a+~b+ ~c = 30.4 ıˆ− 23.3 jˆ .
The magnitude of this result
is
√
30.42 + (−23.3)2 = 38 m.
(b) The two possibilities presented by a simple calculation for the angle between the vector described
in part (a) and the +x direction are tan−1(−23.2/30.4) = −37.5◦, and 180◦ + (−37.5◦) = 142.5◦.
The former possibility is the correct answer since the vector is in the fourth quadrant (indicated
by the signs of its components). Thus, the angle is −37.5◦, which is to say that it is roughly 38◦
clockwise from the +x axis. This is equivalent to 322.5◦ counterclockwise from +x.
(c) We find ~a−~b+~c = (43.3− (−48.3)+35.4) ıˆ− (25− (−12.9)+(−35.4)) jˆ = 127 ıˆ+2.6 jˆ in unit-vector
notation. The magnitude of this result is
√
1272 + 2.62 ≈ 1.3× 102 m.
(d) The angle between the vector described in part (c) and the +x axis is tan−1(2.6/127) ≈ 1◦.
(e) Using unit-vector notation, ~d is given by
~d = ~a+~b− ~c
= −40.4 ıˆ + 47.4 jˆ
which has a magnitude of
√
(−40.4)2 + 47.42 = 62 m.
60 CHAPTER 3.
(f) The two possibilities presented by a simple calculation for the angle between the vector described
in part (e) and the +x axis are tan−1(47.4/(−40.4)) = −50◦, and 180◦+(−50◦) = 130◦. We choose
the latter possibility as the correct one since it indicates that ~d is in the second quadrant (indicated
by the signs of its components).
20. Angles are given in ‘standard’ fashion, so Eq. 3-5 applies directly. We use this to write the vectors in
unit-vector notation before adding them. However, a very different-looking approach using the special
capabilities of most graphical calculators can be imagined. Where the length unit is not displayed in the
solution below, the unit meter should be understood.
(a) Allowing for the different angle units used in the problem statement, we arrive at
~E = 3.73 ıˆ + 4.70 jˆ
~F = 1.29 ıˆ− 4.83 jˆ
~G = 1.45 ıˆ + 3.73 jˆ
~H = −5.20 ıˆ + 3.00 jˆ
~E + ~F + ~G+ ~H = 1.28 ıˆ + 6.60 jˆ .
(b) The magnitude of the vector sum found in part (a) is
√
1.282 + 6.602 = 6.72 m. Its angle measured
counterclockwise from the +x axis is tan−1(6.6/1.28) = 79◦ = 1.38 rad.
21. It should be mentioned that an efficient way to work this vector addition problem is with the cosine
law for general triangles (and since ~a, ~b and ~r form an isosceles triangle, the angles are easy to figure).
However, in the interest of reinforcing the usual systematic approach to vector addition, we note that
the angle ~b makes with the +x axis is 135◦ and apply Eq. 3-5 and Eq. 3-6 where appropriate.
(a) The x component of ~r is 10 cos 30◦ + 10 cos 135◦ = 1.59 m.
(b) The y component of ~r is 10 sin 30◦ + 10 sin 135◦ = 12.1 m.
(c) The magnitude of ~r is
√
1.592 + 12.12 = 12.2 m.
(d) The angle between ~r and the +x direction is tan−1 (12.1/1.59) = 82.5◦.
22. If we wish to use Eq. 3-5 in an unmodified fashion, we should note that the angle between ~C and the
+x axis is 180◦ + 20◦ = 200◦.
(a) The x component of ~B is given by Cx − Ax = 15 cos 200◦ − 12 cos 40◦ = −23.3 m, and the y
component of ~B is given by Cy − Ay = 15 sin200◦ − 12 sin40◦ = −12.8 m. Consequently, its
magnitude is
√
(−23.3)2 + (−12.8)2 = 26.6 m.
(b) The two possibilities presented by a simple calculation for the angle between ~B and the +x axis
are tan−1((−12.8)/(−23.3)) = 28.9◦, and 180◦ + 28.9◦ = 209◦. We choose the latter possibility
as the correct one since it indicates that ~B is in the third quadrant (indicated by the signs of its
components). We note, too, that the answer can be equivalently stated as −151◦.
23. We consider ~A with (x, y) components given by (A cosα,A sinα). Similarly, ~B → (B cosβ,B sinβ). The
angle (measured from the +x direction) for their vector sum must have a slope given by
tan θ =
A sinα+B sinβ
A cosα+B cosβ
.
The problem requires that we now consider the orthogonal direction, where tan θ + 90◦ = − cot θ. If this
(the negative reciprocal of the above expression) is to equal the slope for their vector difference, then we
must have
−A cosα+B cosβ
A sinα+B sinβ
=
A sinα−B sinβ
A cosα−B cosβ .
61
Multiplying both sides by A sinα+B sinβ and doing the same with A cosα−B cosβ yields
A2 cos2 α−B2 cos2 β = A2 sin2 α−B2 sin2 β .
Rearranging, using the cos2 φ+ sin2 φ = 1 identity, we obtain
A2 = B2 =⇒ A = B .
In a later section, the scalar (dot) product of vectors is presented and this result can be revisited with
a more compact derivation.
24. If we wish to use Eq. 3-5 directly, we should note that the angles for ~Q,~R and ~S are 100◦, 250◦ and 310◦,
respectively, if they are measured counterclockwise from the +x axis.
(a) Using unit-vector notation, with the unit meter understood, we have
~P = 10 cos (25◦) ıˆ + 10 sin (25◦) jˆ
~Q = 12 cos (100◦) ıˆ + 12 sin (100◦) jˆ
~R = 8 cos (250◦) ıˆ + 8 sin (250◦) jˆ
~S = 9 cos (310◦) ıˆ + 9 sin (310◦) jˆ
~P + ~Q+ ~R+ ~S = 10.0 ıˆ + 1.6 jˆ
(b) The magnitude of the vector sum is
√
102 + 1.62 = 10.2 m and its angle is tan−1 (1.6/10) ≈ 9.2◦
measured counterclockwise from the +x axis. The appearance of this solution would be quite
different using the vector manipulation capabilities of most modern graphical calculators, although
the principle would be basically the same.
25. Without loss of generality, we assume ~a points along the +x axis, and that ~b is at θ measured counter-
clockwise from ~a. We wish to verify that
r2 = a2 + b2 + 2ab cosθ
where a = |~a| = ax (we’ll call it a for simplicity) and b = |~b| =
√
b2x + b
2
y. Since ~r = ~a +
~b then
r = |~r| =
√
(a+ bx)2 + b2y. Thus, the above expression becomes
(√
(a+ bx)2 + b2y
)2
= a2 +
(√
b2x + b
2
y
)2
+ 2ab cos θ
a2 + 2abx + b
2
x + b
2
y = a
2 + b2x + b
2
y + 2ab cosθ
which makes a valid equality since (the last term) 2ab cos θ is indeed the same as 2abx (on the left-hand
side). In a later section, the scalar (dot) product of vectors is presented and this result can be revisited
with a somewhat different-looking derivation.
26. The vector equation is ~R = ~A + ~B + ~C + ~D. Expressing ~B and ~D in unit-vector notation, we have
1.69 ıˆ + 3.63 jˆ and −2.87 ıˆ + 4.10 jˆ, respectively. Where the length unit is not displayed in the solution
below, the unit meter should be understood.
(a) Adding corresponding components, we obtain ~R = −3.18 ıˆ + 4.72 jˆ.
(b) and (c) Converting this result to polar coordinates (using Eq. 3-6 or functions on a vector-capable
calculator), we obtain
(−3.18, 4.72) −→ (5.69 6 124◦)
which tells us the magnitude is 5.69 m and the angle (measured counterclockwise from +x axis) is
124◦.
62 CHAPTER 3.
27. (a) There are 4 such lines, one from each of the corners on the lower face to the diametrically opposite
corner on the upper face. One is shown on the diagram. Using an xyz coordinate system as shown
(with the origin at the back lower left corner) The position vector of the “starting point” of the
diagonal shown is a ıˆ and the position vector of the ending point is a jˆ + a kˆ, so the vector along
the line is the difference a jˆ + a kˆ− a ıˆ.
The point diametrically opposite
the origin has position vector
a ıˆ + a jˆ + a kˆ and this is the vec-
tor along the diagonal. Another
corner of the bottom face is at
a ıˆ+a jˆ and the diametrically op-
posite corner is at a kˆ, so another
cube diagonal is a kˆ − a ıˆ − a jˆ.
The fourth diagonal runs from a jˆ
to a ıˆ + a kˆ, so the vector along
the diagonal is a ıˆ + a kˆ− a jˆ.
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.
a
a
a
x
y
z
(b) Consider the vector from the back lower left corner to the front upper right corner. It is a ıˆ +
a jˆ + a kˆ. We may think of it as the sum of the vector a ıˆ parallel to the x axis and the vector
a jˆ + a kˆ perpendicular to the x axis. The tangent of the angle between the vector and the x axis
is the perpendicular component divided by the parallel component. Since the magnitude of the
perpendicular component is
√
a2 + a2 = a
√
2 and the magnitude of the parallel component is a,
tan θ =
(
a
√
2
)
/a =
√
2. Thus θ = 54.7◦. The angle between the vector and each of the other two
adjacent sides (the y and z axes) is the same as is the angle between any of the other diagonal
vectors and any of the cube sides adjacent to them.
(c) The length of any of the diagonals is given by
√
a2 + a2 + a2 = a
√
3.
28. Reference to Figure 3-18 (and the accompanyingmaterial in that section) is helpful. If we convert ~B to the
magnitude-angle notation (as ~A already is) we have ~B = (14.4 6 33.7◦) (appropriate notation especially
if we are using a vector capable calculator in polar mode). Where the length unit is not displayed in the
solution, the unit meter should be understood. In the magnitude-angle notation, rotating the axis by
+20◦ amounts to subtracting that angle from the angles previously specified. Thus, ~A = (12.0 6 40.0◦)′
and ~B = (14.4 6 13.7◦)′, where the ‘prime’ notation indicates that the description is in terms of the new
coordinates. Converting these results to (x, y) representations, we obtain
~A = 9.19 ıˆ′ + 7.71 jˆ
′
~B = 14.0 ıˆ′ + 3.41 jˆ
′
with the unit meter understood, as already mentioned.
29. We apply Eq. 3-20 and Eq. 3-27.
(a) The scalar (dot) product of the two vectors is ~a ·~b = ab cosφ = (10)(6.0) cos 60◦ = 30.
(b) The magnitude of the vector (cross) product of the two vectors is |~a×~b| = ab sinφ = (10)(6.0) sin 60◦ =
52.
30. We consider all possible products and then simplify using relations such as ıˆ · kˆ = 0 and ıˆ · ıˆ = 1. Thus,
~a ·~b =
(
ax ıˆ + ay jˆ + az kˆ
)
·
(
bx ıˆ + by jˆ + bz kˆ
)
= axbx ıˆ · ıˆ + axby ıˆ · jˆ + axbz ıˆ · kˆ + aybx jˆ · ıˆ + aybj jˆ · jˆ + · · ·
= axbx(1) + axby(0) + axbz(0) + aybx(0) + aybj(1) + · · ·
63
which is seen to reduce to the desired result (one might wish to show this in two dimensions before
tackling the additional tedium of working with these three-component vectors).
31. Since ab cosφ = axbx + ayby + azbz,
cosφ =
axbx + ayby + azbz
ab
.
The magnitudes of the vectors given in the problem are
a = |~a| =
√
(3.0)2 + (3.0)2 + (3.0)2 = 5.2
b = |~b| =
√
(2.0)2 + (1.0)2 + (3.0)2 = 3.7 .
The angle between them is found from
cosφ =
(3.0)(2.0) + (3.0)(1.0) + (3.0)(3.0)
(5.2)(3.7)
= 0.926 .
The angle is φ = 22◦.
32. We consider all possible products and then simplify using relations such as ıˆ× ıˆ = 0 and the important
fundamental products
ıˆ× jˆ = − jˆ× ıˆ = kˆ
jˆ× kˆ = − kˆ× jˆ = ıˆ
kˆ× ıˆ = − ıˆ× kˆ = jˆ .
Thus,
~a×~b =
(
ax ıˆ + ay jˆ + az kˆ
)
×
(
bx ıˆ + by jˆ + bz kˆ
)
= axbx ıˆ× ıˆ + axby ıˆ× jˆ + axbz ıˆ× kˆ + aybx jˆ× ıˆ +
aybj jˆ× jˆ + · · ·
= axbx(0) + axby( kˆ) + axbz(− jˆ) + aybx(− kˆ) + aybj(0) + · · ·
which is seen to simplify to the desired result.
33. The area of a triangle is half the product of its base and altitude. The base is the side formed by vector
~a. Then the altitude is b sinφ and the area is A = 12ab sinφ =
1
2 |~a×~b|.
34. Applying Eq. 3-23, ~F = q~v × ~B (where q is a scalar) becomes
Fx ıˆ + Fy jˆ + Fz kˆ = q (vyBz − vzBy) ıˆ + q (vzBx − vxBz) jˆ + q (vxBy − vyBx) kˆ
which – plugging in values – leads to three equalities:
4.0 = 2 (4.0Bz − 6.0By)
−20 = 2 (6.0Bx − 2.0Bz)
12 = 2 (2.0By − 4.0Bx)
Since we are told that Bx = By , the third equation leads to By = −3.0. Inserting this value into the
first equation, we find Bz = −4.0. Thus, our answer is
~B = −3.0 ıˆ− 3.0 jˆ− 4.0 kˆ .
35. Both proofs shown below utilize the fact that the vector (cross) product of ~a and ~b is perpendicular to
both ~a and ~b. This is mentioned in the book, and is fundamental to its discussion of the right-hand rule.
64 CHAPTER 3.
(a) (~b × ~a) is a vector that is perpendicular to ~a, so the scalar product of ~a with this vector is zero.
This can also be verified by using Eq. 3-30, and then (with suitable notation changes) Eq. 3-23.
(b) Let ~c = ~b×~a. Then the magnitude of ~c is c = ab sinφ. Since ~c is perpendicular to ~a the magnitude
of ~a×~c is ac. The magnitude of ~a× (~b×~a) is consequently |~a× (~b×~a)| = ac = a2b sinφ. This too
can be verified by repeated application of Eq. 3-30, although it must be admitted that this is much
less intimidating if one is using a math software package such as MAPLE or Mathematica.
36. If a vector capable calculator is used, this makes a good exercise for getting familiar with those features.
Here we briefly sketch the method. Eq. 3-30 leads to
2 ~A× ~B = 2
(
2 ıˆ + 3 jˆ− 4 kˆ
)
×
(
−3 ıˆ + 4 jˆ + 2 kˆ
)
= 44 ıˆ + 16 jˆ + 34 kˆ .
We now apply Eq. 3-23 to evaluate 3~C ·
(
2 ~A× ~B
)
:
3
(
7 ıˆ− 8 jˆ
)
·
(
44 ıˆ + 16 jˆ + 34 kˆ
)
= 3 ((7)(44) + (−8)(16) + (0)(34)) = 540 .
37. From the figure, we note that ~c ⊥ ~b, which implies that the angle between ~c and the +x axis is 120◦.
(a) Direct application of Eq. 3-5 yields the answers for this and the next few parts. ax = a cos 0
◦ =
a = 3.00 m.
(b) ay = a sin 0
◦ = 0.
(c) bx = b cos 30
◦ = (4.00m) cos 30◦ = 3.46 m.
(d) by = b sin 30
◦ = (4.00m) sin30◦ = 2.00 m.
(e) cx = c cos 120
◦ = (10.0m) cos 120◦ = −5.00 m.
(f) cy = c sin 30
◦ = (10.0m) sin 120◦ = 8.66 m.
(g) In terms of components (first x and then y), we must have
−5.00m = p(3.00m) + q(3.46m)
8.66m = p(0) + q(2.00m) .
Solving these equations, we find p = −6.67
(h) and q = 4.33 (note that it’s easiest to solve for q first). The numbers p and q have no units.
38. We apply Eq. 3-20 with Eq. 3-23. Where the length unit is not displayed, the unit meter is understood.
(a) We first note that a = |~a| = √3.22 + 1.62 = 3.58 m and b = |~b| = √0.52 + 4.52 = 4.53 m. Now,
~a ·~b = axbx + ayby = ab cosφ
(3.2)(0.5) + (1.6)(4.5) = (3.58)(4.53) cosφ
which leads to φ = 57◦ (the inverse cosine is double-valued as is the inverse tangent, but we know
this is the right solution since both vectors are in the same quadrant).
(b) Since the angle (measured from +x) for ~a is tan−1 (1.6/3.2) = 26.6◦, we know the angle for ~c
is 26.6◦ − 90◦ = −63.4◦ (the other possibility, 26.6◦ + 90◦ would lead to a cx < 0). Therefore,
cx = c cos−63.4◦ = (5.0)(0.45) = 2.2 m.
(c) Also, cy = c sin−63.4◦ = (5.0)(−0.89) = −4.5 m.
(d) And we know the angle for ~d to be 26.6◦ + 90◦ = 116.6◦, which leads to dx = d cos 116.6◦ =
(5.0)(−0.45) = −2.2 m.
(e) Finally, dy = d sin 116.6
◦ = (5.0)(0.89) = 4.5 m.
65
39. The solution to problem 27 showed that each diagonal has a length given by a
√
3, where a is the length
of a cube edge. Vectors along two diagonals are ~b = a ıˆ + a jˆ + a kˆ and ~c = −a ıˆ + a jˆ + a kˆ. Using
Eq. 3-20 with Eq. 3-23, we find the angle between them:
cosφ =
bxcx + bycy + bzcz
bc
=
−a2 + a2 + a2
3a2
=
1
3
.
The angle is φ = cos−1(1/3) = 70.5◦.
40. (a) The vector equation ~r = ~a−~b−~v is computed as follows: (5.0− (−2.0)+4.0)ˆı+ (4.0− 2.0+3.0)ˆj+
((−6.0)− 3.0 + 2.0)kˆ. This leads to ~r = 11 ıˆ + 5.0 jˆ− 7.0 kˆ.
(b) We find the angle from +z by “dotting” (taking the scalar product) ~r with kˆ. Noting that r =
|~r| =√112 + 52 + (−7)2 = 14, Eq. 3-20 with Eq. 3-23 leads to
~r · kˆ = −7.0 = (14)(1) cosφ =⇒ φ = 120◦ .
(c) To find the component of a vector in a certain direction, it is efficient to “dot” it (take the scalar
product of it) with a unit-vector in that direction. In this case, we make the desired unit-vector by
bˆ =
~b
|~b| =
−2 ıˆ + 2 jˆ + 3 kˆ√
(−2)2 + 22 + 32 .
We therefore obtain
ab = ~a · bˆ = (5)(−2) + (4)(2) + (−6)(3)√
(−2)2 + 22 + 32 = −4.9 .
(d) One approach (if we all we require is the magnitude) is to use the vector cross product, as the
problem suggests; another (which supplies more information) is to subtract the result in part (c)
(multiplied by bˆ) from ~a. We briefly illustrate both methods. We note that if a cos θ (where θ is
the angle between ~a and ~b) gives ab (the component along bˆ) then we expect a sin θ to yield the
orthogonal component:
a sin θ =
|~a×~b|
b
= 7.3
(alternatively, one might compute θ form part (c) and proceed more directly). The second method
proceeds as follows:
~a− ab bˆ = (5.0− 2.35)ˆı + (4.0− (−2.35))ˆj + ((−6.0)− (−3.53))kˆ
= 2.65 ıˆ + 6.35 jˆ− 2.47 kˆ
This describes the perpendicular part of ~a completely. To find the magnitude of this part, we
compute √
2.652 + 6.352 + (−2.47)2 = 7.3
which agrees with the first method.
41. The volume of a parallelepiped is equal to the product of its altitude and the area of its base. Take the
base to be the parallelogram formed by the vectors ~b and ~c. Its area is bc sinφ, where φ is the angle
between ~b and ~c. This is just the magnitude of the vector (cross) product ~b × ~c. The altitude of the
parallelepiped is a cos θ, where θ is the angle between ~a and the normal to the plane of ~b and ~c. Since
~b×~c is normal to that plane, θ is the angle between ~a and ~b×~c. Thus, the volume of the parallelepiped
is V = a|~b× ~c| cos θ = ~a · (~b × ~c).
42. We apply Eq. 3-30 and Eq. 3-23.
(a) ~a×~b = (axby − aybx) kˆ since all other terms vanish, due to the fact that neither ~a nor ~b have any
z components. Consequently, we obtain ((3.0)(4.0)− (5.0)(2.0)) kˆ = 2.0 kˆ .
66 CHAPTER 3.
(b) ~a ·~b = axbx + ayby yields (3)(2) + (5)(4) = 26.
(c) ~a+~b = (3 + 2) ıˆ + (5 + 4) jˆ , so that
(
~a+~b
)
·~b = (5)(2) + (9)(4) = 46.
(d) Several approaches are available. In this solution, we will construct a bˆ unit-vector and “dot” it
(take the scalar product of it) with ~a. In this case, we make the desired unit-vector by
bˆ =
~b
|~b| =
2 ıˆ + 4 jˆ√
22 + 42
.
We therefore obtain
ab = ~a · bˆ = (3)(2) + (5)(4)√
22 + 42
= 5.8 .
43. We apply Eq. 3-30 and Eq.3-23. If a vector capable calculator is used, this makes a good exercise for
getting familiar with those features. Here we briefly sketch the method.
(a) We note that ~b× ~c = −8 ıˆ + 5 jˆ + 6 kˆ. Thus, ~a ·
(
~b× ~c
)
= (3)(−8) + (3)(5) + (−2)(6) = −21.
(b) We note that ~b+ ~c = 1 ıˆ− 2 jˆ + 3 kˆ. Thus, ~a ·
(
~b+ ~c
)
= (3)(1) + (3)(−2) + (−2)(3) = −9.
(c) Finally, ~a×
(
~b+ ~c
)
= ((3)(3)−(−2)(−2)) ıˆ+((−2)(1)−(3)(3)) jˆ+((3)(−2)−(3)(1)) kˆ = 5 ıˆ−11 jˆ−9 kˆ.
44. The components of ~a are ax = 0, ay = 3.20 cos 63
◦ = 1.45, and az = 3.20 sin63◦ = 2.85. The components
of ~b are bx = 1.40 cos 48
◦ = 0.937, by = 0, and bz = 1.40 sin48◦ = 1.04.
(a) The scalar (dot) product is therefore
~a ·~b = axbx + ayby + azbz = (0)(0.937) + (1.45)(0) + (2.85)(1.04) = 2.97 .
(b) The vector (cross) product is
~a×~b = (aybz − azby) ıˆ + (azbx
− axbz) jˆ + (axby − aybx) kˆ
= ((1.45)(1.04)− 0)) ıˆ + ((2.85)(0.937)− 0)) jˆ + (0− (1.45)(0.94)) kˆ
= 1.51 ıˆ + 2.67 jˆ− 1.36 kˆ .
(c) The angle θ between ~a and ~b is given by
θ = cos−1
(
~a ·~b
ab
)
= cos−1
(
2.96
(3.30)(1.40)
)
= 48◦ .
45. We observe that |ˆı× ıˆ| = |ˆı| |ˆı| sin 0◦ vanishes because sin 0◦ = 0. Similarly, jˆ × jˆ = kˆ× kˆ = 0. When the
unit vectors are perpendicular, we have to do a little more work to show the cross product results. First,
the magnitude of the vector ıˆ× jˆ is ∣∣∣ˆı× jˆ∣∣∣ = |ˆı| ∣∣∣ˆj∣∣∣ sin 90◦
which equals 1 because sin 90◦ = 1 and these are all units vectors (each has magnitude equal to 1). This
is consistent with the claim that ıˆ × jˆ = kˆ since the magnitude of kˆ is certainly 1. Now, we use the
right-hand rule to show that ıˆ× jˆ is in the positive z direction. Thus ıˆ× jˆ has the same magnitude and
direction as kˆ, so it is equal to kˆ. Similarly, kˆ × ıˆ = jˆ and jˆ × kˆ = ıˆ. If, however, the coordinate system
is left-handed, we replace kˆ→ −kˆ in the work we have shown above and get
ıˆ× ıˆ = jˆ× jˆ = kˆ× kˆ = 0 .
just as before. But the relations that are different are
ıˆ× jˆ = −kˆ kˆ× ıˆ = −jˆ jˆ× kˆ = −ıˆ .
67
46. (a) By the right-hand rule, ~A× ~B points upward if ~A points north and ~B points west. If ~A and ~B have
magnitude= 1 then, by Eq. 3-27, the result also has magnitude equal to 1.
(b) Since cos 90◦ = 0, the scalar dot product between perpendicular vectors is zero. Thus, ~A · ~B = 0 is
~A points down and ~B points south.
(c) By the right-hand rule, ~A× ~B points south if ~A points east and ~B points up. If ~A and ~B have unit
magnitude then, by Eq. 3-27, the result also has unit magnitude.
(d) Since cos 0◦ = 1, then ~A · ~B = AB (where A is the magnitude of ~A and B is the magnitude of ~B).
If, additionally, we have A = B = 1, then the result is 1.
(e) Since sin 0◦ = 0, ~A× ~B = 0 if both ~A and ~B point south.
47. Let A denote the magnitude of ~A; similarly for the other vectors. The vector equation is ~A + ~B = ~C
where B = 8.0 m and C = 2A. We are also told that the angle (measured in the ‘standard’ sense) for
~A is 0◦ and the angle for ~C is 90◦, which makes this a right triangle (when drawn in a “head-to-tail”
fashion) where B is the size of the hypotenuse. Using the Pythagorean theorem,
B =
√
A2 + C2 =⇒ 8.0 =
√
A2 + 4A2
which leads to A = 8/
√
5 = 3.6 m.
48. We choose +x east and +y north and measure all angles in the “standard” way (positive ones are
counterclockwise from +x). Thus, vector ~d1 has magnitude d1 = 4 (with the unit meter and three
significant figures assumed) and direction θ1 = 225
◦. Also, ~d2 has magnitude d2 = 5 and direction
θ2 = 0
◦, and vector ~d3 has magnitude d3 = 6 and direction θ3 = 60◦.
(a) The x-component of ~d1 is d1 cos θ1 = −2.83 m.
(b) The y-component of ~d1 is d1 sin θ1 = −2.83 m.
(c) The x-component of ~d2 is d2 cos θ2 = 5.00 m.
(d) The y-component of ~d2 is d2 sin θ2 = 0.
(e) The x-component of ~d3 is d3 cos θ3 = 3.00 m.
(f) The y-component of ~d3 is d3 sin θ3 = 5.20 m.
(g) The sum of x-components is −2.83 + 5.00 + 3.00 = 5.17 m.
(h) The sum of y-components is −2.83 + 0 + 5.20 = 2.37 m.
(i) The magnitude of the resultant displacement is
√
5.172 + 2.372 = 5.69 m.
(j) And its angle is θ = tan−1(2.37/5.17) = 24.6◦ which (recalling our coordinate choices) means it
points at about 25◦ north of east.
(k) and (ℓ) This new displacement (the direct line home) when vectorially added to the previous
(net) displacement must give zero. Thus, the new displacement is the negative, or opposite, of the
previous (net) displacement. That is, it has the same magnitude (5.69 m) but points in the opposite
direction (25◦ south of west).
49. Reading carefully, we see that the (x, y) specifications for each “dart” are to be interpreted as (∆x,∆y)
descriptions of the corresponding displacement vectors. We combine the different parts of this problem
into a single exposition. Thus, along the x axis, we have (with the centimeter unit understood)
30.0 + bx − 20.0− 80.0 = −140.0 ,
and along y axis we have
40.0− 70.0 + cy − 70.0 = −20.0 .
Hence, we find bx = −70.0 cm and cy = 80.0 cm. And we convert the final location (−140,−20) into
polar coordinates and obtain (141 6 −172◦), an operation quickly done using a vector capable calculator
in polar mode. Thus, the ant is 141 cm from where it started at an angle of −172◦, which means 172◦
clockwise from the +x axis or 188◦ counterclockwise from the +x axis.
68 CHAPTER 3.
50. We find the components and then add them (as scalars, not vectors). With d = 3.40 km and θ = 35.0◦
we find d cos θ + d sin θ = 4.74 km.
51. We choose +x east and +y north and measure all angles in the “standard” way (positive ones counter-
clockwise from +x, negative ones clockwise). Thus, vector ~d1 has magnitude d1 = 3.66 (with the unit
meter and three significant figures assumed) and direction θ1 = 90
◦. Also, ~d2 has magnitude d2 = 1.83
and direction θ2 = −45◦, and vector ~d3 has magnitude d3 = 0.91 and direction θ3 = −135◦. We add the
x and y components, respectively:
x : d1 cos θ1 + d2 cos θ2 + d3 cos θ3 = 0.651 m
y : d1 sin θ1 + d2 sin θ2 + d3 sin θ3 = 1.723 m .
(a) The magnitude of the direct displacement (the vector sum ~d1+~d2+~d3 ) is
√
0.6512 + 1.7232 = 1.84 m.
(b) The angle (understood in the sense described above) is tan−1(1.723/0.651) = 69◦. That is, the first
putt must aim in the direction 69◦ north of east.
52. (a) We write ~b = b jˆ where b > 0. We are asked to consider
~b
d
=
(
b
d
)
jˆ
in the case d > 0. Since the coefficient of jˆ is positive, then the vector points in the +y direction.
(b) If, however, d < 0, then the coefficient is negative and the vector points in the −y direction.
(c) Since cos 90◦ = 0, then ~a ·~b = 0, using Eq. 3-20.
(d) Since ~b/d is along the y axis, then (by the same reasoning as in the previous part) ~a · (~b/d) = 0.
(e) By the right-hand rule, ~a×~b points in the +z direction.
(f) By the same rule, ~b×~a points in the −z direction. We note that ~b×~a = −~a×~b is true in this case
and quite generally.
(g) Since sin 90◦ = 1, Eq. 3-27 gives |~a×~b| = ab where a is the magnitude of ~a. Also, |~a×~b| = |~b × ~a|
so both results have the same magnitude.
(h) and (i) With d > 0, we find that ~a× (~b/d) has magnitude ab/d and is pointed in the +z direction.
53. (a) With a = 17.0 m and θ = 56.0◦ we find ax = a cos θ = 9.51 m.
(b) And ay = a sin θ = 14.1 m.
(c) The angle relative to the new coordinate system is θ′ = 56−18 = 38◦. Thus, a′x = a cos θ′ = 13.4 m.
(d) And a′y = a sin θ
′ = 10.5 m.
54. Since cos 0◦ = 1 and sin 0◦ = 0, these follows immediately from Eq. 3-20 and Eq. 3-27.
55. (a) The magnitude of the vector ~a = 4.0~d is (4.0)(2.5) = 10 m.
(b) The direction of the vector ~a = 4.0~d is the same as the direction of ~d (north).
(c) The magnitude of the vector ~c = −3.0~d is (3.0)(2.5) = 7.5 m.
(d) The direction of the vector ~c = −3.0~d is the opposite of the direction of ~d. Thus, the direction of ~c
is south.
56. The vector sum of the displacements ~dstorm and ~dnew must give the same result as its originally intended
displacement ~do = 120 jˆ where east is ıˆ, north is jˆ, and the assumed length unit is km. Thus, we write
~dstorm = 100 ıˆ and ~dnew = A ıˆ +B jˆ .
69
(a) The equation ~dstorm + ~dnew = ~do readily yields A = −100 km and B = 120 km. The magnitude of
~dnew is therefore
√
A2 +B2 = 156 km.
(b) And its direction is tan−1(B/A) = −50.2◦ or 180◦+(−50.2◦) = 129.8◦. We choose the latter value
since it indicates a vector pointing in the second quadrant, which is what we expect here. The
answer can be phrased several equivalent ways: 129.8◦ counterclockwise from east, or 39.8◦ west
from north, or 50.2◦ north from west.
57. (a) The height is h = d sin θ, where d = 12.5 m and
θ = 20.0◦. Therefore, h = 4.28 m.
(b) The horizontal distance is d cos θ = 11.7 m.
58. (a) We orient ıˆ eastward, jˆ northward, and kˆ upward. The displacement in meters is consequently
1000 ıˆ + 2000 jˆ− 500 kˆ .
(b) The net displacement is zero since his final position matches his initial position.
59. (a) If we add the equations, we obtain 2~a = 6~c, which leads to ~a = 3~c = 9 ıˆ + 12 jˆ .
(b) Plugging this result back in, we find ~b = ~c = 3 ıˆ + 4 jˆ .
60. (First problem in Cluster 1)
The given angle θ = 130◦ is assumed to be measured counterclockwise from the +x axis. Angles (if
positive) in our results follow the same convention (but if negative are clockwise from +x).
(a) With A = 4.00, the x-component of ~A is A cos θ = −2.57.
(b) The y-component of ~A is A sin θ = 3.06.
(c) Adding ~A and ~B produces a vector we call R with components Rx = −6.43 and Ry = −1.54.
Using Eq. 3-6 (or special functions on a calculator) we present this in magnitude-angle notation:
~R = (6.61 6 − 167◦).
(d) From the discussion in the previous part, it is clear that ~R = −6.43 ıˆ− 1.54 jˆ .
(e) The vector ~C is the difference of ~A and ~B. In unit-vector notation, this becomes
~C = ~A− ~B =
(
−2.57 ıˆ− 3.06 jˆ
)
−
(
−3.86 ıˆ− 4.60 jˆ
)
which yields ~C = 1.29 ıˆ + 7.66 jˆ .
(f) Using Eq. 3-6 (or special functions on a calculator) we present this in magnitude-angle notation:
~C = (7.77 6 80.5◦).
(g) We note that ~C is the “constant” in all six pictures. Remembering that the negative of a vector
simply reverses it, then we see that in form or another, all six pictures express the relation ~C = ~A− ~B.
61. (Second problem in Cluster 1)
(a) The dot (scalar) product of 3 ~A and ~B is found using Eq. 3-23:
3 ~A · ~B = 3 (4.00 cos130◦) (−3.86) + 3 (4.00 sin130◦) (−4.60) = −12.5 .
(b) We call the result ~D and combine the scalars ((3)(4) = 12). Thus, ~D = (4 ~A)× (3 ~B) becomes, using
Eq. 3-30,
12 ~A× ~B = 12 ((4.00 cos130◦) (−4.60)− (4.00 sin130◦) (−3.86)) kˆ
which yields ~D = 284 kˆ .
(c) Since ~D has magnitude 284 and points in the +z direction, it has radial coordinate 284 and angle-
measured-from-z-axis equal to 0◦. The angle measured in the xy plane does not have a well-defined
value (since this vector does not have a component in that plane)
70 CHAPTER 3.
(d) Since ~A is in the xy plane, then it is clear that ~A ⊥ ~D. The angle between them is 90◦.
(e) Calling this new result ~G we have
~G = (4.00 cos130◦) ıˆ + (4.00 sin130◦) jˆ + (3.00)kˆ
which yields ~G = −2.57 ıˆ + 3.06ˆj + 3.00 kˆ .
(f) It is straightforward using a vector-capable calculator to convert the above into spherical coordi-
nates. We, however, proceed “the hard way”, using the notation in Fig. 3-44 (where θ is in the xy
plane and φ is measured from the z axis):
|~G| = r =
√
(−2.57)2 + 3.062 + 3.002 = 5.00
φ = tan−1(4.00/3.00) = 53.1◦
θ = 130◦ given in problem 60 .
62. (Third problem in Cluster 1)
(a) Looking at the xy plane in Fig. 3-44, it is clear that the angle to ~A (which is the vector lying
in the plane, not the one rising out of it, which we called ~G in the previous problem) measured
counterclockwise from the −y axis is 90◦+130◦ = 220◦. Had we measured this clockwise we would
obtain (in absolute value) 360◦ − 220◦ = 140◦.
(b) We found in part (b) of the previous problem that ~A× ~B points along the z axis, so it is perpendicular
to the −y direction.
(c) Let ~u = −jˆ represent the −y direction, and ~w = 3 kˆ is the vector being added to ~B in this problem.
The vector being examined in this problem (we’ll call it ~Q) is, using Eq. 3-30 (or a vector-capable
calculator),
~Q = ~A×
(
~B + ~w
)
= 9.19 1ˆ + 7.71 jˆ + 23.7 kˆ
and is clearly in the first octant (since all components are positive); using Pythagorean theorem,
its magnitude is Q = 26.52. From Eq. 3-23, we immediately find ~u · ~Q = −7.71. Since ~u has unit
magnitude, Eq. 3-20 leads to
cos−1
(
~u · ~Q
Q
)
= cos−1
(−7.71
26.52
)
which yields a choice of angles 107◦ or −107◦. Since we have already observed that ~Q is in the first
octant, the the angle measured counterclockwise (as observed by someone high up on the +z axis)
from the −y axis to ~Q is 107◦.
Chapter 4
1. Where the length unit is not specified (in this solution), the unit meter should be understood.
(a) The position vector, according to Eq. 4-1, is ~r = −5.0 ıˆ + 8.0 jˆ (in meters).
(b) The magnitude is |~r| =
√
x2 + y2 + z2 = 9.4 m.
(c) Many calculators have polar ↔ rectangular conversion capabilities which make this computation
more efficient than what is shown below. Noting that the vector lies in the xy plane, we are using
Eq. 3-6:
tan−1
(
8.0
−5.0
)
= −58◦ or 122◦
where we choose the latter possibility (122◦ measured counterclockwise from the +x direction) since
the signs of the components imply the vector is in the second quadrant.
(d) In the interest of saving space, we omit the sketch. The vector is 32◦ counterclockwise from the
+y direction, where the +y direction is assumed to be (as is standard) +90◦ counterclockwise from
+x, and the +z direction would therefore be “out of the paper.”
(e) The displacement is ∆~r = ~r ′ − ~r where ~r is given in part (a) and ~r ′ = 3.0 ıˆ. Therefore, ∆~r =
8.0 ıˆ− 8.0 jˆ (in meters).
(f) The magnitude of the displacement is |∆~r| =√82 + (−8)2 = 11 m.
(g) The angle for the displacement, using Eq. 3-6, is found from
tan−1
(
8.0
−8.0
)
= −45◦ or 135◦
where we choose the former possibility (−45◦, which means 45◦ measured clockwise from +x, or
315◦ counterclockwise from +x) since the signs of the components imply the vector is in the fourth
quadrant.
2. (a) The magnitude of ~r is
√
5.02 + (−3.0)2 + 2.02 = 6.2 m.
(b) A sketch is shown. The coordinate values are in meters.
+y
+z
�
�
�
�
+x
�����
~r
5
−3 2
71
72 CHAPTER 4.
3. Where the unit is not specified, the unit meter is understood. We use Eq. 4-2 and Eq. 4-3.
(a) With the initial position vector as ~r1 and the later vector as ~r2, Eq. 4-3 yields
∆r = ((−2)− 5) ıˆ + (6 − (−6)) jˆ + (2 − 2) kˆ = −7.0 ıˆ + 12 jˆ
for the displacement vector in unit-vector notation (in meters).
(b) Since there is no z component (that is, the coefficient of kˆ is zero), the displacement vector is in
the xy plane.
4. We use a coordinate system with +x eastward and +y upward. We note that 123◦ is the angle between
the initial position and later position vectors, so that the angle from +x to the later position vector is
40◦ + 123◦ = 163◦. In unit-vector notation, the position vectors are
~r1 = 360 cos(40
◦) ıˆ + 360 sin(40◦) jˆ = 276 ıˆ + 231 jˆ
~r2 = 790 cos(163
◦) ıˆ + 790 sin(163◦) jˆ = −755 ıˆ + 231 jˆ
respectively (in meters). Consequently, we plug into Eq. 4-3
∆r = ((−755)− 276) ıˆ + (231− 231) jˆ
and find the displacement vector is horizontal (westward) with a length of 1.03 km. If unit-vector
notation is not a priority in this problem, then the computation can be approached in a variety of ways
– particularly in view of the fact that a number of vector capable calculators are on the market which
reduce this problem to a very few keystrokes (using magnitude-angle notation throughout).
5. The average velocity is given by Eq. 4-8. The total displacement ∆~r is the sum of three displacements,
each result of a (constant) velocity during a given time. We use a coordinate system with +x East and
+y North. In unit-vector notation, the first displacement is given by
∆~r1 =
(
60
km
h
)(
40min
60min/h
)
ıˆ = 40 ıˆ
in kilometers. The second displacement has a magnitude of 60 kmh · 20min60min/h = 20 km, and its direction
is 40◦ north of east. Therefore,
∆~r2 = 20 cos(40
◦) ıˆ + 20 sin(40◦) jˆ = 15.3 ıˆ + 12.9 jˆ
in kilometers. And the third displacement is
∆~r3 = −
(
60
km
h
)(
50min
60min/h
)
ıˆ = −50 ıˆ
in kilometers. The total displacement is
∆~r = ∆~r1 +∆~r2 +∆~r3
= 40 ıˆ + 15.3 ıˆ + 12.9 jˆ− 50 ıˆ
= 5.3 ıˆ + 12.9 jˆ (km) .
The time for the trip is 40 + 20 + 50 = 110 min, which is equivalent to 1.83 h. Eq. 4-8 then yields
~vavg =
(
5.3 km
1.83 h
)
ıˆ +
(
12.9 km
1.83 h
)
jˆ = 2.90 ıˆ + 7.01 jˆ
in kilometers-per-hour. If it is desired to express this in magnitude-angle notation, then this is equivalent
to a vector of magnitude
√
2.92 + 7.012 = 7.59 km/h, which is inclined 67.5◦ north of east (or, 22.5◦
east of north). If unit-vector notation is not a priority in this problem, then the computation can
be approached in a variety of ways – particularly in view of the fact that a number of vector capable
calculators are on the market which reduce this problem to a very few keystrokes (using magnitude-angle
notation throughout).
73
6. Using Eq. 4-3 and Eq. 4-8, we have
~vavg =
(−2.0 ıˆ + 8.0 jˆ− 2.0 kˆ)− (5.0 ıˆ− 6.0 jˆ + 2.0 kˆ)
10
= −0.70 ıˆ + 1.40 jˆ− 0.40 kˆ
in meters-per-second.
7. To emphasize the fact that the velocity is a function of time, we adopt the notation v(t) for dxdt .
(a) Eq. 4-10 leads to
v(t) =
d
dt
(
3.00t ıˆ− 4.00t2 jˆ + 2.00 kˆ
)
= 3.00 ıˆ− 8.00t jˆ
in meters-per-second.
(b) Evaluating this result at t = 2 s produces ~v = 3.0 ıˆ− 16.0 jˆ m/s.
(c) The speed at t = 2 s is v = |~v| =√32 + (−16)2 = 16.3 m/s.
(d) And the angle of ~v at that moment is one of the possibilities
tan−1
(−16
3
)
= −79.4◦ or 101◦
where we choose the first possibility (79.4◦ measured clockwise from the +x direction, or 281◦
counterclockwise from +x) since the signs of the components imply the vector is in the fourth
quadrant.
8. On the one hand, we could perform the vector addition of the displacements with a vector capable
calculator in polar mode ((75 6 37◦) + (65 6 − 90◦) = (63 6 − 18◦)), but in keeping with Eq. 3-5 and
Eq. 3-6 we will show the details in unit-vector notation. We use a ‘standard’ coordinate system with +x
East and +y North. Lengths are in kilometers and times are in hours.
(a) We perform the vector addition of individual displacements to find the net displacement of the
camel.
∆~r1 = 75 cos(37
◦) ıˆ + 75 sin(37◦) jˆ
∆~r2 = −65 jˆ
∆~r1 +∆~r2 = 60 ıˆ− 20 jˆ km .
If it is desired to express this in magnitude-angle notation, then this is equivalent to a vector of
length
√
602 + (−20)2 = 63 km, which is directed at 18◦ south of east.
(b) We use the result from part (a) in Eq. 4-8 along with the fact that ∆t = 90 h. In unit vector
notation, we obtain
~vavg =
60 ıˆ− 20 jˆ
90
= 0.66 ıˆ− 0.22 jˆ
in kilometers-per-hour. This result in magnitude-angle notation is ~vavg = 0.70 km/h at 18
◦ south
of east.
(c) Average speed is distinguished from the magnitude of average velocity in that it depends on the
total distance as opposed to the net displacement. Since the camel travels 140 km, we obtain
140/90 = 1.56 km/h.
(d) The net displacement is required to be the 90 km East from A to B. The displacement from the
resting place to B is denoted ~r3. Thus, we must have (in kilometers)
~r1 + ~r2 + ~r3 = 90 ıˆ
74 CHAPTER 4.
which produces ~r3 = 30 ıˆ + 20 jˆ in unit-vector notation, or (36 6 33◦) in magnitude-angle notation.
Therefore, using Eq. 4-8 we obtain
|~vavg | = 36 km
120− 90 h = 1.2 km/h
and the direction of this vector is the same as ~r3 (that is, 33
◦ north of east).
9. We apply Eq. 4-10 and Eq. 4-16.
(a) Taking the derivative of the position vector with respect to time, we have
~v =
d
dt
(
ıˆ + 4t2 jˆ + t kˆ
)
= 8t jˆ + kˆ
in SI units (m/s).
(b) Taking another derivative with respect to time leads to
~a =
d
dt
(
8t jˆ + kˆ
)
= 8 jˆ
in SI units (m/s2).
10. We use Eq. 4-15 with ~v1 designating the initial velocity and ~v2 designating the later one.
(a) The average acceleration during the ∆t = 4 s interval is
~aavg =
(
−2 ıˆ− 2 jˆ + 5 kˆ
)
−
(
4 ıˆ− 22 jˆ + 3 kˆ
)
4
= −1.5 ıˆ + 0.5 kˆ
in SI units (m/s2).
(b) The magnitude of ~aavg is
√
(−1.5)2 + 0.52 = 1.6 m/s2. Its angle in the xz plane (measured from
the +x axis) is one of these possibilities:
tan−1
(
0.5
−1.5
)
= −18◦ or 162◦
where we settle on the second choice since the signs of its components imply that it is in the second
quadrant.
11. In parts (b) and (c), we use Eq. 4-10 and Eq. 4-16. For part (d), we find the direction of the velocity
computed in part (b), since that represents the asked-for tangent line.
(a) Plugging into the given expression, we obtain
~r
∣∣∣
t=2
= (2(8)− 5(2)) ıˆ + (6 − 7(16)) jˆ = 6.00 ıˆ− 106 jˆ
in meters.
(b) Taking the derivative of the given expression produces
~v(t) =
(
6.00t2 − 5.00) ıˆ + 28.0t3 jˆ
where we have written v(t) to emphasize its dependence on time. This becomes, at t = 2.00 s, ~v =
19.0 ıˆ− 224 jˆ m/s.
(c) Differentiating the ~v(t) found above, with respect to t produces 12.0t ıˆ− 84.0t2 jˆ, which yields ~a =
24.0 ıˆ− 336 jˆ m/s2 at t = 2.00 s.
75
(d) The angle of ~v, measured from +x, is either
tan−1
(−224
19.0
)
= −85.2◦ or 94.8◦
where we settle on the first choice (−85.2◦, which is equivalent to 275◦) since the signs of its
components imply that it is in the fourth quadrant.
12. Noting that ~v2 = 0, then, using Eq. 4-15, the average acceleration is
~aavg =
∆~v
∆t
=
0− (6.30 ıˆ− 8.42 jˆ)
3
= −2.1 ıˆ + 2.8 jˆ
in SI units (m/s2).
13. Constant acceleration in both directions (x and y) allows us to use Table 2-1 for the motion along each
direction. This can be handled individually (for ∆x and ∆y) or together with the unit-vector notation
(for ∆r). Where units are not shown, SI units are to be understood.
(a) The velocity of the particle at any time t is given by ~v = ~v0 + ~at, where ~v0 is the initial velocity
and ~a is the (constant) acceleration. The x component is vx = v0x + axt = 3.00− 1.00t, and the y
component is vy = v0y + ayt = −0.500t since v0y = 0. When the particle reaches its maximum x
coordinate at t = tm, we must have vx = 0. Therefore, 3.00 − 1.00tm = 0 or tm = 3.00 s. The y
component of the velocity at this time is vy = 0−0.500(3.00) = −1.50 m/s; this is the only nonzero
component of ~v at tm.
(b) Since it started at the origin, the coordinates of the particle at any time t are given by ~r = ~v0t+
1
2~at
2.
At t = tm this becomes
(3.00 ıˆ)(3.00) +
1
2
(−1.00 ıˆ− 0.50 jˆ)(3.00)2 = 4.50 ıˆ− 2.25 jˆ
in meters.
14. (a) Using Eq. 4-16, the acceleration as a function of time is
~a =
d~v
dt
=
d
dt
(
(6.0t− 4.0t2)ˆı + 8.0 jˆ
)
= (6.0− 8.0t)ˆı
in SI units. Specifically, we find the acceleration vector at t = 3.0 s to be (6.0 − 8.0(3.0))ˆı =
−18 ıˆ m/s2.
(b) The equation is ~a = (6.0− 8.0t)ˆı = 0; we find t = 0.75 s.
(c) Since the y component of the velocity, vy = 8.0 m/s, is never zero, the velocity cannot vanish.
(d) Since speed is the magnitude of the velocity, we have v = |~v| = √(6.0t− 4.0t2)2 + (8.0)2 = 10 in
SI units (m/s). We solve for t as follows:
squaring (6t− 4t2)2 + 64 = 100
rearranging (6t− 4t2)2 = 36
taking square root 6t− 4t2 = ±6
rearranging 4t2 − 6t± 6 = 0
using quadratic formula t =
6±√36− 4(4)(±6)
2(8)
where the requirement of a real positive result leads to the unique answer: t = 2.2 s.
15. Since the x and y components of the acceleration are constants, then we can use Table 2-1 for the motion
along both axes. This can be handled individually (for ∆x and ∆y) or together with the unit-vector
notation (for ∆r). Where units are not shown, SI units are to be understood.
76 CHAPTER 4.
(a) Since ~r0 = 0, the position vector of the particle is (adapting Eq. 2-15)
~r = ~v0t+
1
2
~at2
= (8.0 jˆ)t+
1
2
(4.0 ıˆ + 2.0 jˆ)t2
= (2.0t2) ıˆ + (8.0t+ 1.0t2) jˆ .
Therefore, we find when
x = 29 m, by solving 2.0t2 = 29, which leads to t = 3.8 s. The y coordinate
at that time is y = 8.0(3.8) + 1.0(3.8)2 = 45 m.
(b) Adapting Eq. 2-11, the velocity of the particle is given by
~v = ~v0 + ~at .
Thus, at t = 3.8 s, the velocity is
~v = 8.0 jˆ + (4.0 ıˆ + 2.0 jˆ)(3.8) = 15.2 ıˆ + 15.6 jˆ
which has a magnitude of
v =
√
v2x + v
2
y =
√
15.22 + 15.62 = 22 m/s .
16. The acceleration is constant so that use of Table 2-1 (for both the x and y motions) is permitted. Where
units are not shown, SI units are to be understood. Collision between particles A and B requires two
things. First, the y motion of B must satisfy (using Eq. 2-15 and noting that θ is measured from the y
axis)
y =
1
2
ayt
2 =⇒ 30 = 1
2
(0.40 cos θ) t2 .
Second, the x motions of A and B must coincide:
vt =
1
2
axt
2 =⇒ 3.0t = 1
2
(0.40 sin θ) t2 .
We eliminate a factor of t in the last relationship and formally solve for time:
t =
3
0.2 sin θ
.
This is then plugged into the previous equation to produce
30 =
1
2
(0.40 cos θ)
(
3
0.2 sin θ
)2
which, with the use of sin2 θ = 1− cos2 θ, simplifies to
30 =
9
0.2
cos θ
1− cos2 θ =⇒ 1− cos
2 θ =
9
(0.2)(30)
cos θ .
We use the quadratic formula (choosing the positive root) to solve for cos θ:
cos θ =
−1.5 +√1.52 − 4(1)(−1)
2
=
1
2
which yields
θ = cos−1
(
1
2
)
= 60◦ .
77
17. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are
directly applicable.
(a) With the origin at the firing point, the y coordinate of the bullet is given by y = − 12gt2. If t is the
time of flight and y = −0.019 m indicates where the bullet hits the target, then
t =
√
2(0.019)
9.8
= 6.2× 10−2 s .
(b) The muzzle velocity is the initial (horizontal) velocity of the bullet. Since x = 30 m is the horizontal
position of the target, we have x = v0t. Thus,
v0 =
x
t
=
30
6.3× 10−2 = 4.8× 10
2 m/s .
18. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are
directly applicable.
(a) With the origin at the initial point (edge of table), the y coordinate of the ball is given by y = − 12gt2.
If t is the time of flight and y = −1.20 m indicates the level at which the ball hits the floor, then
t =
√
2(1.20)
9.8
= 0.495 s .
(b) The initial (horizontal) velocity of the ball is ~v = v0 ıˆ. Since x = 1.52 m is the horizontal position
of its impact point with the floor, we have x = v0t. Thus,
v0 =
x
t
=
1.52
0.495
= 3.07 m/s .
19. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are
directly applicable. The initial velocity is horizontal so that v0y = 0 and v0x = v0 = 161 km/h.
Converting to SI units, this is v0 = 44.7 m/s.
(a) With the origin at the initial point (where the ball leaves the pitcher’s hand), the y coordinate
of the ball is given by y = − 12gt2, and the x coordinate is given by x = v0t. From the latter
equation, we have a simple proportionality between horizontal distance and time, which means the
time to travel half the total distance is half the total time. Specifically, if x = 18.3/2 m, then
t = (18.3/2)/44.7 = 0.205 s.
(b) And the time to travel the next 18.3/2 m must also be 0.205 s. It can be useful to write the
horizontal equation as ∆x = v0∆t in order that this result can be seen more clearly.
(c) From y = − 12gt2, we see that the ball has reached the height of − 12 (9.8)(0.205)2 = −0.205 m at the
moment the ball is halfway to the batter.
(d) The ball’s height when it reaches the batter is − 12 (9.8)(0.409)2 = −0.820 m, which, when subtracted
from the previous result, implies it has fallen another 0.615 m. Since the value of y is not simply
proportional to t, we do not expect equal time-intervals to correspond to equal height-changes; in
a physical sense, this is due to the fact that the initial y-velocity for the first half of the motion is
not the same as the “initial” y-velocity for the second half of the motion.
20. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are
directly applicable. The initial velocity is horizontal so that v0y = 0 and v0x = v0 = 10 m/s.
(a) With the origin at the initial point (where the dart leaves the thrower’s hand), the y coordinate of
the dart is given by y = − 12gt2, so that with y = −PQ we have PQ = 12 (9.8)(0.19)2 = 0.18 m.
(b) From x = v0t we obtain x = (10)(0.19) = 1.9 m.
78 CHAPTER 4.
21. Since this problem involves constant downward acceleration of magnitude a, similar to the projectile
motion situation, we use the equations of §4-6 as long as we substitute a for g. We adopt the positive
direction choices used in the textbook so that equations such as Eq. 4-22 are directly applicable. The
initial velocity is horizontal so that v0y = 0 and v0x = v0 = 1.0× 109 cm/s.
(a) If ℓ is the length of a plate and t is the time an electron is between the plates, then ℓ = v0t, where
v0 is the initial speed. Thus
t =
ℓ
v0
=
2.0 cm
1.0× 109 cm/s = 2.0× 10
−9 s .
(b) The vertical displacement of the electron is
y = −1
2
at2 = −1
2
(
1.0× 1017 cm/s2
) (
2.0× 10−9 s)2 = −0.20 cm .
(c) and (d) The x component of velocity does not change: vx = v0 = 1.0 × 109 cm/s, and the y
component is
vy = ayt =
(
1.0× 1017 cm/s2
) (
2.0× 10−9 s) = 2.0× 108 cm/s .
22. We use Eq. 4-26
Rmax =
(
v20
g
sin 2θ0
)
max
=
v20
g
=
(9.5m/s)2
9.80m/s
2 = 9.21 m
to compare with Powell’s long jump; the difference from Rmax is only ∆R = 9.21− 8.95 = 0.26 m.
23. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are
directly applicable. The coordinate origin is throwing point (the stone’s initial position). The x com-
ponent of its initial velocity is given by v0x = v0 cos θ0 and the y component is given by v0y = v0 sin θ0,
where v0 = 20 m/s is the initial speed and θ0 = 40.0
◦ is the launch angle.
(a) At t = 1.10 s, its x coordinate is
x = v0t cos θ0 = (20.0m/s)(1.10 s) cos 40.0
◦ = 16.9m
(b) Its y coordinate at that instant is
y = v0t sin θ0 − 1
2
gt2 = (20.0m/s)(1.10 s) sin 40◦ − 1
2
(9.80m/s
2
)(1.10 s)2 = 8.21m .
(c) At t′ = 1.80 s, its x coordinate is
x = (20.0m/s)(1.80 s) cos 40.0◦ = 27.6m
(d) Its y coordinate at t′ is
y = (20.0m/s)(1.80 s) sin 40◦ − 1
2
(
9.80m/s
2
)
(1.80 s)2 = 7.26 m .
(e) and (f) The stone hits the ground earlier than t = 5.0 s. To find the time when it hits the ground
solve y = v0t sin θ0 − 12gt2 = 0 for t. We find
t =
2v0
g
sin θ0 =
2(20.0m/s)
9.8m/s
2 sin 40
◦ = 2.62 s .
Its x coordinate on landing is
x = v0t cos θ0 = (20.0m/s)(2.62 s) cos 40
◦ = 40.2 m
(or Eq. 4-26 can be used). Assuming it stays where it lands, its coordinates at t = 5.00 s are
x = 40.2m and y = 0.
79
24. In this projectile motion problem, we have v0 = vx = constant, and what is plotted is v =
√
v2x + v
2
y .
We infer from the plot that at t = 2.5 s, the ball reaches its maximum height, where vy = 0. Therefore,
we infer from the graph that vx = 19 m/s.
(a) During t = 5 s, the horizontal motion is x− x0 = vxt = 95 m.
(b) Since
√
192 + v02y = 31 m/s (the first point on the graph), we find v0y = 24.5 m/s. Thus, with
t = 2.5 s, we can use ymax − y0 = v0yt − 12gt2 or v2y = 0 = v02y − 2g(ymax − y0), or ymax − y0 =
1
2
(
vy + v0y
)
t to solve. Here we will use the latter:
ymax − y0 = 1
2
(
vy + v0y
)
t =⇒ ymax = 1
2
(0 + 24.5)(2.5) = 31 m
where we have taken y0 = 0 as the ground level.
25. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are
directly applicable. The coordinate origin is at the end of the rifle (the initial point for the bullet
as it
begins projectile motion in the sense of §4-5), and we let θ0 be the firing angle. If the target is a distance
d away, then its coordinates are x = d, y = 0. The projectile motion equations lead to d = v0t cos θ0 and
0 = v0t sin θ0 − 12gt2. Eliminating t leads to 2v20 sin θ0 cos θ0 − gd = 0. Using sin θ0 cos θ0 = 12 sin(2θ0),
we obtain
v20 sin(2θ0) = gd =⇒ sin(2θ0) =
gd
v20
=
(9.8)(45.7)
4602
which yields sin(2θ0) = 2.12 × 10−3 and consequently θ0 = 0.0606◦. If the gun is aimed at a point a
distance ℓ above the target, then tan θ0 = ℓ/d so that
ℓ = d tan θ0 = 45.7 tan0.0606
◦ = 0.0484m = 4.84 cm .
26. The figure offers many interesting points to analyze, and others are easily inferred (such as the point of
maximum height). The focus here, to begin with, will be the final point shown (1.25 s after the ball is
released) which is when the ball returns to its original height. In English units, g = 32 ft/s2.
(a) Using x− x0 = vxt we obtain vx = (40 ft)/(1.25 s) = 32 ft/s. And y − y0 = 0 = v0yt− 12gt2 yields
v0y =
1
2 (32)(1.25) = 20 ft/s. Thus, the initial speed is
v0 = |~v0| =
√
322 + 202 = 38 ft/s .
(b) Since vy = 0 at the maximum height and the horizontal velocity stays constant, then the speed at
the top is the same as vx = 32 ft/s.
(c) We can infer from the figure (or compute from vy = 0 = v0y − gt) that the time to reach the top
is 0.625 s. With this, we can use y − y0 = v0yt − 12gt2 to obtain 9.3 ft (where y0 = 3 ft has been
used). An alternative approach is to use v2y = v
2
0y − 2g(y − y0).
27. Taking the y axis to be upward and placing the origin at the firing point, the y coordinate is given by
y = v0t sin θ0 − 12gt2 and the y component of the velocity is given by vy = v0 sin θ0 − gt. The maximum
height occurs when vy = 0. Thus, t = (v0/g) sin θ0 and
y = v0
(
v0
g
)
sin θ0 sin θ0 − 1
2
g(v0 sin θ0)
2
g2
=
(v0 sin θ0)
2
2g
.
28. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are
directly applicable. The coordinate origin is at the release point (the initial position for the ball as it
begins projectile motion in the sense of §4-5), and we let θ0 be the angle of throw (shown in the figure).
Since the horizontal component of the velocity of the ball is vx = v0 cos 40.0
◦, the time it takes for the
ball to hit the wall is
t =
∆x
vx
=
22.0
25.0 cos40.0◦
= 1.15 s .
80 CHAPTER 4.
(a) The vertical distance is
∆y = (v0 sin θ0)t− 1
2
gt2
= (25.0 sin40.0◦)(1.15)− 1
2
(9.8)(1.15)2 = 12.0 m .
(b) The horizontal component of the velocity when it strikes the wall does not change from its initial
value: vx = v0 cos 40.0
◦ = 19.2 m/s, while the vertical component becomes (using Eq. 4-23)
vy = v0 sin θ0 − gt = 25.0 sin40.0◦ − (9.8)(1.15) = 4.80 m/s .
(c) Since vy > 0 when the ball hits the wall, it has not reached the highest point yet.
29. We designate the given velocity ~v = 7.6 ıˆ + 6.1 jˆ (SI units understood) as ~v1 – as opposed to the velocity
when it reaches the max height ~v2 or the velocity when it returns to the ground ~v3 – and take ~v0 as the
launch velocity, as usual. The origin is at its launch point on the ground.
(a) Different approaches are available, but since it will be useful (for the rest of the problem) to first
find the initial y velocity, that is how we will proceed. Using Eq. 2-16, we have
v21 y = v
2
0 y − 2g∆y
6.12 = v20 y − 2(9.8)(9.1)
which yields v0 y = 14.7 m/s. Knowing that v2 y must equal 0, we use Eq. 2-16 again but now with
∆y = h for the maximum height:
v22 y = v
2
0 y − 2gh
0 = 14.72 − 2(9.8)h
which yields h = 11 m.
(b) Recalling the derivation of Eq. 4-26, but using v0 y for v0 sin θ0 and v0x for v0 cos θ0, we have
0 = v0 yt− 1
2
gt2
R = v0xt
which leads to R =
2 v0 x v0 y
g . Noting that v0 x = v1 x = 7.6 m/s, we plug in values and obtain
R = 2(7.6)(14.7)/9.8 = 23 m.
(c) Since v3 x = v1 x = 7.6 m/s and v3 y = −v0 y = −14.7 m/s, we have
v3 =
√
v23 x + v
2
3 y =
√
(−14.7)2 + 7.62 = 17 m/s .
(d) The angle (measured from horizontal) for ~v3 is one of these possibilities:
tan−1
(−14.7
7.6
)
= −63◦ or 117◦
where we settle on the first choice (−63◦, which is equivalent to 297◦) since the signs of its compo-
nents imply that it is in the fourth quadrant.
30. We apply Eq. 4-21, Eq. 4-22 and Eq. 4-23.
(a) From ∆x = v0xt, we find v0x = 40/2 = 20 m/s.
(b) From ∆y = v0yt− 12gt2, we find v0y = (53 + 12 (9.8)(2)2)/2 = 36 m/s.
81
(c) From vy = v0y−gt′ with vy = 0 as the condition for maximum height, we obtain t′ = 36/9.8 = 3.7 s.
During that time the x-motion is constant, so x′ − x0 = (20)(3.7) = 74 m.
31. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are
directly applicable. The coordinate origin is at the the initial position for the football as it begins
projectile motion in the sense of §4-5), and we let θ0 be the angle of its initial velocity measured from
the +x axis.
(a) x = 46m and y = −1.5m are the coordinates for the landing point; it lands at time t = 4.5 s. Since
x = v0xt,
v0 x =
x
t
=
46m
4.5 s
= 10.2 m/s .
Since y = v0yt− 12gt2,
v0 y =
y + 12gt
2
t
=
(−1.5m) + 12 (9.8m/s2)(4.5 s)2
4.5 s
= 21.7 m/s .
The magnitude of the initial velocity is
v0 =
√
v20 x + v
2
0 y =
√
(10.2m/s)2 + (21.7m/s)2 = 24 m/s .
(b) The initial angle satisfies tan θ0 = v0 y/v0x. Thus, θ0 = tan
−1(21.7/10.2) = 64.8◦.
32. The initial velocity has no vertical component – only an x component equal to +2.00 m/s. Also,
y0 = +10.0 m if the water surface is established as y = 0.
(a) x− x0 = vxt readily yields x− x0 = 1.60 m.
(b) Using y − y0 = v0yt− 12gt2, we obtain y = 6.86 m when t = 0.800 s.
(c) With t unknown and y = 0, the equation y − y0 = v0yt − 12gt2 leads to t =
√
2(10)/9.8 = 1.43 s.
During this time, the x-displacement of the diver is x− x0 = (2.00 m/s)(1.43 s) = 2.86 m.
33. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are
directly applicable. The coordinate origin is at ground level directly below the release point. We write
θ0 = −30◦ since the angle shown in the figure is measured clockwise from horizontal. We note that
the initial speed of the decoy is the plane’s speed at the moment of release: v0 = 290 km/h, which we
convert to SI units: (290)(1000/3600) = 80.6 m/s.
(a) We use Eq. 4-12 to solve for the time:
∆x = (v0 cos θ0) t =⇒ t = 700
(80.6) cos−30◦ = 10.0 s .
(b) And we use Eq. 4-22 to solve for the initial height y0:
y − y0 = (v0 sin θ0) t− 1
2
gt2
0− y0 = (−40.3)(10.0)− 1
2
(9.8)(10.0)2
which yields y0 = 897 m.
34. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are
directly applicable. The coordinate origin is at its initial position (where it is launched). At maximum
height, we observe vy = 0 and denote vx = v (which is also equal to v0 x). In this notation, we have
v0 = 5v .
82 CHAPTER 4.
Next, we observe v0 cos θ0 = v0x = v, so that we arrive at an equation (where v 6= 0 cancels) which can
be solved for θ0:
(5v) cos θ0 = v =⇒ θ0 = cos−1 1
5
= 78◦ .
35. We denote h as the height of a step and w as the width. To hit step n, the ball must fall a distance nh
and travel horizontally a distance between (n− 1)w and nw. We take the origin of a coordinate system
to be at the point where the ball leaves the top of the stairway, and we choose the y axis to be positive
in the upward direction. The coordinates of the ball at time t are given by x = v0 xt and y = − 12gt2
(since v0 y = 0). We equate y to −nh and solve for the time to reach the level of step n:
t =
√
2nh
g
.
The x coordinate then is
x = v0 x
√
2nh
g
= (1.52m/s)
√
2n(0.203m)
9.8m/s2
= (0.309m)
√
n .
The
method is to try values of n until we find one for which x/w is less than n but greater than n−1. For
n = 1, x = 0.309m and x/w = 1.52, which is greater than n. For n = 2, x = 0.437m and x/w = 2.15,
which is also greater than n. For n = 3, x = 0.535m and x/w = 2.64. Now, this is less than n and
greater than n− 1, so the ball hits the third step.
36. Although we could use Eq. 4-26 to find where it lands, we choose instead to work with Eq. 4-21 and
Eq. 4-22 (for the soccer ball) since these will give information about where and when and these are also
considered more fundamental than Eq. 4-26. With ∆y = 0, we have
∆y = (v0 sin θ0) t− 1
2
gt2 =⇒ t = (19.5) sin 45
◦
1
2 (9.8)
= 2.81 s .
Then Eq. 4-21 yields ∆x = (v0 cos θ0) t = 38.3 m. Thus, using Eq. 4-8 and SI units, the player must
have an average velocity of
~vavg =
∆~r
∆t
=
38.3 ıˆ− 5 ıˆ
2.81
= −5.8 ıˆ
which means his average speed (assuming he ran in only one direction) is 5.8 m/s.
37. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are
directly applicable. The coordinate origin is at ground level directly below the release point. We write
θ0 = −37◦ for the angle measured from +x, since the angle given in the problem is measured from the
−y direction. We note that the initial speed of the projectile is the plane’s speed at the moment of
release.
(a) We use Eq. 4-22 to find v0 (SI units are understood).
y − y0 = (v0 sin θ0) t− 1
2
gt2
0− 730 = v0 sin(−37◦) (5.00)− 1
2
(9.8)(5.00)2
which yields v0 = 202 m/s.
(b) The horizontal distance traveled is x = v0t cos θ0 = (202)(5.00) cos−37.0◦ = 806 m.
(c) The x component of the velocity (just before impact) is vx = v0 cos θ0 = (202) cos−37.0◦ = 161 m/s.
(d) The y component of the velocity (just before impact) is vy = v0 sin θ0 − gt = (202) sin(−37◦) −
(9.80)(5.00) = −171 m/s.
83
38. We assume the ball’s initial velocity is perpendicular to the plane of the net. We choose coordinates so
that (x0, y0) = (0, 3.0) m, and vx > 0 (note that v0y = 0).
(a) To (barely) clear the net, we have
y − y0 = v0yt− 12gt
2 =⇒ 3.0− 2.24 = 0− 1
2
(9.8)t2
which gives t = 0.39 s for the time it is passing over the net. This is plugged into the x-equation to
yield the (minimum) initial velocity vx = (8.0 m)/(0.39 s) = 20.3 m/s.
(b) We require y = 0 and find t from y − y0 = v0yt − 12gt2. This value (t =
√
2(3.0)/9.8 = 0.78
s) is plugged into the x-equation to yield the (maximum) initial velocity vx = (17.0 m)/(0.78 s)
= 21.7 m/s.
39. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are
directly applicable. The coordinate origin is at ground level directly below impact point between bat
and ball. The Hint given in the problem is important, since it provides us with enough information to
find v0 directly from Eq. 4-26.
(a) We want to know how high the ball is from the ground when it is at x = 97.5 m, which requires
knowing the initial velocity. Using the range information and θ0 = 45
◦, we use Eq. 4-26 to solve
for v0:
v0 =
√
g R
sin 2θ0
=
√
(9.8)(107)
1
= 32.4 m/s .
Thus, Eq. 4-21 tells us the time it is over the fence:
t =
x
v0 cos θ0
=
97.5
(32.4) cos 45◦
= 4.26 s .
At this moment, the ball is at a height (above the ground) of
y = y0 + (v0 sin θ0) t− 1
2
gt2 = 9.88 m
which implies it does indeed clear the 7.32 m high fence.
(b) At t = 4.26 s, the center of the ball is 9.88− 7.32 = 2.56 m above the fence.
40. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are
directly applicable. The coordinate origin is at ground level directly below the point where the ball was
hit by the racquet.
(a) We want to know how high the ball is above the court when it is at x = 12 m. First, Eq. 4-21 tells
us the time it is over the fence:
t =
x
v0 cos θ0
=
12
(23.6) cos 0◦
= 0.508 s .
At this moment, the ball is at a height (above the court) of
y = y0 + (v0 sin θ0) t− 1
2
gt2 = 1.103 m
which implies it does indeed clear the 0.90 m high fence.
(b) At t = 0.508 s, the center of the ball is 1.103− 0.90 = 0.20 m above the net.
(c) Repeating the computation in part (a) with θ0 = −5◦ results in t = 0.510 s and y = 0.04 m, which
clearly indicates that it cannot clear the net.
84 CHAPTER 4.
(d) In the situation discussed in part (c), the distance between the top of the net and the center of the
ball at t = 0.510 s is 0.90− 0.04 = 0.86 m.
41. We adopt the positive direction choices used in the textbook so that equations such as Eq. 4-22 are
directly applicable. The coordinate origin is at the point where the ball is kicked. Where units are not
displayed, SI units are understood. We use x and y to denote the coordinates of ball at the goalpost,
and try to find the kicking angle(s) θ0 so that y = 3.44 m when x = 50 m. Writing the kinematic
equations for projectile motion: x = v0t cos θ0 and y = v0t sin θ0 − 12gt2, we see the first equation gives
t = x/v0 cos θ0, and when this is substituted into the second the result is
y = x tan θ0 − gx
2
2v20 cos
2 θ0
.
One may solve this by trial and error: systematically trying values of θ0 until you find the two that
satisfy the equation. A little manipulation, however, will give an algebraic solution:
Using the trigonometric identity 1/ cos2 θ0 = 1 + tan
2 θ0, we obtain
1
2
gx2
v20
tan2 θ0 − x tan θ0 + y + 1
2
gx2
v20
= 0
which is a second-order equation for tan θ0. To simplify writing the solution, we denote c =
1
2gx
2/v20 =
1
2 (9.80)(50)
2/(25)2 = 19.6 m. Then the second-order equation becomes c tan2 θ0 − x tan θ0 + y + c = 0.
Using the quadratic formula, we obtain its solution(s).
tan θ0 =
x±√x2 + 4(y + c)c
2c
=
50±√502 − 4(3.44 + 19.6)(19.6)
2(19.6)
.
The two solutions are given by tan θ0 = 1.95 and tan θ0 = 0.605. The corresponding (first-quadrant)
angles are θ0 = 63
◦ and θ0 = 31◦. If kicked at any angle between these two, the ball will travel above
the cross bar on the goalposts.
42. The magnitude of the acceleration is
a =
v2
r
=
(10m/s)2
25m
= 4.0m/s
2
.
43. We apply Eq. 4-33 to solve for speed v and Eq. 4-32 to find acceleration a.
(a) Since the radius of Earth is 6.37×106m, the radius of the satellite orbit is 6.37×106m+640×103m
= 7.01× 106m. Therefore, the speed of the satellite is
v =
2πr
T
=
2π(7.01× 106m)
(98.0min)(60 s/min)
= 7.49× 103m/s .
(b) The magnitude of the acceleration is
a =
v2
r
=
(7.49× 103m/s)2
7.01× 106m = 8.00m/s
2
.
44. We note that the period of revolution is (1200 rev/min)
−1
= 8.3× 10−4 min which becomes, in SI units,
T = 0.050 s.
(a) The circumference is c = 2πr = 2π(0.15) = 0.94 m.
(b) The speed is v = c/T = (0.94)/(0.050) = 19 m/s. This is equivalent to using Eq. 4-33.
85
(c) The magnitude of the acceleration is a = v2/r = 192/0.15 = 2.4× 103 m/s2.
(d) As noted above, T = 50 ms.
45. We apply Eq. 4-32 to solve for speed v and Eq. 4-33 to find the period T .
(a) We obtain
v =
√
ra =
√
(5.0m)(7.0)(9.8m/s2) = 19 m/s .
(b) The time to go around once (the period) is T = 2πr/v = 1.7 s. Therefore, in one minute (t = 60 s),
the astronaut executes
t
T
=
60
1.7
= 35
revolutions. Thus, 35 rev/min is needed to produce a centripetal acceleration of 7g when the radius
is 5.0 m.
(c) As noted above, T = 1.7 s.
46. The magnitude of centripetal acceleration (a = v2/r) and its direction (towards the center of the circle)
form the basis of this problem.
(a) If a passenger at this location experiences ~a = 1.83 m/s2 east, then the center of the circle is east
of this location. And the distance is r = v2/a = (3.662)/(1.83) = 7.32 m. Thus, relative to the
center, the passenger at that moment is located 7.32 m toward the west.
(b) We see the distance is the same, but now the direction of ~a experienced by the passenger is south –
indicating that the center of the merry-go-round is south of him. Therefore, relative to the center,
the passenger at that moment located 7.32 m toward the north.
47. The radius of Earth may be found in Appendix C.
(a) The speed of a person at Earth’s equator is v = 2πR/T , where R is the radius of Earth (6.37×106m)
and T is the length of a day (8.64 × 104 s): v = 2π(6.37 × 106m)/(8.64 × 104 s) = 463m/s. The
magnitude of the acceleration is given by
a =
v2
R
=
(463m/s)2
6.37× 106m = 0.034 m/s
2
.
(b) If T is the period, then v = 2πR/T is the speed and a = v2/R = 4π2R2/T 2R = 4π2R/T 2 is the
magnitude of the acceleration. Thus
T = 2π
√
R
a
= 2π
√
6.37× 106m
9.8m/s
2 = 5.1× 103 s = 84 min .
48. Eq. 4-32 describes an inverse proportionality between r and a, so that a large acceleration results from
a small radius. Thus, an upper limit for a corresponds to a lower limit for r.
(a) The minimum turning radius of the train is given by
rmin =
v2
amax
=
(216 km/h)2
(0.050)(9.8m/s2)
= 7.3× 103m .
(b) The speed of the train must be reduced to no more than
v =
√
amax r =
√
0.050(9.8)(1.00× 103) = 22 m/s
which is roughly 80 km/h.
86 CHAPTER 4.
49. (a) Since the wheel completes 5 turns each minute, its period is one-fifth of a minute, or 12 s.
(b) The magnitude of the centripetal acceleration is given by a = v2/R, where R is the radius of
the wheel, and v is the speed of the passenger. Since the passenger goes a distance 2πR for each
revolution, his speed is
v =
2π(15m)
12 s
= 7.85m/s
and his centripetal acceleration is
a =
(7.85m/s)2
15m
= 4.1m/s
2
.
When the passenger is at the highest point, his centripetal acceleration is downward, toward the
center of the orbit.
(c) At the lowest point, the centripetal acceleration vector points up, toward the center of the orbit.
It has the same magnitude as in part (b).
50. We apply Eq. 4-33 to solve for speed v and Eq. 4-32 to find centripetal acceleration a.
(a) v = 2πr/T = 2π(20 km)/1.0 s = 1.3× 105 km/s.
(b)
a =
v2
r
=
(126 km/s)2
20 km
= 7.9× 105m/s2 .
(c) Clearly, both v and a will increase if T is reduced.
51. To calculate the centripetal acceleration of the stone, we need to know its speed during its circular
motion (this is also its initial speed when it flies off). We use the kinematic equations of projectile
motion (discussed in §4-6) to find that speed. Taking the +y direction to be upward and placing the
origin at the point where the stone leaves its circular orbit, then the coordinates of the stone during
its motion as a projectile are given by x = v0t and y = − 12gt2 (since v0 y = 0). It hits the ground at
x = 10m and y = −2.0m. Formally solving the second equation for the time, we obtain t = √−2y/g,
which we substitute into the first equation:
v0 = x
√
− g
2y
= (10m)
√
− 9.8m/s
2
2(−2.0m) = 15.7 m/s .
Therefore, the magnitude of the centripetal acceleration is
a =
v2
r
=
(15.7m/s)2
1.5m
= 160 m/s2 .
52. We write our magnitude-angle results in the form (R 6 θ) with SI units for the magnitude understood (m
for distances, m/s for speeds, m/s2 for accelerations). All angles θ are measured counterclockwise from
+x, but we will occasionally refer to angles φ which are measured counterclockwise from the vertical
line between the circle-center and the coordinate origin and the line drawn from the circle-center to
the particle location (see r in the figure). We note that the speed of the particle is v = 2πr/T where
r = 3.00 m and T = 20.0 s; thus, v = 0.942 m/s. The particle is moving counterclockwise in Fig. 4-37.
(a) At t = 5.00 s, the particle has traveled a fraction of
t
T
=
5.00
20.0
=
1
4
of a full revolution around the circle (starting at the origin). Thus, relative to the circle-center, the
particle is at
φ =
1
4
(360◦) = 90◦
87
measured from vertical (as explained above). Referring to Fig. 4-37, we see that this position (which
is the “3 o’clock” position on the circle) corresponds to x = 3.00 m and y = 3.00 m relative to the
coordinate origin. In our magnitude-angle notation, this is expressed as (R 6 θ) = (4.24 6 45◦).
Although this position is easy to analyze without resorting to trigonometric relations, it is useful
(for the computations below) to note that these values of x and y relative to coordinate origin
can be gotten from the angle φ from the relations x = r sinφ and y = r − r cosφ. Of course,
R =
√
x2 + y2 and θ comes from choosing the appropriate possibility from tan−1 (y/x) (or by using
particular functions of vector capable calculators).
(b) At t = 7.50 s, the particle has traveled a fraction of 7.50/20.0 = 3/8 of a revolution around the circle
(starting at the origin). Relative to the circle-center, the particle is therefore at φ = 3/8 (360◦) =
135◦ measured from vertical in the manner discussed above. Referring to Fig. 4-37, we compute
that this position corresponds to x = 3.00 sin135◦ = 2.12 m and y = 3.00− 3.00 cos135◦ = 5.12 m
relative to the coordinate origin. In our magnitude-angle notation, this is expressed as (R 6 θ) =
(5.54 6 67.5◦).
(c) At t = 10.0 s, the particle has traveled a fraction of 10.0/20.0 = 1/2 of a revolution around
the circle. Relative to the circle-center, the particle is at φ = 180◦ measured from vertical (see
explanation, above). Referring to Fig. 4-37, we see that this position corresponds to x = 0 and
y = 6.00 m relative to the coordinate origin. In our magnitude-angle notation, this is expressed as
(R 6 θ) = (6.00 6 90.0◦).
(d) We subtract the position vector in part (a) from the position vector in part (c): (6.00 6 90.0◦) −
(4.24 6 45◦) = (4.24 6 135◦) using magnitude-angle notation (convenient when using vector capable
calculators). If we wish instead to use unit-vector notation, we write
∆~R = (0− 3) ıˆ + (6− 3) jˆ = −3 ıˆ + 3 jˆ
which leads to |∆~R| = 4.24 m and θ = 135◦.
(e) From Eq. 4-8, we have
~vavg =
∆~R
∆t
where ∆t = 5.00 s
which produces −0.6 ıˆ + 0.6 jˆ m/s in unit-vector notation or (0.849 6 135◦) in magnitude-angle
notation.
(f) The speed has already been noted (v = 0.942 m/s), but its direction is best seen by referring again
to Fig. 4-37. The velocity vector is tangent to the circle at its “3 o’clock position” (see part (a)),
which means ~v is vertical. Thus, our result is (0.942 6 90◦).
(g) Again, the speed has been noted above (v = 0.942 m/s), but its direction is best seen by referring
to Fig. 4-37. The velocity vector is tangent to the circle at its “12 o’clock position” (see part (c)),
which means ~v is horizontal. Thus, our result is (0.942 6 180◦).
(h) The acceleration has magnitude v2/r = 0.296 m/s2, and at this instant (see part (a)) it is horizontal
(towards the center of the circle). Thus, our result is (0.296 6 180◦).
(i) Again, a = v2/r = 0.296 m/s2, but at this instant (see part (c)) it is vertical (towards the center of
the circle). Thus, our result is (0.296 6 270◦).
53. We use Eq. 4-15 first using velocities relative to the truck (subscript t) and then using velocities relative
to the ground (subscript g). We work with SI units, so 20 km/h → 5.6 m/s, 30 km/h → 8.3 m/s, and
45 km/h → 12.5 m/s. We choose east as the + ıˆ direction.
(a) The velocity of the cheetah (subscript c) at the end of the 2.0 s interval is (from Eq. 4-42)
~vc t = ~vc g − ~vt g = 12.5 ıˆ− (−5.6 ıˆ) = 18.1 ıˆ m/s
relative to the truck. The (average) acceleration vector relative to the cameraman (in the truck) is
~aavg =
18.1 ıˆ− (−8.3 ıˆ)
2.0
= 13 ıˆ m/s2 .
88 CHAPTER 4.
(b) The velocity of the cheetah at the start of the 2.0 s interval is (from Eq. 4-42)
~v0 c g = ~v0 c t + ~v0 t g = (−8.3 ıˆ) + (−5.6 ıˆ) = −13.9 ıˆ m/s
relative to the ground. The
(average) acceleration vector relative to the crew member (on the
ground) is
~aavg =
12.5 ıˆ− (−13.9 ıˆ)
2.0
= 13 ıˆ m/s
2
identical to the result of part (a).
54. We choose upstream as the + ıˆ direction, and use Eq. 4-42.
(a) The subscript b is for the boat, w is for the water, and g is for the ground.
~vb g = ~vbw + ~vw g = (14 km/h) ıˆ + (−9 km/h) ıˆ = (5 km/h) ıˆ
(b) And we use the subscript c for the child.
~vc g = ~vc b + ~vb g = (−6 km/h) ıˆ + (5 km/h) ıˆ = (−1 km/h) ıˆ
55. When the escalator is stalled the speed of the person is vp = ℓ/t, where ℓ is the length of the escalator
and t is the time the person takes to walk up it. This is vp = (15m)/(90 s) = 0.167m/s. The escalator
moves at ve = (15m)/(60 s) = 0.250m/s. The speed of the person walking up the moving escalator is
v = vp + ve = 0.167m/s+ 0.250m/s = 0.417m/s and the time taken to move the length of the escalator
is
t = ℓ/v = (15m)/(0.417m/s) = 36 s .
If the various times given are independent of the escalator length, then the answer does not depend on
that length either. In terms of ℓ (in meters) the speed (in meters per second) of the person walking on the
stalled escalator is ℓ/90, the speed of the moving escalator is ℓ/60, and the speed of the person walking
on the moving escalator is v = (ℓ/90) + (ℓ/60) = 0.0278ℓ. The time taken is t = ℓ/v = ℓ/0.0278ℓ = 36 s
and is independent of ℓ.
56. We denote the velocity of the player with ~v1 and the relative velocity between the player and the ball be
~v2. Then the velocity ~v of the ball relative to the field is given by ~v = ~v1 + ~v2. The smallest angle θmin
corresponds to the case when
~v ⊥ ~v1. Hence,
θmin = 180
◦ − cos−1
( |~v1|
|~v2|
)
= 180◦ − cos−1
(
4.0m/s
6.0m/s
)
≈ 130◦ .
goal
θmin-
~v1
6
~v
HH
HH
HHY ~v2
57. Relative to the car the velocity of the snowflakes has a vertical component of 8.0m/s and a horizontal
component of 50 km/h = 13.9m/s. The angle θ from the vertical is found from
tan θ = vh/vv = (13.9m/s)/(8.0m/s) = 1.74
which yields θ = 60◦.
58. We denote the police and the motorist with subscripts p and m, respectively. The coordinate system is
indicated in Fig. 4-38.
89
(a) The velocity of the motorist with respect to the police car is
~vmp = ~vm − ~vp = −60 jˆ− (−80 ıˆ) = 80 ıˆ− 60 jˆ (km/h) .
(b) ~vmp does happen to be along the line of sight. Referring to Fig. 4-38, we find the vector pointing
from car to another is ~r = 800 ıˆ− 600 jˆ m (from M to P ). Since the ratio of components in ~r is the
same as in ~vmp, they must point the same direction.
(c) No, they remain unchanged.
59. Since the raindrops fall vertically relative to the train, the horizontal component of the velocity of
a raindrop is vh = 30m/s, the same as the speed of the train. If vv is the vertical component of
the velocity and θ is the angle between the direction of motion and the vertical, then tan θ = vh/vv.
Thus vv = vh/ tan θ = (30m/s)/ tan 70
◦ = 10.9m/s. The speed of a raindrop is v =
√
v2h + v
2
v =√
(30m/s)2 + (10.9m/s)2 = 32m/s.
60. Here, the subscript W refers to the water. Our coordinates are chosen with +x being east and +y being
north. In these terms, the angle specifying east would be 0◦ and the angle specifying south would be
−90◦ or 270◦. Where the length unit is not displayed, km is to be understood.
(a) We have ~vAW = ~vAB + ~vBW , so that ~vAB = (22 6 − 90◦) − (40 6 37◦) = (56 6 − 125◦) in
the magnitude-angle notation (conveniently done with a vector capable calculator in polar mode).
Converting to rectangular components, we obtain
~vAB = −32 ıˆ− 46 jˆ km/h .
Of course, this could have been done in unit-vector notation from the outset.
(b) Since the velocity-components are constant, integrating them to obtain the position is straightfor-
ward (~r − ~r0 =
∫
~v dt)
~r = (2.5− 32t) ıˆ + (4.0− 46t) jˆ
with lengths in kilometers and time in hours.
(c) The magnitude of this ~r is
r =
√
(2.5− 32t)2 + (4.0− 46t)2
We minimize this by taking a derivative and requiring it to equal zero – which leaves us with an
equation for t
dr
dt
=
1
2
6286t− 528√
(2.5− 32t)2 + (4.0− 46t)2 = 0
which yields t = 0.084 h.
(d) Plugging this value of t back into the expression for the distance between the ships (r), we obtain
r = 0.2 km. Of course, the calculator offers more digits (r = 0.225...), but they are not significant;
in fact, the uncertainties implicit in the given data, here, should make the ship captains worry.
61. The velocity vector (relative to the shore) for ships A and B are given by
~vA = − (vA cos 45◦) ıˆ + (vA sin 45◦) jˆ
and
~vB = − (vB sin 40◦) ıˆ− (vB cos 40◦) jˆ
respectively (where vA = 24 knots and vB = 28 knots). We are taking East as + ıˆ and North as jˆ.
90 CHAPTER 4.
(a) Their relative velocity is
~vAB = ~vA − ~vB = (vB sin 40◦ − vA cos 45◦) ıˆ + (vB cos 40◦ + vA sin 45◦) jˆ
the magnitude of which is |~vAB| =
√
1.02 + 38.42 ≈ 38 knots. The angle θ which ~vAB makes with
North is given by
θ = tan−1
(
vAB,x
vAB,y
)
= tan−1
(
1.0
38.4
)
= 1.5◦
which is to say that ~vAB points 1.5
◦ east of north.
(b) Since they started at the same time, their relative velocity describes at what rate the distance
between them is increasing. Because the rate is steady, we have
t =
|∆rAB|
|~vAB | =
160
38
= 4.2 h .
(c) The velocity ~vAB does not change with time in this problem, and ~rAB is in the same direction as
~vAB since they started at the same time. Reversing the points of view, we have ~vAB = −~vBA so
that ~rAB = −~rBA (i.e., they are 180◦ opposite to each other). Hence, we conclude that B stays at
a bearing of 1.5◦ west of south relative to A during the journey (neglecting the curvature of Earth).
62. The (box)car has velocity ~vc g = v1 ıˆ relative to the ground, and the bullet has velocity
~v0 b g = v2 cos θ ıˆ + v2 sin θ jˆ
relative to the ground before entering the car (we are neglecting the effects of gravity on the bullet).
While in the car, its velocity relative to the outside ground is ~vb g = 0.8v2 cos θ ıˆ + 0.8v2 sin θ jˆ (due to
the 20% reduction mentioned in the problem). The problem indicates that the velocity of the bullet in
the car relative to the car is (with v3 unspecified) ~vb c = v3 jˆ. Now, Eq. 4-42 provides the condition
~vb g = ~vb c + ~vc g
0.8v2 cos θ ıˆ + 0.8v2 sin θ jˆ = v3 jˆ + v1 ıˆ
so that equating x components allows us to find θ. If one wished to find v3 one could also equate the y
components, and from this, if the car width were given, one could find the time spent by the bullet in the
car, but this information is not asked for (which is why the width is irrelevant). Therefore, examining
the x components in SI units leads to
θ = cos−1
(
v1
0.8v2
)
= cos−1
(
85
(
1000
3600
)
0.8(650)
)
which yields 87◦ for the direction of ~vb g (measured from ıˆ, which is the direction of motion of the
car). The problem asks, “from what direction was it fired?” – which means the answer is not 87◦ but
rather its supplement 93◦ (measured from the direction of motion). Stating this more carefully, in the
coordinate system we have adopted in our solution, the bullet velocity vector is in the first quadrant, at
87◦ measured counterclockwise from the +x direction (the direction of train motion), which means that
the direction from which the bullet came (where the sniper is) is in the third quadrant, at −93◦ (that
is, 93◦ measured clockwise from +x).
63. We construct a right triangle starting from the clearing on the south
91
bank, drawing a line (200 m long)
due north (upward in our sketch)
across the river, and then a line
due west (upstream, leftward in
our sketch) along the north bank
for a distance (82m)+(1.1m/s)t,
where the t-dependent contribu-
tion is the distance that the river
will carry the boat downstream
during time t. ................
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The hypotenuse of this right triangle (the arrow in our sketch) also depends on t and on the boat’s speed
(relative to the water), and we set it equal to the Pythagorean “sum” of the triangle’s sides:
(4.0)t =
√
2002 + (82 + 1.1t)
2
which leads to a quadratic equation for t
46724 + 180.4t− 14.8t2 = 0 .
We solve this and find a positive value: t = 62.6 s. The angle between the northward (200 m) leg of the
triangle and the hypotenuse (which is measured “west of north”) is then given by
θ = tan−1
(
82 + 1.1t
200
)
= tan−1
(
151
200
)
= 37◦ .
64. (a) We compute the coordinate pairs (x, y) from x = v0 cos θt and x = v0 sin θt− 12gt2 for t = 20 s and
the speeds and angles given in the problem. We obtain (in kilometers)
(xA , yA ) = (10.1, 0.56) (xB , yB ) = (12.1, 1.51)
(xC , yC ) = (14.3, 2.68) (xD , yD ) = (16.4, 3.99)
and (xE , yE ) = (18.5, 5.53) which we plot in the next part.
(b) The vertical (y) and horizontal (x) axes are in kilometers. The graph does not start at the origin.
The curve to “fit” the data is not shown, but is easily imagined (forming the “curtain of death”).
1
2
3
4
5
10 12 14 16 18
65. We denote ~vPG as the velocity of the plane relative to the ground, ~vAG as the velocity of the air relative
to the ground, and ~vPA be the velocity of the plane relative to the air.
(a) The vector diagram is shown below. ~vPG = ~vPA + ~vAG. Since the magnitudes vPG and vPA are
equal the triangle is isosceles, with two sides of equal length. Consider either of the right triangles
92 CHAPTER 4.
formed when the bisector of θ
is drawn (the dashed line). It
bisects ~vAG, so
sin(θ/2) =
vAG
2vPG
=
70.0mi/h
2(135mi/h)
which leads to θ = 30.1◦.
Now ~vAG makes the same an-
gle with the E-W line as the
dashed line does with the N-
S line. The wind is blowing
in the direction 15◦ north of
west. Thus, it is blowing from
75◦ east of south.
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...........................
...
...
...
...
...
...
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...
...
...
...
...
...
...
...
...
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...
...
...
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...
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...
...
................
.....
....
...
.
...........................
.
θ
2
θ
2
φ
~vPG
~vPA
~vAG
N
S
EW
(b) The plane is headed along ~vPA, in the direction 30
◦ east of north. There is another solution, with
the plane headed 30◦ west of north and the wind blowing 15◦ north of east (that is, from 75◦ west
of south).
66. (a) The ball must increase in height by ∆y = 0.193 m and cover a horizontal distance ∆x = 0.910 m
during a very short time t0 = 1.65 × 10−2 s. The statement that the “initial curvature of the
ball’s path can be neglected” is essentially the same as saying the average velocity for 0 ≤ t ≤ t0
may be taken equal to the instantaneous initial velocity ~v0. Thus, using Eq. 4-8 to figure its two
components, we have
tan θ0 =
v0y
v0x
=
∆y
t0
∆x
t0
=
∆y
∆x
so that θ0 = tan
−1(0.193/0.910) = 12◦.
(b) The magnitude of ~v0 is
√
v20x + v
2
0y =
√(
∆x
t0
)2
+
(
∆y
t0
)2
=
√
∆x2 +∆y2
t0
which yields v0 = 56.4 m/s.
(c) The range is given by Eq. 4-26:
R =
v20
g
sin 2θ0 = 132 m .
(d) Partly because of its dimpled surface (but other air-flow related effects are important here) the golf
ball travels farther than one would expect based on the simple projectile-motion analysis done here.
67. (a) Since the performer returns to the original level, Eq. 4-26 applies. With R = 4.0 m and θ0 = 30
◦,
the initial speed (for the projectile motion) is consequently
v0 =
√
gR
sin 2θ0
= 6.7 m/s .
This is, of course, the final speed v for the Air Ramp’s acceleration process (for which the initial
speed is taken to be zero) Then, for that process, Eq. 2-11 leads to
a =
v
t
=
6.7
0.25
= 27 m/s2 .
We express this as a multiple of g by setting up a ratio: a = (27/9.8)g = 2.7g.
93
(b) Repeating the above steps for R = 12 m, t = 0.29 s and θ0 = 45
◦ gives a = 3.8g.
68. The initial position vector ~ro satisfies ~r − ~ro = ∆~r, which results in
~ro = ~r −∆~r = (3 jˆ− 4 kˆ)− (2 ıˆ− 3 jˆ + 6 kˆ) = −2.0 ıˆ + 6.0 jˆ− 10 kˆ
where the understood unit is meters.
69. We adopt a coordinate system with ıˆ pointed east and jˆ pointed north; the coordinate origin is the
flagpole. With SI units understood, we “translate” the given information into unit-vector notation as
follows:
~ro = 40 ıˆ and ~vo = −10 jˆ
~r = 40 jˆ and ~v = 10 ıˆ .
(a) Using Eq. 4-2, the displacement ∆~r is
~r − ~ro = (56.6 6 135◦)
where we have expressed the result in magnitude-angle notation. The displacement has magnitude
40
√
2 = 56.6 m and points due Northwest.
(b) Eq. 4-8 shows that ~vavg points in the same direction as ∆~r, and that its magnitude is simply the
magnitude of the displacement divided by the time (∆t = 30 s). Thus, the average velocity has
magnitude 56.6/30 = 1.89m/s and points due Northwest.
(c) Using Eq. 4-15, we have
~aavg =
~v − ~vo
∆t
= 0.333 ıˆ + 0.333 jˆ
in SI units. The magnitude of the average acceleration vector is therefore 0.333
√
2 = 0.471m/s2,
and it points due Northeast.
70. The velocity of Larry is v1 and that of Curly is v2. Also, we denote the length of the corridor by L. Now,
Larry’s time of passage is t1 = 150 s (which must equal L/v1 ), and Curly’s time of passage is t2 = 70 s
(which must equal L/v2 ). The time Moe takes is therefore
t =
L
v1 + v2
=
1
v1/L+ v2/L
=
1
1
150 +
1
70
= 48 s .
71. We choose horizontal x and vertical y axes such that both components of ~v0 are positive. Positive angles
are counterclockwise from +x and negative angles are clockwise from it. In unit-vector notation, the
velocity at each instant during the projectile motion is
~v = v0 cos θ0 ıˆ + (v0 sin θ0 − gt) jˆ .
(a) With v0 = 30 m/s and θ0 = 60
◦, we obtain ~v = 15 ıˆ + jˆ in m/s, for t = 2.0 s. Converting to
magnitude-angle notation, this is ~v = (16 6 23◦) with the magnitude in m/s.
(b) Now with t = 5.0 s, we find ~v = (27 6 − 57◦).
72. (a) The helicopter’s speed is v′ = 6.2 m/s. From the discussions in §4-9 we see that the speed of the
package is v0 = 12− v′ = 5.8 m/s, relative to the ground.
(b) Letting +x be in the direction of ~v0 for the package and +y be downward, we have (for the motion
of the package)
∆x = v0t and ∆y =
1
2
gt2
where ∆y = 9.5 m. From these, we find t = 1.39 s and ∆x = 8.08 m for the package, while ∆x′ (for
the helicopter, which is moving in the opposite direction) is −v′ t = −8.63 m. Thus, the horizontal
separation between them is 8.08− (−8.63) = 16.7 m.
94 CHAPTER 4.
(c) The components of ~v at the moment of impact are (vx, vy) = (5.8, 13.6) in SI units. The vertical
component has been computed using Eq. 2-11. The angle (which is below horizontal) for this vector
is tan−1(13.6/5.8) = 67◦.
73. (a) By symmetry, y = H occurs at x = R/2 (taking the coordinate origin to be at the launch point).
Substituting this into Eq. 4-25 gives
H =
R
2
tan θ0 − gR
2/4
2v20 cos
2 θ0
which leads immediately to
H
R
=
1
2
(
tan θ0 − gR
4v20 cos
2 θ0
)
.
In the far right term, we substitute from Eq. 4-26 for the range:
H
R
=
1
2
(
tan θ0 −
g
(
v20 sin(2θ0)/g
)
4v20 cos
2 θ0
)
which, upon setting sin 2θ0 = 2 sin θ0 cos θ0 and simplifying that last term, yields
H
R
=
1
2
(
tan θ0 − sin θ0
2 cos θ0
)
which clearly leads to the relation we wish to prove.
(b) Setting H/R = 1 in that relation, we have θ0 = tan
−1(4) = 76◦.
74. (a) The tangent of the angle φ is found from the ratio of y to x coordinates of the highest point (taking
the coordinate origin to be at the launch point). Using the same notation as in problem 73, we
have
φ = tan−1
(
H
1
2R
)
tan−1
(
2
H
R
)
.
Substituting H/R = 14 tan θ0 from problem 73, we obtain the relation
tan−1
(
1
2
tan θ0
)
.
(b) Since tan 45◦ = 1, then φ = tan−1
(
1
2
)
= 27◦.
75. The initial velocity has magnitude v0 and because it is horizontal, it is equal to vx the horizontal
component of velocity at impact. Thus, the speed at impact is√
v20 + v
2
y = 3v0 where vy =
√
2gh
where we use Eq. 2-16 with ∆x replaced with the h = 20 m to obtain that second equality. Squaring
both sides of the first equality and substituting from the second, we find
v20 + 2gh = (3v0)
2
which leads to gh = 4v20 and therefore to v0 =
√
(9.8)(20)/2 = 7.0 m/s.
76. (a) The magnitude of the displacement vector ∆~r is given by
|∆~r| =
√
21.52 + 9.72 + 2.882 = 23.8 km .
Thus,
|~vavg| = |∆~r|
∆t
=
23.8
3.50
= 6.79 km/h .
95
(b) The angle θ in question is given by
θ = tan−1
(
2.88√
21.52 + 9.72
)
= 6.96◦ .
77. With no acceleration in the x direction yet a constant acceleration of 1.4 m/s2 in the y direction, the
position (in meters) as a function of time (in seconds) must be
~r = (6.0t)ˆı +
(
1
2
(1.4)t2
)
jˆ
and ~v is its derivative with respect to t.
(a) At t = 3.0 s, therefore, ~v = 6.0ˆı + 4.2ˆj m/s.
(b) At t = 3.0 s, the position is ~r = 18 ıˆ + 6.3 jˆ m.
78. We choose a coordinate system with origin at the clock center and +x rightward (towards the “3:00”
position) and +y upward (towards “12:00”).
(a) In unit-vector notation, we have (in centimeters) ~r1 = 10 ıˆ and ~r2 = −10 jˆ. Thus, Eq. 4-2 gives
∆~r = ~r2 − ~r1 = −10 ıˆ− 10 jˆ −→
(14 6 − 135◦)
where we have switched to magnitude-angle notation in the last step.
(b) In this case, ~r1 = −10 jˆ and ~r2 = 10 jˆ, and ∆~r = 20 jˆ cm.
(c) In a full-hour sweep, the hand returns to its starting position, and the displacement is zero.
79. We let gp denote the magnitude of the gravitational acceleration on the planet. A number of the points on
the graph (including some “inferred” points – such as the max height point at x = 12.5 m and t = 1.25 s)
can be analyzed profitably; for future reference, we label (with subscripts) the first ((x0, y0) = (0, 2) at
t0 = 0) and last (“final”) points ((xf , yf ) = (25, 2) at tf = 2.5), with lengths in meters and time in
seconds.
(a) The x-component of the initial velocity is found from xf − x0 = v0xtf . Therefore, v0x = 25/2.5 =
10 m/s. And we try to obtain the y-component from yf − y0 = 0 = v0ytf − 12gpt2f . This gives us
v0y = 1.25gp , and we see we need another equation (by analyzing another point, say, the next-to-
last one) y−y0 = v0yt− 12gpt2 with y = 6 and t = 2; this produces our second equation v0y = 2+gp
. Simultaneous solution of these two equations produces results for v0y and gp (relevant to part
(b)). Thus, our complete answer for the initial velocity is ~v = 10ˆı + 10ˆj m/s.
(b) As a by-product of the part (a) computations, we have gp = 8.0m/s
2.
(c) Solving for tg (the time to reach the ground) in yg = 0 = y0 + v0ytg − 12gpt2g leads to a positive
answer: tg = 2.7 s.
(d) With g = 9.8 m/s2, the method employed in part (c) would produce the quadratic equation−4.9t2g+
10tg + 2 = 0 and then the positive result tg = 2.2 s.
80. At maximum height, the y-component of a projectile’s velocity vanishes, so the given 10 m/s is the
(constant) x-component of velocity.
(a) Using v0y to denote the y-velocity 1.0 s before reaching the maximum height, then (with vy = 0)
the equation vy = v0y − gt leads to v0y = 9.8 m/s. The magnitude of the velocity vector at that
moment (also known as the speed) is therefore√
v2x + v0y
2 =
√
102 + 9.82 = 14 m/s .
96 CHAPTER 4.
(b) It is clear from the symmetry of the problem that the speed is the same 1.0 s after reaching the
top, as it was 1.0 s before (14 m/s again). This may be verified by using vy = v0y − gt again
but now “starting the clock” at the highest point so that v0y = 0 (and t = 1.0 s). This leads to
vy = −9.8 m/s and ultimately to
√
102 + (−18)2 = 14 m/s.
(c) With v0y denoting the y-component of velocity one second before the top of the trajectory – as
in part (a) – then we have y = 0 = y0 + v0yt − 12gt2 where t = 1.0 s. This yields y0 = −4.9 m.
Alternatively, Eq. 2-18 could have been used, with vy = 0 to the same end. The x0 value more
simply results from x = 0 = x0+(10m/s)(1.0 s). Thus, the coordinates (in meters) of the projectile
one second before reaching maximum height is (−10,−4.9).
(d) It is clear from symmetry that the coordinate one second after the maximum height is reached
is (10,−4.9) (in meters). But this can be verified by considering t = 0 at the top and using
y − y0 = v0yt− 12gt2 where y0 = v0y = 0 and t = 1 s. And by using x− x0 = (10m/s)(1.0 s) where
x0 = 0. Thus, x = 10 m and y = −4.9 m is obtained.
81. With gB = 9.8128 m/s
2 and gM = 9.7999 m/s
2, we apply Eq. 4-26:
RM −RB = v
2
0 sin 2θ0
gM
− v
2
0 sin 2θ0
gB
=
v20 sin 2θ0
gB
(
gB
gM
− 1
)
which becomes
RM −RB = RB
(
9.8128
9.7999
− 1
)
and yields (upon substituting RB = 8.09 m) RM −RB = 0.01 m.
82. (a) Using the same coordinate system assumed in Eq. 4-25, we rearrange that equation to solve for the
initial speed:
v0 =
x
cos θ0
√
g
2 (x tan θ0 − y)
which yields v0 = 255.5 ≈ 2.6× 102 m/s for x = 9400 m, y = −3300 m, and θ0 = 35◦.
(b) From Eq. 4-21, we obtain the time of flight:
t =
x
v0 cos θ0
=
9400
255.5 cos35◦
= 45 s .
(c) We expect the air to provide resistance but no appreciable lift to the rock, so we would need a
greater launching speed to reach the same target.
83. (a) Using the same coordinate system assumed in Eq. 4-22, we solve for y = h:
h = y0 + v0 sin θ0 − 1
2
gt2
which yields h = 51.8 m for y0 = 0, v0 = 42 m/s, θ0 = 60
◦ and t = 5.5 s.
(b) The horizontal motion is steady, so vx = v0x = v0 cos θ0 , but the vertical component of velocity
varies according to Eq. 4-23. Thus, the speed at impact is
v =
√
(v0 cos θ0 )
2 + (v0 sin θ0 − gt)2 = 27 m/s .
(c) We use Eq. 4-24 with vy = 0 and y = H :
H =
(v0 sin θ0 )
2
2g
= 67.5 m .
97
84. (a) Using the same coordinate system assumed in Eq. 4-25, we find
y = x tan θ0 − gx
2
2 (v0 cos θ0)
2 = −
gx2
2v20
if θ0 = 0 .
Thus, with v0 = 3 × 106 m/s and x = 1 m, we obtain y = −5.4 × 10−13 m which is not practical
to measure (and suggests why gravitational processes play such a small role in the fields of atomic
and subatomic physics).
(b) It is clear from the above expression that |y| decreases as v0 is reduced.
85. (a) Using the same coordinate system assumed in Eq. 4-21, we obtain the time of flight
t =
∆x
v0 cos θ0
=
20
15 cos 35◦
= 1.63 s .
(b) At that moment, its height of above the ground (taking y0 = 0) is
y = (v0 sin θ0) t− 1
2
gt2 = 1.02 m .
Thus, the ball is 18 cm below the center of the circle; since the circle radius is 15 cm, we see that
it misses it altogether.
(c) The horizontal component of velocity (at t = 1.63 s) is the same as initially:
vx = v0x = v0 cos θ0 = 15 cos 35
◦ = 12.3 m/s .
The vertical component is given by Eq. 4-23:
vy = v0 sin θ0 − gt = 15 sin 35◦ − (9.8)(1.63) = −7.3 m/s .
Thus, the magnitude of its speed at impact is
√
v2x + v
2
y = 14.3 m/s.
(d) As we saw in the previous part, the sign of vy is negative, implying that it is now heading down
(after reaching its max height).
86. (a) From Eq. 4-22 (with θ0 = 0), the time of flight is
t =
√
2h
g
=
√
2(45)
9.8
= 3.03 s .
(b) The horizontal distance traveled is given by Eq. 4-21:
∆x = v0t = (250)(3.03) = 758 m .
(c) And from Eq. 4-23, we find
|vy| = gt = (9.80)(3.03) = 29.7 m/s .
87. Using the same coordinate system assumed in Eq. 4-25, we find x for the elevated cannon from
y = x tan θ0 − gx
2
2 (v0 cos θ0)
2 where y = −30 m.
Using the quadratic formula (choosing the positive root), we find
x = v0 cos θ0

v0 sin θ0 +
√
(v0 sin θ0)
2 − 2gy
g


which yields x = 715 m for v0 = 82 m/s (from Sample Problem 4-7) and θ0 = 45
◦. This is 29 m longer
than the 686 m found in that Sample Problem. The “9” in 29 m is not reliable, considering the low level
of precision in the given data.
98 CHAPTER 4.
88. (a) With r = 0.15 m and a = 3.0× 10614 m/s2, Eq. 4-32 gives
v =
√
ra = 6.7× 106 m/s .
(b) The period is given by Eq. 4-33:
T =
2πr
v
= 1.4× 10−7 s .
89. The type of acceleration involved in steady-speed circular motion is the centripetal acceleration a = v2/r
which is at each moment directed towards the center of the circle. The radius of the circle is r = 122/3 =
48 m. Thus, the car is at the present moment 48 m west of the center of its circular path; this is equally
true in part (a) and part (b).
90. (a) With v = c/10 = 3× 107 m/s and a = 20g = 196 m/s2, Eq. 4-32 gives
r =
v2
a
= 4.6× 1012 m .
(b) The period is given by Eq. 4-33:
T =
2πr
v
= 9.6× 105 s .
Thus, the time to make a quarter-turn is T/4 = 2.4× 105 s or about 2.8 days.
91. (a) Using the same coordinate system assumed in Eq. 4-21 and Eq. 4-22 (so that θ0 = −20.0◦), we use
v0 = 15.0 m/s and find the horizontal displacement of the ball at t = 2.30 s:
∆x = (v0 cos θ0) t = 32.4 m .
(b) And we find the vertical displacement:
∆y = (v0 sin θ0) t− 1
2
gt2 = −37.7 m .
92. This is a classic problem involving two-dimensional relative motion; see §4-9. The steps in Sample
Problem 4-11 in the textbook are similar to
those used here. We align our coordinates so that east
corresponds to +x and north corresponds to +y. We write the vector addition equation as ~vBG =
~vBW+~vWG. We have ~vWG = (2.0 6 0◦) in the magnitude-angle notation (with the unit m/s understood),
or ~vWG = 2.0 ıˆ in unit-vector notation. We also have ~vBW = (8.0 6 120◦) where we have been careful
to phrase the angle in the ‘standard’ way (measured counterclockwise from the +x axis), or ~vBW =
−4.0 ıˆ + 6.9 jˆ .
(a) We can solve the vector addition equation for ~vBG:
~vBG = ~vBW + ~vWG = (2.0 6 0◦) + (8.0 6 120◦) = (7.2 6 106◦)
which is very efficiently done using a vector capable calculator in polar mode. Thus |~vBG| = 7.2 m/s,
and its direction is 16◦ west of north, or 74◦ north of west.
(b) The velocity is constant, and we apply y − y0 = vyt in a reference frame. Thus, in the ground
reference frame, we have 200 = 7.2 sin(106◦)t → t = 29 s. Note: if a student obtains “28 s”, then
the student has probably neglected to take the y component properly (a common mistake).
93. The topic of relative motion (with constant velocity motion) in a two-dimensional setting is covered in
§4-9. We note that
~vPG = ~vPA + ~vAG
99
describes a right triangle, with one leg being ~vPG (east), another leg being ~vAG (magnitude = 20,
direction = south), and the hypotenuse being ~vPA (magnitude = 70). Lengths are in kilometers and
time is in hours. Using the Pythagorean theorem, we have
|~vPA| =
√
|~vPG|2 + |~vAG|2 =⇒ 70 =
√
|~vPG|2 + 202
which is easily solved for the ground speed: |~vPG| = 67 km/h.
94. Our coordinate system has ıˆ pointed east and jˆ pointed north. All distances are in kilometers, times in
hours, and speeds in km/h. The first displacement is ~rAB = 483 ıˆ and the second is ~rBC = −966 jˆ.
(a) The net displacement is
~rAC = ~rAB + ~rBC = 483 ıˆ− 966 jˆ −→ (1080 6 − 63.4◦)
where we have expressed the result in magnitude-angle notation in the last step. We observe that
the angle can be alternatively expressed as 63.4◦ south of east, or 26.6◦ east of south.
(b) Dividing the magnitude of ~rAC by the total time (2.25 h) gives the magnitude of ~vavg and its
direction is the same as in part (a). Thus, ~vavg = (480 6 −63.4◦) in magnitude-angle notation (with
km/h understood).
(c) Assuming the AB trip was a straight one, and similarly for the BC trip, then |~rAB| is the distance
traveled during the AB trip, and |~rBC | is the distance traveled during the BC trip. Since the
average speed is the total distance divided by the total time, it equals
483 + 966
2.25
= 644 km/h .
95. We take the initial (x, y) specification to be (0.000, 0.762) m, and the positive x direction to be towards
the “green monster.” The components of the initial velocity are (33.53 6 55◦) → (19.23, 27.47) m/s.
(a) With t = 5.00 s, we have x = x0 + vxt = 96.2 m.
(b) At that time, y = y0 + v0yt− 12gt2 = 15.59 m, which is 4.31 m above the wall.
(c) The moment in question is specified by t = 4.50 s. At that time, x − x0 = (19.23)(4.5) = 86.5 m,
and y = y0 + v0yt− 12gt2 = 25.1 m.
96. The displacement of the one-way trip is the same as the displacement, which has magnitudeD = 4350 km
for the flight (we are in a frame of reference that rotates with the earth). The velocity of the flight relative
to the earth is
~vfe = ~va+ ~ae
where ~ae is the velocity of the (eastward) jet stream (with magnitude v > 0), and ~ae is the velocity of
the plane relative to the air (with magnitude u = 966 m/s). And the magnitudes of the eastward flight
velocity (relative to earth) and of the westward flight velocity (primed) are, respectively,
|~vfe| = D
t
and
∣∣~v ′fe∣∣ = Dt′ .
The time difference (5/6 of an hour) is therefore
t′ − t = D∣∣∣~v ′fe∣∣∣ −
D
|~vfe|
∆t =
D
u− v −
D
u+ v
.
Using the quadratic formula to solve for v, we obtain
v =
−D +√D2 + u2(∆t)2
∆t
= 89 km/h .
100 CHAPTER 4.
97. Using the same coordinate system assumed in Eq. 4-25, we rearrange that equation to solve for the initial
speed:
v0 =
x
cos θ0
√
g
2 (x tan θ0 − y)
which yields v0 = 23 ft/s for g = 32 ft/s
2, x = 13 ft, y = 3 ft and θ0 = 55
◦.
98. We establish coordinates with ıˆ pointing to the far side of the river (perpendicular to the current) and
jˆ pointing in the direction of the current. We are told that the magnitude (presumed constant) of the
velocity of the boat relative to the water is |~vbw | = u = 6.4 km/h. Its angle, relative to the x axis is θ.
With km and h as the understood units, the velocity of the water (relative to the ground) is ~vwg = 3.2 jˆ .
(a) To reach a point “directly opposite” means that the velocity of her boat relative to ground must
be ~bg = v ıˆ where v > 0 is unknown. Thus, all jˆ components must cancel in the vector sum
~vbw + ~vwg = ~vbg
which means the u sin θ = −3.2, so θ = sin−1(−3.2/6.4) = −30◦.
(b) Using the result from part (a), we find v = u cos θ = 5.5 km/h. Thus, traveling a distance of
ℓ = 6.4 km requires a time of 6.4/5.5 = 1.15 h or 69 min.
(c) If her motion is completely along the y axis (as the problem implies) then with vw = 3.2 km/h (the
water speed) we have
ttotal =
D
u+ vw
+
D
u− vw = 1.33 h
where D = 3.2 km. This is equivalent to 80 min.
(d) Since
D
u+ vw
+
D
u− vw =
D
u− vw +
D
u+ vw
the answer is the same as in the previous part.
(e) The case of general θ leads to
~vbg = ~vbw + ~vwg = u cos θ ıˆ + (u sin θ + vw) jˆ
where the x component of ~vbg must equal ell/t. Thus,
t =
ℓ
u cos θ
which can be minimized using dt/dθ = 0 (though, of course, an easier way is to appeal to either
physical or mathematical intuition – concluding that the shortest-time path should have θ = 0).
Then t = 6.4/6.4 = 1.0 h, or 60 min.
99. With v0 = 30 m/s and R = 20 m, Eq. 4-26 gives
sin 2θ0 =
gR
v20
= 0.218 .
Because sin (φ) = sin (180◦ − φ), there are two roots of the above equation:
2θ0 = sin
−1(0.218) = 12.6◦ and 167.4◦ .
Therefore, the two possible launch angles that will hit the target (in the absence of air friction and
related effects) are θ0 = 6.3
◦ and θ0 = 83.7◦. An alternative approach to this problem in terms of
Eq. 4-25 (with y = 0 and 1/ cos2 = 1 + tan2) is possible – and leads to a quadratic equation for tan θ0
with the roots providing these two possible θ0 values.
101
100. (a) The time available before the train arrives at the impact spot is
ttrain =
40m
30m/s
= 1.33 s
(the train does not reduce its speed). We interpret the phrase “distance between the car and the
center of the crossing” to refer to the distance from the front bumper of the car to that point. In
which case, the car needs to travel a total distance of ∆x = 40 + 5 + 1.5 = 46.5 m in order for its
rear bumper and the edge of the train not to collide (the distance from the center of the train to
either edge of the train is 1.5 m). With a starting velocity of v0 = 30 m/s and an acceleration of
a = 1.5 m/s2, Eq. 2-15 leads to
∆x = v0t+
1
2
at2 =⇒ t = −v0 ±
√
v20 + 2a∆x
a
which yields (upon taking the positive root) a time tcar = 1.49 s needed for the car to make it.
Recalling our result for ttrain we see the car doesn’t have enough time available to make it across.
(b) The difference is tcar − ttrain = 0.16 s. We note that at t = ttrain the front bumper of the car is
v0t+
1
2at
2 = 41.33 m from where it started, which means it is 1.33 m past the center of the track
(but the edge of the track is 1.5 m from the center). If the car was coming from the south, then the
point P on the car impacted by the southern-most corner of the front of the train is 2.83 m behind
the front bumper (or 2.17 m in front of the rear bumper). The motion of P is what is plotted below
(the top graph – looking like a
line instead of a parabola be-
cause the final speed of the car
is not much different than its
initial
speed). Since the posi-
tion of the train is on an en-
tirely different axis than that
of the car, we plot the dis-
tance (in meters) from P to
“south” rail of the tracks (the
top curve shown), and the dis-
tance of the “south” front cor-
ner of the train to the line-of-
motion of the car (the bottom
line shown).
0
10
20
30
40
0.2 0.4 0.6 0.8 1 1.2t
101. (a) With v0 = 6.3 m/s and R = 0.40 m, Eq. 4-26 gives
sin 2θ0 =
gR
v20
= 0.0988 .
Because sin (φ) = sin (180◦ − φ), there are two roots of the above equation:
2θ0 = sin
−1(0.0988) = 5.7◦ and 174.3◦ .
Therefore, the two possible launch angles that will hit the target (in the absence of air friction and
related effects) are θ0 = 2.8
◦ and θ0 = 87.1◦. But the juggler is trying to achieve a visual effect by
having a relatively high trajectory for the balls, so θ0 = 87.1
◦ is the result he should choose.
(b) We do not show the graph here. It would be very much like the higher parabola shown in Fig. 4-51.
(c) , (d) and (e) The problem requests that the student work with his graphs, here, but we – for
doublechecking purposes – use Eq. 4-26 to calculate R− 0.40 m for θ0 − 87.1◦ = −2◦,−1◦, 1◦, and
2◦. We obtain the respective values (in meters) 0.28, 0.14,−0.14, and −0.28.
102 CHAPTER 4.
102. (First problem in Cluster 1)
Using the coordinate system employed in §4-5 and §4-6, we have v0x = vx > 0 and v0y = 0. Also,
y0 = h > 0, x0 = 0, y = 0 (when it hits the ground at t = 3.00)), and x = 150, with lengths in meters
and time in seconds.
(a) The equation y − y0 = v0yt− 12gt2 becomes −h = − 12 (9.8)(3.00)2, so that h = 44.1 m.
(b) The equation vy = v0y − gt gives the y-component of the “final” velocity as vy = −(9.8)(3.00) =
29.4 m/s. The x-component of velocity (which is constant) is computed from vx = (x − x0)/t =
150/3.00 = 50.0 m/s. Therefore,
|~v| =
√
v2x + v
2
y =
√
502 + 29.42 = 58.0 m/s .
103. (Second problem in Cluster 1)
Using the coordinate system employed in §4-5 and §4-6, we have v0x = v0 cos 30◦ > 0 and v0y =
v0 sin 30
◦ > 0. Also, y0 = 0 (corresponding to the dashed line in the figure), x0 = 0, y = h > 0 (where
it lands at t = 3.00), and x = 100, with lengths in meters and time in seconds.
(a) The x-equation determines v0
x− x0 = v0 cos (30)t =⇒ 100 = v0(0.866)(3.00)
which leads to v0 = 38.5 m/s. The y-equation y−y0 = v0yt− 12gt2 becomes h = (38.5)(sin 30)(3.00)−
1
2 (9.8)(3.00)
2 = 13.6 m.
(b) As a byproduct of part (a)’s computation, we found v0 = 38.5 m/s.
(c) Although a somewhat easier method will be found in the energy chapter (especially Chapter 8),
we will find the “final” velocity components with the methods of §4-6. We have vx = v0x =
38.5 cos 30 = 33.3 m/s. And vy = v0y − gt = 38.5 sin30− (9.8)(3.00) = −10.2 m/s. Therefore,
|~v| =
√
v2x + v
2
y =
√
(33.3)2 + (−10.2)2 = 34.8 m/s .
104. (Third problem in Cluster 1)
Following the hint, we have the time-reversed problem with the ball thrown from the roof, towards the
left, at 60◦ measured clockwise from a leftward axis. We see in this time-reversed situation that it is
convenient to take +x as leftward with positive angles measured clockwise. Lengths are in meters and
time is in seconds.
(a) With y0 = 20.0, and y = 0 at t = 4.00, we have y − y0 = v0yt − 12gt2 where v0y = v0 sin 60◦.
This leads to v0 = 16.9 m/s. This plugs into the x-equation (with x0 = 0 and x = d) to produce
d = (16.9 cos 60◦)(4.00) = 33.7 m.
(b) Although a somewhat easier method will be found in the energy chapter (especially Chapter 8), we
will find the “final” velocity components with the methods of §4-6. Note that we’re still working
the time-reversed problem; this “final” ~v is actually the velocity with which it was thrown. We have
vx = v0x = 16.9 cos 60
◦ = 8.43 m/s. And vy = v0y − gt = 16.9 sin 60◦ − (9.8)(4.00) = −24.6 m/s.
We convert from rectangular components to polar (that is, magnitude-angle) representation:
~v = (8.43,−24.6) −→ (26.0 6 − 71.1◦) .
and we now interpret our result (“undoing” the time reversal) as an initial velocity of magnitude
26 m/s with angle (up from rightward) of 71◦.
105. (Fourth problem in Cluster 1)
Following the hint, we have the time-reversed problem with the ball thrown from the ground, towards
the right, at 60◦ measured counterclockwise from a rightward axis. We see in this time-reversed situation
that it is convenient to use the familiar coordinate system with +x as rightward and with positive angles
measured counterclockwise. Lengths are in meters and time is in seconds.
103
(a) The x-equation (with x0 = 0 and x = 25.0) leads to 25 = (v0 cos 60
◦)(1.50), so that v0 = 33.3 m/s.
And with y0 = 0, and y = h > 0 at t = 1.50, we have y − y0 = v0yt− 12gt2 where v0y = v0 sin 60◦.
This leads to h = 32.3 m.
(b) Although a somewhat easier method will be found in the energy chapter (especially Chapter 8), we
will find the “final” velocity components with the methods of §4-6. Note that we’re still working
the time-reversed problem; this “final” ~v is actually the velocity with which it was thrown. We have
vx = v0x = 33.3 cos 60
◦ = 16.7 m/s. And vy = v0y − gt = 33.3 sin 60◦− (9.8)(1.50) = 14.2 m/s. We
convert from rectangular to polar in terms of the magnitude-angle notation:
~v = (16.7, 14.2) −→ (21.9 6 40.4◦) .
We now interpret this result (“undoing” the time reversal) as an initial velocity (from the edge of
the building) of magnitude 22 m/s with angle (down from leftward) of 40◦.
106. (Fifth problem in Cluster 1)
Let y0 = 1.0 m at x0 = 0 when the ball is hit. Let y1 = h (the height of the wall) and x1 describe the
point where it first rises above the wall one second after being hit; similarly, y2 = h and x2 describe the
point where it passes back down behind the wall four seconds later. And yf = 1.0 m at xf = R is where
it is caught. Lengths are in meters and time is in seconds.
(a) Keeping in mind that vx is constant, we have x2 − x1 = 50.0 = v1x(4.00), which leads to v1x =
12.5 m/s. Thus, applied to the full six seconds of motion: xf − x0 = R = vx(6.00) = 75.0 m.
(b) We apply y − y0 = v0yt− 12gt2 to the motion above the wall.
y2 − y1 = 0 = v1y(4.00)−
1
2
g(4.00)2
leads to v1y = 19.6 m/s. One second earlier, using v1y = v0y − g(1.00), we find v0y = 29.4 m/s.
We convert from (x, y) to magnitude-angle (polar) representation:
~v0 = (16.7, 14.2) −→ (31.9 6 66.9◦) .
We interpret this result as a velocity of magnitude 32 m/s, with angle (up from rightward) of 67◦.
(c) During the first 1.00 s of motion, y = y0+v0yt− 12gt2 yields h = 1.0+(29.4)(1.00)− 12 (9.8)(1.00)2 =
25.5 m.
107. (First problem in Cluster 2)
(a) Since v2y = v0
2
y − 2g∆y, and vy = 0 at the target, we obtain v0y =
√
2(9.8)(5.00) = 9.90 m/s. Since
v0 sin θ0 = v0y, with v0 = 12 m/s, we find θ0 = 55.6
◦.
(b) Now, vy = v0y − gt gives t = 9.90/9.8 = 1.01 s. Thus, ∆x = (v0 cos θ0)t = 6.85 m.
(c) The velocity at the target has only the vx component, which is equal to v0x = v0 cos θ0 = 6.78 m/s.
108. (Second problem in Cluster 2)
(a) The magnitudes of the components are equal at point A, but in terms of the coordinate system
usually employed in projectile motion problems, we have vx > 0 and vy = −vx. The problem gives
v0 which is related to its components by v
2
0 = v0
2
x + v0
2
y which suggests that we look at the pair of
equations
v2y = v0
2
y − 2g∆y
v2x = v0
2
x
104 CHAPTER 4.
which we can add to obtain 2v2x = v
2
0−2g∆y (this is closely related to the type of reasoning that will
be employed in some Chapter 8 problems). Therefore, we find vx = −vy = 6.53 m/s. Therefore,
∆y = vyt+
1
2gt
2 (Eq. 2-16) can be used to find t.
3.00 = (−6.53)t+ 1
2
(9.8)t2 =⇒ t = 1.69 or − 0.36
from the quadratic formula or or with a polynomial solver available with some calculators. We
choose the positive root: t = 1.69 s. Finally,
we obtain
∆x = vxt = 11.1 m .
(b) The speed is v =
√
v2x + v
2
y = 9.23 m/s.
109. (Third problem in Cluster 2)
(a) Eq. 4-25, which assumes (x0, y0) = (0, 0), gives
y = 5.00 = (tan θ0)x− gx
2
2 (v0 cos θ0)
2
where x = 30.0 (lengths are in meters and time is in seconds). Using the trig identity suggested in
the problem and letting u stand for tan θ0, we have a second-degree equation for u (its two roots
leading to the values θ0min, and θ0max) parameterized by the initial speed v0.
4410
v20
u2 − 30.0u+
(
4410
v20
+ 5.00
)
= 0
where numerical simplifications have already been made. To see these steps written with the
variables x, y, v0 and g made explicit, see the solution to problem 111, below. Now, we solve for u
using the quadratic formula, and then find the angles:
θ0 = tan
−1
(
1
294
v20 ±
1
294
√
v40 − 98 v20 − 86436
)
where the plus is chosen for θ0max and the negative is chosen for θ0min .
(b) These angles are plotted (in degrees) versus v0 (in m/s) as follows. There are no (real) solutions of
the above equations for 18.0 ≤ v0 ≤ 18.6 m/s (this is further discussed in the next problem).
105
30
40
50
60
70
19 20 21 22 23 24 25
v
110. (Fourth problem in Cluster 2)
Following the hint in the problem (regarding analytic solution), we equate the square root expression,
above, to zero: √
v40 − 98 v20 − 86436 = 0 =⇒ v0 = 18.6 m/s .
That solution can be obtained either with the quadratic formula (by writing the equation, first, in terms
of w = v20) or with a polynomial solver built into many calculators; in the latter approach, this is
straightforwardly handled as a fourth degree polynomial. Note that the other root (v0 = 15.8 m/s) is
dismissed since we are finding where the real solutions for angle disappear as one decreases the initial
speed from roughly 20 m/s. In case this problem was assigned without assigning Problem 109 first, then
this (the choice of root) might be a confusing point. Plugging v0 = 18.6 m/s into
θ0 = tan
−1
(
1
294
v20 ±
1
294
√
v40 − 98 v20 − 86436
)
(which is unambiguous since the square root factor is zero) provides the launch angle: θ0 = 49.7
◦ in this
“critical” case.
111. (Fifth problem in Cluster 2)
(a) This builds directly on the solutions of the previous two problems. If we return to the solution of
problem 109 without plugging in the data for x, y, and g, we obtain the following expression for the
θ0 roots.
θ0 = tan
−1
(
v20
gx
(
1±
√
1− g
v20
(
2y +
gx2
v20
)))
And for the “critical case” of maximum distance for a given launch-speed, we set the square root
expression to zero (as in the previous problem) and solve for xmax.
xmax =
v20
g
√
1− 2gy
v20
106 CHAPTER 4.
which one might wish to check for the “straight-up” case (where x = 0, and the familiar result
ymax =
1
2v
2
0/g is obtained) and for the “range” case (where y = 0 and this then agrees with Eq. 4-
26 where θ0 = 45
◦). In the problem at hand, we have y = 5.00 m, and v0 = 15.0 m/s. This leads
to xmax = 17.2 m.
(b) When the square root term vanishes, the expression for θ0 becomes
θ0 = tan
−1
(
v20
gx
)
= 53.1◦
using x = xmax from part (a).
Chapter 5
1. We apply Newton’s second law (specifically, Eq. 5-2).
(a) We find the x component of the force is
Fx = max = ma cos 20
◦ = (1.00 kg)(2.00m/s2) cos 20◦ = 1.88 N .
(b) The y component of the force is
Fy = may = ma sin 20
◦ = (1.0 kg)(2.00m/s2) sin 20◦ = 0.684 N .
(c) In unit-vector notation, the force vector (in Newtons) is
~F = Fx ıˆ + Fy jˆ = 1.88 ıˆ + 0.684 jˆ .
2. We apply Newton’s second law (Eq. 5-1 or, equivalently, Eq. 5-2). The net force applied on the chopping
block is ~Fnet = ~F1 + ~F2 , where the vector addition is done using unit-vector notation. The acceleration
of the block is given by ~a =
(
~F1 + ~F2
)
/m.
(a) In the first case
~F1 + ~F2 =
(
(3.0N)ˆı + (4.0N)ˆj
)
+
(
(−3.0N)ˆı + (−4.0N)ˆj
)
= 0
so ~a = 0.
(b) In the second case, the acceleration ~a equals
~F1 + ~F2
m
=
(
(3.0N) ıˆ + (4.0N)ˆj
)
+
(
(−3.0N) ıˆ + (4.0N)ˆj
)
2.0 kg
= 4.0 jˆ m/s
2
.
(c) In this final situation, ~a is
~F1 + ~F2
m
=
(
(3.0N) ıˆ + (4.0N)ˆj
)
+
(
(3.0N)ˆı + (−4.0N)ˆj
)
2.0 kg
= 3.0 ıˆ m/s2 .
3. We are only concerned with horizontal forces in this problem (gravity plays no direct role). We take East
as the +x direction and North as +y. This calculation is efficiently implemented on a vector capable
calculator, using magnitude-angle notation (with SI units understood).
~a =
~F
m
=
(9.0 6 0◦) + (8.0 6 118◦)
3.0
= (2.9 6 53◦)
Therefore, the acceleration has a magnitude of 2.9 m/s2.
107
108 CHAPTER 5.
4. Since ~v =constant, we have ~a = 0, which implies
~Fnet = ~F1 + ~F2 = m~a = 0 .
Thus, the other force must be
~F2 = −~F1 = −2ˆı + 6 jˆ N .
5. Since the velocity of the particle does not change, it undergoes no acceleration and must therefore be
subject to zero net force. Therefore,
~Fnet = ~F1 + ~F2 + ~F3 = 0 .
Thus, the third force ~F3 is given by
~F3 = −~F1 − ~F2
= −
(
2 ıˆ + 3 jˆ − 2 kˆ
)
−
(
−5 ıˆ + 8 jˆ − 2 kˆ
)
= 3 ıˆ − 11 jˆ + 4 kˆ
in Newtons. The specific value of the velocity is not used in the computation.
6. The net force applied on the chopping block is ~Fnet = ~F1 + ~F2 + ~F3 , where the vector addition is done
using unit-vector notation. The acceleration of the block is given by ~a =
(
~F1 + ~F2 + ~F3
)
/m.
(a) The forces exerted by the three astronauts can be expressed in unit-vector notation as follows:
~F1 = 32(cos 30
◦ ıˆ + sin 30◦ jˆ)
= 27.7 ıˆ + 16 jˆ
~F2 = 55(cos 0
◦ ıˆ + sin 0◦ jˆ)
= 55 ıˆ
in Newtons, and
~F3 = 41
(
cos(−60◦)ˆı + sin(−60◦)ˆj
)
= 20.5 ıˆ− 35.5 jˆ
in Newtons. The resultant acceleration of the asteroid of mass m = 120 kg is therefore
~a =
(27.7 ıˆ + 16 jˆ) + (55 ıˆ) + (20.5 ıˆ− 35.5 jˆ)
120
= 0.86 ıˆ− 0.16 jˆ m/s2 .
(b) The magnitude of the acceleration vector is
|~a| =
√
a2x + a
2
y =
√
0.862 + (−0.16)2 = 0.88 m/s2 .
(c) The vector ~a makes an angle θ with the +x axis, where
θ = tan−1
(
ay
ax
)
= tan−1
(−0.16
0.86
)
= −11◦ .
7. We denote the two forces ~F1 and ~F2. According to Newton’s second law, ~F1+ ~F2 = m~a, so ~F2 = m~a− ~F1.
(a) In unit vector notation ~F1 = (20.0N)ˆı and
~a = −(12 sin30◦ m/s2)ˆı− (12 cos 30◦ m/s2)ˆj = −(6.0m/s2)ˆı− (10.4m/s2)ˆj .
Therefore,
~F2 = (2.0 kg)
(
−6.0m/s2
)
ıˆ + (2.0 kg)
(
−10.4m/s2
)
jˆ− (20.0N)ˆı
= (−32N)ˆı− (21N)ˆj .
109
(b) The magnitude of ~F2 is ∣∣∣~F2∣∣∣ =√F 22x + F 22y =√(−32)2 + (−21)2 = 38 N .
(c) The angle that ~F2 makes with the positive x axis is found from tan θ = F2y/F2x = 21/32 = 0.656 .
Consequently, the angle is either 33◦ or 33◦+180◦ = 213◦. Since both the x and y components are
negative, the correct result is 213◦.
8. The goal is to arrive at the least magnitude of ~Fnet , and as long as the magnitudes of ~F2 and ~F3 are (in
total) less than or equal to |~F1| then we should orient them opposite to the direction of ~F1 (which is the
+x direction).
(a) We orient both ~F2 and ~F3 in the −x direction. Then, the magnitude of the net force is 50−30−20 =
0, resulting in zero acceleration for the tire.
(b) We again orient ~F2 and ~F3 in the negative x direction. We obtain an acceleration along the +x
axis with magnitude
a =
F1 − F2 − F3
m
=
50N− 30N− 10N
12 kg
= 0.83 m/s2 .
(c) In this case, the forces ~F2 and ~F3 are collectively strong enough to have y components (one positive
and one negative) which cancel each other and still have enough x contributions (in the−x direction)
to cancel ~F1 . Since |~F2| = |~F3|, we see that the angle above the −x axis to one of them should
equal the
angle below the −x axis to the other one (we denote this angle θ). We require
−50N = ~F2x + ~F3x
= −(30N) cos θ − (30N) cos θ
which leads to
θ = cos−1
(
50N
60N
)
= 34◦ .
9. In all three cases the scale is not accelerating, which means that the two cords exert forces of equal
magnitude on it. The scale reads the magnitude of either of these forces. In each case the tension force
of the cord attached to the salami must be the same in magnitude as the weight of the salami because
the salami is not accelerating. Thus the scale reading is mg, where m is the mass of the salami. Its
value is (11.0 kg)(9.8m/s
2
) = 108N.
10. Three vertical forces are acting on the block: the earth pulls down on the block with gravitational force
3.0 N; a spring pulls up on the block with elastic force 1.0 N; and, the surface pushes up on the block
with normal force N . There is no acceleration, so∑
Fy = 0 = N + (1.0 N) + (−3.0 N)
yields N = 2.0 N. By Newton’s third law, the force exerted by the block on the surface has that same
magnitude but opposite direction: 2.0 N down.
11. We apply Eq. 5-12.
(a) The mass is m = W/g = (22N)/(9.8m/s
2
) = 2.2 kg. At a place where g = 4.9m/s
2
, the mass is
still 2.2 kg but the gravitational force is Fg = mg = (2.2 kg)(4.9m/s
2
) = 11N.
(b) As noted, m = 2.2 kg.
(c) At a place where g = 0 the gravitational force is zero.
(d) The mass is still 2.2 kg.
110 CHAPTER 5.
12. We use Wp = mgp, where Wp is the weight of an object of mass m on the surface of a certain planet p,
and gp is the acceleration of gravity on that planet.
(a) The weight of the space ranger on Earth isWe = mge which we compute to be (75 kg)
(
9.8m/s
2
)
=
7.4× 102N.
(b) The weight of the space ranger on Mars isWm = mgm which we compute to be (75 kg)(3.8m/s
2
) =
2.9× 102N.
(c) The weight of the space ranger in interplanetary space is zero, where the effects of gravity are
negligible.
(d) The mass of the space ranger remains the same (75 kg) at all the locations.
13. According to Newton’s second law, the magnitude of the force is given by F = ma, where a is the
magnitude of the acceleration of the neutron. We use kinematics (Table 2-1) to find the acceleration
that brings the neutron to rest in a distance d. Assuming the acceleration is constant, then v2 = v20+2ad
produces the value of a:
a =
(v2 − v20)
2d
=
−(1.4× 107m/s)2
2(1.0× 10−14m) = −9.8× 10
27m/s
2
.
The magnitude of the force is consequently
F = m|a| = (1.67× 10−27 kg)(9.8 × 1027m/s2) = 16 N .
14. The child-backpack is in static equilibrium while he waits, so Newton’s second law applies with
∑
~F = 0.
Since students sometimes confuse this with Newton’s third law, we phrase our results carefully.
(a) The magnitude of the normal force ~N exerted upward by the sidewalk is equal, in this situation, to
the total weight of the child-backpack, as a result of
∑ ~F = 0. Thus, ~N = (33.5 kg)(9.8 m/s2) =
328 N and is directed up; this is ~Fsc – the force of the sidewalk exerted up on the child’s feet. By
Newton’s third law, the force exerted down (at the child’s feet) on the sidewalk is ~Fcs = 328 N
downward.
(b) Except for an entirely negligible gravitation attraction between the child and the concrete, there is
no force exerted on the sidewalk by the child when the child is not in contact with it.
(c) Earth pulls gravitationally on the child, and the child pulls equally in the opposite direction on
Earth. This force is the previously computed weight (29.0)(9.8) = 284 N. The gravitational force
on Earth exerted by the child is 284 N up. But the contact force exerted by the child on the
sidewalk (hence, on Earth) is (see part (a)) 328 N downward. Thus, the net force exerted by the
child on Earth is zero.
(d) Here the answer is simply the gravitational interaction: 284 N up.
15. We note that the free-body diagram is shown in Fig. 5-18 of the text.
(a) Since the acceleration of the block is zero, the components of the Newton’s second law equation
yield T −mg sin θ = 0 and N−mg cos θ = 0. Solving the first equation for the tension in the string,
we find
T = mg sin θ = (8.5 kg)(9.8m/s
2
) sin 30◦ = 42 N .
(b) We solve the second equation in part (a) for the normal force N :
N = mg cos θ = (8.5 kg)(9.8m/s
2
) cos 30◦ = 72 N .
111
(c) When the string is cut, it no longer exerts a force on the block and the block accelerates. The x
component of the second law becomes −mg sin θ = ma, so the acceleration becomes
a = −g sin θ = −9.8 sin 30◦ = −4.9
in SI units. The negative sign indicates the acceleration is down the plane. The magnitude of the
acceleration is 4.9 m/s2.
16. An excellent analysis of the accelerating elevator is given in Sample Problem 5-8 in the textbook.
(a) From Newton’s second law
N −mg = ma where a = amax = 2.0 m/s2
we obtain N = 590 N upward, for m = 50 kg.
(b) Again, we use Newton’s second law
N −mg = ma where a = amax = −3.0 m/s2 .
Now, we obtain N = 340 N upward.
(c) Returning to part (a), we use Newton’s third law, and conclude that the force exerted by the
passenger on the floor is ~FPF = 590 N downward.
17. (a) The acceleration is
a =
F
m
=
20N
900 kg
= 0.022 m/s2 .
(b) The distance traveled in 1 day (= 86400 s) is
s =
1
2
at2 =
1
2
(
0.0222m/s
2
)
(86400 s)2 = 8.3× 107 m .
(c) The speed it will be traveling is given by
v = at = (0.0222m/s
2
)(86400 s) = 1.9× 103 m/s .
18. Some assumptions (not so much for realism but rather in the interest of using the given information
efficiently) are needed in this calculation: we assume the fishing line and the path of the salmon are
horizontal. Thus, the weight of the fish contributes only (via Eq. 5-12) to information about its mass
(m =W/g = 8.7 kg). Our +x axis is in the direction of the salmon’s velocity (away from the fisherman),
so that its acceleration (“deceleration”) is negative-valued and the force of tension is in the −x direction:
~T = −T . We use Eq. 2-16 and SI units (noting that v = 0).
v2 = v20 + 2a∆x =⇒ a = −
v20
2∆x
= − 2.8
2
2(0.11)
which yields a = −36 m/s2. Assuming there are no significant horizontal forces other than the tension,
Eq. 5-1 leads to
~T = m~a =⇒ −T = (8.7 kg)
(
−36m/s2
)
which results in T = 3.1× 102 N.
19. In terms of magnitudes, Newton’s second law is F = ma, where F represents |~Fnet|, a represents |~a|
(which it does not always do; note the use of a in the previous solution), and m is the (always positive)
mass. The magnitude of the acceleration can be found using constant acceleration kinematics (Table 2-
1). Solving v = v0 + at for the case where it starts from rest, we have a = v/t (which we interpret
in terms of magnitudes, making specification of coordinate directions unnecessary). The velocity is
v = (1600 km/h)(1000m/km)/(3600 s/h) = 444m/s, so
F = (500 kg)
444m/s
1.8 s
= 1.2× 105N .
112 CHAPTER 5.
20. The stopping force ~F and the path of the car are horizontal. Thus, the weight of the car contributes
only (via Eq. 5-12) to information about its mass (m =W/g = 1327 kg). Our +x axis is in the direction
of the car’s velocity, so that its acceleration (“deceleration”) is negative-valued and the stopping force
is in the −x direction: ~F = −F .
(a) We use Eq. 2-16 and SI units (noting that v = 0 and v0 = 40(1000/3600) = 11.1 m/s).
v2 = v20 + 2a∆x =⇒ a = −
v20
2∆x
= −11.1
2
2(15)
which yields a = −4.12 m/s2. Assuming there are no significant horizontal forces other than the
stopping force, Eq. 5-1 leads to
~F = m~a =⇒ −F = (1327 kg)
(
−4.12m/s2
)
which results in F = 5.5× 103 N.
(b) Eq. 2-11 readily yields t = −v0/a = 2.7 s.
(c) Keeping F the same means keeping a the same, in which case (since v = 0) Eq. 2-16 expresses a
direct proportionality between ∆x and v20 . Therefore, doubling v0 means quadrupling
∆x. That is,
the new over the old stopping distances is a factor of 4.0.
(d) Eq. 2-11 illustrates a direct proportionality between t and v0 so that doubling one means doubling
the other. That is, the new time of stopping is a factor of 2.0 greater than the one found in part (c).
21. The acceleration of the electron is vertical and for all practical purposes the only force acting on it is
the electric force. The force of gravity is negligible. We take the +x axis to be in the direction of the
initial velocity and the +y axis to be in the direction of the electrical force, and place the origin at the
initial position of the electron. Since the force and acceleration are constant, we use the equations from
Table 2-1: x = v0t and
y =
1
2
at2 =
1
2
(
F
m
)
t2 .
The time taken by the electron to travel a distance x (= 30mm) horizontally is t = x/v0 and its deflection
in the direction of the force is
y =
1
2
F
m
(
x
v0
)2
=
1
2
(
4.5× 10−16
9.11× 10−31
)(
30× 10−3
1.2× 107
)2
= 1.5× 10−3 m .
22. The stopping force ~F and the path of the passenger are horizontal. Our +x axis is in the direction of
the passenger’s motion, so that the passenger’s acceleration (“deceleration”) is negative-valued and the
stopping force is in the −x direction: ~F = −F . We use Eq. 2-16 and SI units (noting that v = 0 and
v0 = 53(1000/3600) = 14.7 m/s).
v2 = v20 + 2a∆x =⇒ a = −
v20
2∆x
= − 14.7
2
2(0.65)
which yields a = −167 m/s2. Assuming there are no significant horizontal forces other than the stopping
force, Eq. 5-1 leads to
~F = m~a =⇒ −F = (41 kg)
(
−167m/s2
)
which results in F = 6.8× 103 N.
23. We note that The rope is 22◦ from vertical – and therefore 68◦ from horizontal.
113
(a) With T = 760 N, then its components are
~T = T cos 68◦ ıˆ + T sin 68◦ jˆ = 285 ıˆ + 705 jˆ
understood to be in newtons.
(b) No longer in contact with the cliff, the only other force on Tarzan is due to earth’s gravity (his
weight). Thus,
~Fnet = ~T + ~W = 285 ıˆ + 705 jˆ− 820 jˆ = 285 ıˆ− 115 jˆ
again understood to be in newtons.
(c) In a manner that is efficiently implemented on a vector capable calculator, we convert from rect-
angular (x, y) components to magnitude-angle notation:
~Fnet = (285,−115) −→ (307 6 − 22◦)
so that the net force has a magnitude of 307 N.
(d) The angle (see part (c)) has been found to be 22◦ below horizontal (away from cliff)
(e) Since ~a = ~Fnet /m where m =W/g = 84 kg, we obtain ~a = 3.67 m/s
2
(f) Eq. 5-1 requires that ~a ‖ ~Fnet so that it is also directed at 22◦ below horizontal (away from cliff).
24. The analysis of coordinates and forces (the free-body diagram) is exactly as in the textbook in Sample
Problem 5-7 (see Fig. 5-18(b) and (c)).
(a) Constant velocity implies zero acceleration, so the “uphill” force must equal (in magnitude) the
“downhill” force: T = mg sin θ. Thus, with m = 50 kg and θ = 8.0◦, the tension in the rope equals
68 N.
(b) With an uphill acceleration of 0.10 m/s
2
, Newton’s second law (applied to the x axis shown in
Fig. 5-18(b)) yields
T −mg sin θ = ma =⇒ T − (50)(9.8) sin 8.0◦ = (50)(0.10)
which leads to T = 73 N.
25. (a) Since friction is negligible the force of the girl is the only horizontal force on the sled. The vertical
forces (the force of gravity and the normal force of the ice) sum to zero. The acceleration of the
sled is
as =
F
ms
=
5.2N
8.4 kg
= 0.62 m/s
2
.
(b) According to Newton’s third law, the force of the sled on the girl is also 5.2N. Her acceleration is
ag =
F
mg
=
5.2N
40 kg
= 0.13 m/s
2
.
(c) The accelerations of the sled and girl are in opposite directions. Assuming the girl starts at the
origin and moves in the +x direction, her coordinate is given by xg =
1
2agt
2. The sled starts at
x0 = 1.5m and moves in the −x direction. Its coordinate is given by xs = x0 − 12ast2. They meet
when
xg = xs
1
2
agt
2 = x0 − 1
2
ast
2 .
This occurs at time
t =
√
2x0
ag + as
.
114 CHAPTER 5.
By then, the girl has gone the distance
xg =
1
2
agt
2 =
x0ag
ag + as
=
(15)(0.13)
0.13 + 0.62
= 2.6 m .
26. We assume the direction of motion is +x and assume the refrigerator starts from rest (so that the speed
being discussed is the velocity v which results from the process). The only force along the x axis is the
x component of the applied force ~F .
(a) Since v0 = 0, the combination of Eq. 2-11 and Eq. 5-2 leads simply to
Fx = m
(v
t
)
=⇒ vi =
(
F cos θi
m
)
t
for i = 1 or 2 (where we denote θ1 = 0 and θ2 = θ for the two cases). Hence, we see that the ratio
v2 over v1 is equal to cos θ.
(b) Since v0 = 0, the combination of Eq. 2-16 and Eq. 5-2 leads to
Fx = m
(
v2
2∆x
)
=⇒ vi =
√
2
(
F cos θi
m
)
∆x
for i = 1 or 2 (again, θ1 = 0 and θ2 = θ is used for the two cases). In this scenario, we see that the
ratio v2 over v1 is equal to
√
cos θ.
27. We choose up as the +y direction, so ~a = −3.00 m/s2 jˆ (which, without the unit-vector, we denote
as a since this is a 1-dimensional problem in which Table 2-1 applies). From Eq. 5-12, we obtain the
firefighter’s mass: m =W/g = 72.7 kg.
(a) We denote the force exerted by the pole on the firefighter ~F f p = F jˆ and apply Eq. 5-1 (using SI
units).
~Fnet = m~a
F − Fg = ma
F − 712 = (72.7)(−3.00)
which yields F = 494 N. The fact that the result is positive means ~F f p points up.
(b) Newton’s third law indicates ~F f p = −~Fp f , which leads to the conclusion that ~Fp f = 494 N down.
28. The coordinate choices are made in the problem statement.
(a) We write the velocity of the armadillo as ~v = vx ıˆ + vy jˆ . Since there is no net force exerted on it
in the x direction, the x component of the velocity of the armadillo is a constant: vx = 5.0m/s. In
the y direction at t = 3.0 s, we have (using Eq. 2-11 with v0 y = 0)
vy = v0 y + ayt = v0 y +
(
Fy
m
)
t =
(
17
12
)
(3.0) = 4.3
in SI units. Thus
~v = 5.0 ıˆ + 4.3 jˆ m/s .
(b) We write the position vector of the armadillo as ~r = rx ıˆ + ry jˆ . At t = 3.0 s we have rx =
(5.0)(3.0) = 15 and (using Eq. 2-15 with v0 y = 0)
ry = v0 y t+
1
2
ay t
2 =
1
2
(
Fy
m
)
t2 =
1
2
(
17
12
)
(3.0)2 = 6.4
in SI units. The position vector at t = 3.0 s is therefore
~r = 15 ıˆ + 6.4 jˆ m .
115
29. The solutions to parts (a) and (b) have been combined here. The free-body diagram is shown below,
with the tension of the string ~T , the force of gravity m~g, and the force of the air ~F . Our coordinate
system is shown. The x component of the net force is T sin θ − F and the y component is T cos θ −mg,
where θ = 37◦.
Since the sphere is motionless the
net force on it is zero. We answer
the questions in the reverse order.
Solving T cos θ − mg = 0 for the
tension, we obtain T = mg/ cos θ =
(3.0 × 10−4)(9.8)/ cos 37◦ = 3.7 ×
10−3N. Solving T sin θ−F = 0 for
the force of the air: F = T sin θ =
(3.7×10−3) sin 37◦ = 2.2×10−3N.
............................................................................................................................
...
...
...
...
..
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..
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..
..
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.
~T
m~g
~F
θ
• +x
+y
30. We label the 40 kg skier “m” which is represented as a block in the
figure shown. The force of the
wind is denoted ~Fw and might be
either “uphill” or “downhill” (it is
shown uphill in our sketch). The
incline angle θ is 10◦. The +x di-
rection is downhill.
HH
HH
HH
HH
HH
HH
HH
HH
HH
�
�
�
�
�
�
HH
H
HHHHHjmg sin θ
�
�
�
��� mg cos θ�
�
�
���
~N
HHY ~Fw
(a) Constant velocity implies zero acceleration; thus, application of Newton’s second law along the x
axis leads to
mg sin θ − Fw = 0 .
This yields Fw = 68 N (uphill).
(b) Given our coordinate choice, we have a = +1.0 m/s
2
. Newton’s second law
mg sin θ − Fw = ma
now leads to Fw = 28 N (uphill).
(c) Continuing with the forces as shown in our figure, the equation
mg sin θ − Fw = ma
will lead to Fw = −12 N when a = +2.0 m/s2. This simply tells us that the wind is opposite to
the direction shown in our sketch; in other words, ~Fw = 12 N downhill.
31. The free-body diagrams for part (a) are shown below. ~F is the applied force and ~f is the force exerted
by block 1 on block 2. We note that ~F is applied directly to block 1 and that block 2 exerts the force
−~f on block 1 (taking Newton’s third law into account).
•.................
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.............................................................................................................................................................................
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..
..
..
..
...............................................................................................
...
...
...
...
..
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..
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..
..
............................
~F~f
m1~g
~N1
•.................
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.......................................................................................................................
....
...
...
...
...
..
..
..
..
..
..
~f
m2~g
~N2
116 CHAPTER 5.
(a) Newton’s second law for block 1 is F − f = m1a, where a is the acceleration. The second law for
block 2 is f = m2a. Since the blocks move together they have the same acceleration and the same
symbol is used in both equations. From the second equation we obtain the expression a = f/m2,
which we substitute into the first equation to get F − f = m1f/m2. Therefore,
f =
Fm2
m1 +m2
=
(3.2N)(1.2 kg)
2.3 kg + 1.2 kg
= 1.1N .
(b) If ~F is applied to block 2 instead of block 1 (and in the opposite direction), the force of contact
between the blocks is
f =
Fm1
m1 +m2
=
(3.2N)(2.3 kg)
2.3 kg + 1.2 kg
= 2.1N .
(c) We note that the acceleration of the blocks is the same in the two cases. In part (a), the force f
is the only horizontal force on the block of mass m2 and in part (b) f is the only horizontal force
on the block with m1 > m2 . Since f = m2a in part (a) and f = m1a in part (b), then for the
accelerations to be the same, f must be larger in part (b).
32. The additional “apparent weight” experienced during upward acceleration is well treated in Sample
Problem 5-8. The discussion in the textbook surrounding Eq. 5-13 is also relevant to this.
(a) When ~Fnet = 3F −mg = 0, we have
F =
1
3
mg =
1
3
(1400 kg)
(
9.8m/s2
)
= 4.6× 103 N
for the force exerted by each bolt on the engine.
(b) The force on each bolt now satisfies 3F −mg = ma, which yields
F =
1
3
m(g + a) =
1
3
(1400)(9.8 + 2.6) = 5.8× 103 N .
33. The free-body diagram is shown below. ~T is the tension of the cable and m~g is the force of gravity. If
the upward direction is positive, then Newton’s second law is T −mg = ma, where a is the acceleration.
Thus, the tension is T = m(g + a). We use
con-
stant acceleration kinematics (Table 2-1) to find
the acceleration (where v = 0 is the final ve-
locity, v0 = −12m/s is the initial velocity, and
y = −42m is the coordinate at the stopping
point). Consequently, v2 = v20 + 2ay leads to
a = −v20/2y = −(−12)2/2(−42) = 1.71m/s2. We
now return to calculate the tension:
T = m(g + a)
= (1600 kg)(9.8m/s
2
+ 1.71m/s
2
)
= 1.8× 104 N .
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~T
m~g
•
34. First, we consider all the penguins (1 through 4, counting left to right) as one system, to which we apply
Newton’s second law:
Fnet = (m1 +m2 +m3 +m4)a
222N = (20 kg + 15 kg +m3 + 12 kg)a .
117
Second, we consider penguins 3 and 4 as one system, for which we have
F ′net = (m3 +m4)a
111N = (m3 + 12 kg)a .
We solve these two equations for m3 to obtain m3 = 23 kg. The solution step can be made a little easier,
though, by noting that the net force on penguins 1 and 2 is also 111 N and applying Newton’s law to
them as a single system to solve first for a.
35. We take the down to be the +y direction.
(a) The first diagram (below) is the free-body diagram for the person and parachute, considered as a
single object with a mass of 80 kg + 5 kg = 85 kg. ~Fa is the force of the air on the parachute and
m~g is the force of gravity. Application of Newton’s second law produces mg−Fa = ma, where a is
the acceleration. Solving for Fa we find
Fa = m(g − a) = (85 kg)(9.8m/s2 − 2.5m/s2) = 620 N .
•.................
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~Fa
m~g
•.................
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~Fa
mp~g
~Fp
(b) The second diagram (above) is the free-body diagram for the parachute alone. ~Fa is the force of
the air, mp~g is the force of gravity, and ~Fp is the force of the person. Now, Newton’s second law
leads to mpg + Fp − Fa = mpa. Solving for Fp, we obtain
Fp = mp(a− g) + Fa = (5.0)(2.5− 9.8) + 620 = 580 N .
36. We apply Newton’s second law first to the three blocks as a single system and then to the individual
blocks. The +x direction is to the right in Fig. 5-37.
(a) With msys = m1 +m2 +m3 = 67.0 kg, we apply Eq. 5-2 to the x motion of the system – in which
case, there is only one force ~T3 = +T3 ıˆ .
T3 = msys a
65.0N = (67.0 kg)a
which yields a = 0.970 m/s2 for the system (and for each of the blocks individually).
(b) Applying Eq. 5-2 to block 1, we find
T1 = m1a = (12.0 kg)
(
0.970 m/s
2
)
= 11.6 N .
(c) In order to find T2 , we can either analyze the forces on block 3 or we can treat blocks 1 and 2 as a
system and examine its forces. We choose the latter.
T2 = (m1 +m2) a = (12.0 + 24.0)(0.970) = 34.9 N .
37. We use the notation g as the acceleration due to gravity near the surface of Callisto, m as the mass of
the landing craft, a as the acceleration of the landing craft, and F as the rocket thrust. We take down
to be the positive direction. Thus, Newton’s second
law takes the form mg − F = ma. If the thrust is
F1 (= 3260N), then the acceleration is zero, so mg − F1 = 0. If the thrust is F2 (= 2200N), then the
acceleration is a2 (= 0.39m/s
2), so mg − F2 = ma2.
118 CHAPTER 5.
(a) The first equation gives the weight of the landing craft: mg = F1 = 3260N.
(b) The second equation gives the mass:
m =
mg − F2
a2
=
3260N− 2200N
0.39m/s
2 = 2.7× 103 kg .
(c) The weight divided by the mass gives the acceleration due to gravity: g = (3260N)/(2.7×103 kg) =
1.2m/s
2
.
38. Although the full specification of ~Fnet = m~a in this situation involves both x and y axes, only the
x-application is needed to find what this particular problem asks for. We note that ay = 0 so that there
is no ambiguity denoting ax simply as a. We choose +x to the right and +y up, in Fig. 5-38. We also
note that the x component of the rope’s tension (acting on the crate) is Tx = +450 cos38
◦ = 355 N, and
the resistive force (pointing in the −x direction) has magnitude f = 125 N.
(a) Newton’s second law leads to
Tx − f = ma =⇒ a = 355− 125
310
= 0.74 m/s
2
.
(b) In this case, we use Eq. 5-12 to find the mass: m = W/g = 31.6 kg. Now, Newton’s second law
leads to
Tx − f = ma =⇒ a = 355− 125
31.6
= 7.3 m/s
2
.
39. The force diagrams in Fig. 5-18 are helpful to refer to. In adapting Fig. 5-18(b) to this problem, the
normal force ~N and the tension ~T should be labeled Fm,ry and Fm,rx , respectively, and thought of as the
y and x components of the force ~Fm,r exerted by the motorcycle on the rider. We adopt the coordinates
used in Fig. 5-18 and note that they are not the usual horizontal and vertical axes.
(a) Since the net force equalsma, then the magnitude of the net force on the rider is (60.0 kg)(3.0 m/s2) =
1.8× 102 N.
(b) We apply Newton’s second law to the x axis:
Fm,rx −mg sin θ = ma
where m = 60.0 kg, a = 3.0 m/s
2
, and θ = 10◦. Thus, Fm,rx = 282 N. Applying it to the y axis
(where there is no acceleration), we have
Fm,ry −mg cos θ = 0
which produces Fm,ry = 579 N. Using the Pythagorean theorem, we find√
Fm,r
2
x + Fm,r
2
y = 644 N .
Now, the magnitude of the force exerted on the rider by the motorcycle is the same magnitude of
force exerted by the rider on the motorcycle, so the answer is 6.4× 102 N.
40. Referring to Fig. 5-10(c) is helpful. In this case, viewing the man-rope-sandbag as a system means that
we should be careful to choose a consistent positive direction of motion (though there are other ways
to proceed – say, starting with individual application of Newton’s law to each mass). We take down
as positive for the man’s motion and up as positive for the sandbag’s motion and, without ambiguity,
denote their acceleration as a. The net force on the system is the difference between the weight of the
man and that of the sandbag. The system mass is msys = 85 + 65 = 150 kg. Thus, Eq. 5-1 leads to
(85)(9.8)− (65)(9.8) = msys a
119
which yields a = 1.3 m/s2. Since the system starts from rest, Eq. 2-16 determines the speed (after
traveling ∆y = 10 m) as follows:
v =
√
2a∆y =
√
2(1.3)(10) = 5.1 m/s .
41. (a) The links are numbered from bottom to top. The forces on the bottom link are the force of
gravity m~g, downward, and the force ~F2on1 of link 2, upward. Take the positive direction to be
upward. Then Newton’s second law for this link is F2on1 −mg = ma. Thus F2on1 = m(a + g) =
(0.100 kg)(2.50m/s2 + 9.8m/s2) = 1.23N.
(b) The forces on the second link are the force of gravity m~g, downward, the force ~F1on2 of link 1,
downward, and the force ~F3on2 of link 3, upward. According to Newton’s third law ~F1on2 has the
same magnitude as ~F2on1. Newton’s second law for the second link is F3on2 − F1on2 −mg = ma, so
F3on2 = m(a+ g) + F1on2 = (0.100 kg)(2.50m/s
2 + 9.8m/s2) + 1.23N = 2.46N.
(c) Newton’s second for link 3 is F4on3−F2on3−mg = ma, so F4on3 = m(a+g)+F2on3 = (0.100N)(2.50m/s2+
9.8m/s
2
) + 2.46N = 3.69N, where Newton’s third law implies F2on3 = F3on2 (since these are mag-
nitudes of the force vectors).
(d) Newton’s second law for link 4 is F5on4 − F3on4 − mg = ma, so F5on4 = m(a + g) + F3on4 =
(0.100 kg)(2.50m/s2+9.8m/s2)+3.69N = 4.92N, where Newton’s third law implies F3on4 = F4on3.
(e) Newton’s second law for the top link is F − F4on5 − mg = ma, so F = m(a + g) + F4on5 =
(0.100 kg)(2.50m/s2 + 9.8m/s2) + 4.92N = 6.15N, where F4on5 = F5on4 by Newton’s third law.
(f) Each link has the same mass and the same acceleration, so the same net force acts on each of them:
Fnet = ma = (0.100 kg)(2.50m/s
2) = 0.25N.
42. The mass of the jet is m =W/g = 2.36× 104 kg. Its acceleration is found from Eq. 2-16:
v2 = v20 + 2a∆x =⇒ a =
852
2(90)
= 40 m/s
2
.
Thus, Newton’s second law provides the needed force F from the catapult.
F + Fthrust = ma =⇒ F =
(
2.36× 104) (40)− 107× 103
which yields F = 8.4× 105 N.
43. The free-body diagram for each block is shown below. T is the tension in the cord and θ = 30◦ is the
angle of the incline. For block 1, we take the +x direction to be up the incline and the +y direction
to be in the direction of the normal force ~N that the plane exerts on the block. For block 2, we take
the +y direction to be down. In this way, the accelerations of the two blocks can be represented by the
same symbol a, without ambiguity. Applying Newton’s second law to the x and y axes for block 1 and
to the y axis of block 2, we obtain
T −m1g sin θ = m1a
N −m1g cos θ = 0
m2g − T = m2a
respectively. The first and third of these equations provide a simultaneous set for obtaining values of a
and T . The second equation is not needed in this problem, since the normal force is neither asked for
nor is it needed as part of some further computation (such as can occur in formulas for friction).
120 CHAPTER 5.
•..................................
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~T
(+x)
m1~g
~N
θ
•.................
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~T
m2~g
(+y)
(a) We add the first and third equations above: m2g −m1g sin θ = m1a+m2a. Consequently, we find
a =
(m2 −m1 sin θ)g
m1 +m2
=
(2.30 kg)− 3.70 sin30.0◦) (9.8)
3.70 + 2.30
= 0.735 m/s
2
.
(b) The result for a is positive, indicating that the acceleration of block 1 is indeed up the incline and
that the acceleration of block 2 is vertically down.
(c) The tension in the cord is
T = m1a+m1g sin θ = (3.70)(0.735) + (3.70)(9.8) sin 30
◦ = 20.8 N .
44. For convenience, we have labeled the 2.0 kg mass m and the 3.0 kg mass M . The +x direction for
m is “downhill” and the +x direction for M is rightward; thus, they accelerate with the same sign.
@
@
@
@
@
@
@@
��
��
@
@
m@@I
~T
���
~Nm @@R mg sin 30◦��	
mg cos 30◦
6
~NM
-~T-~F
?
M~g
M
30◦
(a) We apply Newton’s second law to each block’s x axis:
mg sin 30◦ − T = ma
F + T = Ma
Adding the two equations allows us to solve for the acceleration. With F = 2.3 N, we have
a = 1.8 m/s
2
. We plug back in to find the tension T = 3.1 N.
(b) We consider the “critical” case where the F has reached themax value, causing the tension to vanish.
The first of the equations in part (a) shows that a = g sin 30◦ in this case; thus, a = 4.9 m/s2. This
implies (along with T = 0 in the second equation in part (a)) that F = (3.0)(4.9) = 14.7 N in the
critical case.
45. The free-body diagram is shown below. ~N is the normal force of the
121
plane on the block and m~g is the
force of gravity on the block. We
take the +x direction to be down
the incline, in the direction of the
acceleration, and the +y direc-
tion to be in the direction of the
normal force exerted by the in-
cline on the block. The x com-
ponent of Newton’s second law is
then mg sin θ = ma; thus, the ac-
celeration is a = g sin θ.
•......
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.~N
(+x)
m~g
θ
(a) Placing the origin at the bottom of the plane, the kinematic equations (Table 2-1) for motion along
the x axis which we will use are v2 = v20 +2ax and v = v0+ at. The block momentarily stops at its
highest point, where v = 0; according to the second equation, this occurs at time t = −v0/a. The
position where it stops is
x = −1
2
v20
a
= −1
2
(
(−3.50m/s)2
(9.8m/s2) sin 32.0◦
)
= −1.18 m .
(b) The time is
t = −v0
a
= − v0
g sin θ
= − −3.50m/s
(9.8m/s
2
) sin 32.0◦
= 0.674 s .
(c) That the return-speed is identical to the initial speed is to be expected since there are no dissipative
forces in this problem. In order to prove this, one approach is to set x = 0 and solve x = v0t+
1
2at
2
for the total time (up and back down) t. The result is
t = −2v0
a
= − 2v0
g sin θ
= − 2(−3.50m/s)
(9.8m/s2) sin 32.0◦
= 1.35 s .
The velocity when it returns is therefore
v = v0 + at = v0 + gt sin θ = −3.50 + (9.8)(1.35) sin 32◦ = 3.50 m/s .
46. We write the length unit light-month as c·month in this solution.
(a) The magnitude of the required acceleration is given by
a =
∆v
∆t
=
(0.10)(3.0× 108m/s)
(3.0 days)(86400 s/day)
= 1.2× 102 m/s2 .
(b) The acceleration in terms of g is
a =
(
a
g
)
g =
(
1.2× 102m/s2
9.8m/s2
)
g = 12g .
(c) The force needed is
F = ma =
(
1.20× 106) (1.2× 102) = 1.4× 108 N .
122 CHAPTER 5.
(d) The spaceship will travel a distance d = 0.1 c·month during one month. The time it takes for the
spaceship to travel at constant speed for 5.0 light-months is
t =
d
v
=
5.0 c ·months
0.1c
= 50 months
which is about 4.2 years.
47. We take +y to be up for both the monkey and the package.
(a) The force the monkey pulls downward
on the rope has magnitude F . According to Newton’s third
law, the rope pulls upward on the monkey with a force of the same magnitude, so Newton’s second
law for forces acting on the monkey leads to F − mmg = mmam, where mm is the mass of the
monkey and am is its acceleration. Since the rope is massless F = T is the tension in the rope.
The rope pulls upward on the package with a force of magnitude F , so Newton’s second law for the
package is F +N −mpg = mpap , where mp is the mass of the package, ap is its acceleration, and
N is the normal force exerted by the ground on it. Now, if F is the minimum force required to lift
the package, then N = 0 and ap = 0. According to the second law equation for the package, this
means F = mpg. Substituting mpg for F in the equation for the monkey, we solve for am:
am =
F −mmg
mm
=
(mp −mm) g
mm
=
(15− 10)(9.8)
10
= 4.9 m/s
2
.
(b) As discussed, Newton’s second law leads to F−mpg = mpap for the package and F−mmg = mmam
for the monkey. If the acceleration of the package is downward, then the acceleration of the monkey
is upward, so am = −ap. Solving the first equation for F
F = mp(g + ap) = mp(g − am)
and substituting this result into the second equation, we solve for am:
am =
(mp −mm) g
mp +mm
=
(15− 10)(9.8)
15 + 10
= 2.0 m/s
2
.
(c) The result is positive, indicating that the acceleration of the monkey is upward.
(d) Solving the second law equation for the package, we obtain
F = mp (g − am) = (15)(9.8− 2.0) = 120 N .
48. The direction of motion (the direction of the barge’s acceleration) is +ıˆ, and +jˆ is chosen so that the
pull ~Fh from the horse is in the first quadrant. The components of the unknown force of the water are
denoted simply Fx and Fy .
(a) Newton’s second law applied to the barge, in the x and y directions, leads to
(7900N) cos18◦ + Fx = ma
(7900N) sin18◦ + Fy = 0
respectively. Plugging in a = 0.12 m/s2 and m = 9500 kg, we obtain Fx = 6.4 × 103 N and
Fy = −2.4× 103 N. The magnitude of the force of the water is therefore
Fwater =
√
F 2x + F
2
y = 6.8× 103 N .
(b) Its angle measured from +ıˆ is either
tan−1
(
Fy
Fx
)
= −21◦ or 159◦.
The signs of the components indicate the former is correct, so ~Fwater is at 21
◦ measured clockwise
from the line of motion.
123
49. The force diagram (not to scale) for the block is shown below. ~N is the normal force exerted by the
floor and m~g is the force of gravity.
(a) The x component of Newton’s second law is F cos θ = ma, where m is the mass of block and a is
the x component of its acceleration. We obtain
a =
F cos θ
m
=
(12.0N) cos 25.0◦
5.00 kg
= 2.18m/s2 .
This is its acceleration provided it remains in contact with the
floor. Assuming it does, we find
the value of N (and if N is
positive, then the assumption is
true but if N is negative then
the block leaves the floor). The
y component of Newton’s sec-
ond law becomes N + F sin θ −
mg = 0, so N = mg − F sin θ =
(5.00)(9.8) − (12.0) sin 25.0◦ =
43.9N. Hence the block remains
on the floor and its acceleration
is a = 2.18m/s2.
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.
~N
m~g
~F
+x
+y
θ
(b) If F is the minimum force for which the block leaves the floor, then N = 0 and the y component of
the acceleration vanishes. The y component of the second law becomes F sin θ −mg = 0, so
F =
mg
sin θ
=
(5.00)(9.8)
sin 25.0◦
= 116 N .
(c) The acceleration is still in the x direction and is still given by the equation developed in part (a):
a =
F cos θ
m
=
116 cos 25◦
5.00
= 21.0 m/s
2
.
50. The motion of the man-and-chair is positive if upward.
(a) When the man is grasping the rope, pulling with a force equal to the tension T in the rope, the
total upward force on the man-and-chair due its two contact points with the rope is 2T . Thus,
Newton’s second law leads to
2T −mg = ma
so that when a = 0, the tension is T = 466 N.
(b) When a = +1.3 m/s2 the equation in part (a) predicts that the tension will be T = 527 N.
(c) When the man is not holding the rope (instead, the co-worker attached to the ground is pulling on
the rope with a force equal to the tension T in it), there is only one contact point between the rope
and the man-and-chair, and Newton’s second law now leads to
T −mg = ma
so that when a = 0, the tension is T = 931 N.
(d) When a = +1.3 m/s2 the equation in part (c) predicts that the tension will be T = 1.05× 103 N.
(e) The rope comes into contact (pulling down in each case) at the left edge and the right edge of the
pulley, producing a total downward force of magnitude 2T on the ceiling. Thus, in part (a) this
gives 2T = 931 N.
(f) In part (b) the downward force on the ceiling has magnitude 2T = 1.05× 103 N.
124 CHAPTER 5.
(g) In part (c) the downward force on the ceiling has magnitude 2T = 1.86× 103 N.
(h) In part (d) the downward force on the ceiling has magnitude 2T = 2.11× 103 N.
51. (a) A small segment of the rope has mass and is pulled down by the gravitational force of the Earth.
Equilibrium is reached because neighboring portions of the rope pull up sufficiently on it. Since
tension is a force along the rope, at least one of the neighboring portions must slope up away from
the segment we are considering. Then, the tension has an upward component which means the rope
sags.
(b) The only force acting with a horizontal component is the applied force ~F . Treating the block
and rope as a single
object, we write Newton’s second law for it: F = (M +m)a, where a is the
acceleration and the positive direction is taken to be to the right. The acceleration is given by
a = F/(M +m).
(c) The force of the rope Fr is the only force with a horizontal component acting on the block. Then
Newton’s second law for the block gives
Fr =Ma =
MF
M +m
where the expression found above for a has been used.
(d) Treating the block and half the rope as a single object, with mass M + 12m, where the horizontal
force on it is the tension Tm at the midpoint of the rope, we use Newton’s second law:
Tm = (M +
1
2
m)a =
(M + 12m)F
(M +m)
=
(2M +m)F
2(M +m)
.
52. The coordinate system we wish to use is shown in Fig. 5-18(c) in the textbook, so we resolve this
horizontal force into appropriate components.
��
��
��
��
��
��
��
��
��
A
A
A A
A
A��
�
-
~FA
AAU
Fy = F sin θ
��
��*
Fx = F cos θ
θ = 30◦
(a) Referring to Fig. 5-18 in the textbook, we see that Newton’s second law applied to the x axis
produces
F cos θ −mg sin θ = ma .
For a = 0, this yields F = 566 N.
(b) Applying Newton’s second law to the y axis (where there is no acceleration), we have
N − F sin θ −mg cos θ = 0
which yields the normal force N = 1.13× 103 N.
53. The forces on the balloon are the force of gravity m~g (down) and the force of the air ~Fa (up). We take
the +y to be up, and use a to mean the magnitude of the acceleration (which is not its usual use in
this chapter). When the mass is M (before the ballast is thrown out) the acceleration is downward
and Newton’s second law is Fa −Mg = −Ma. After the ballast is thrown out, the mass is M − m
(where m is the mass of the ballast) and the acceleration is upward. Newton’s second law leads to
125
Fa − (M −m)g = (M −m)a. The earlier equation gives Fa = M(g − a), and this plugs into the new
equation to give
M(g − a)− (M −m)g = (M −m)a =⇒ m = 2Ma
g + a
.
54. The free-body diagram is shown below. Newton’s second law for the
mass m for the x direction leads
to
T1 − T2 −mg sin θ = ma
which gives the difference in the
tension in the pull cable:
T1 − T2 = m(g sin θ + a)
= (2800) (9.8 sin 35◦ + 0.81)
= 1.8× 104 N .
•......
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.
~N
~T2
~T1
+x
m~g
θ
55. (a) The mass of the elevator is m = 27800/9.8 = 2837 kg and (with +y upward) the acceleration is
a = +1.22 m/s2. Newton’s second law leads to
T −mg = ma =⇒ T = m(g + a)
which yields T = 3.13× 104 N for the tension.
(b) The term “deceleration” means the acceleration vector is in the direction opposite to the velocity
vector (which the problem tells us is upward). Thus (with +y upward) the acceleration is now
a = −1.22 m/s2, so that the tension T = m(g + a) turns out to be T = 2.43× 104 N in this case.
56. (a) The term “deceleration” means the acceleration vector is in the direction opposite to the velocity
vector (which the problem tells us is downward). Thus (with +y upward) the acceleration is
a = +2.4 m/s2. Newton’s second law leads to
T −mg = ma =⇒ m = T
g + a
which yields m = 7.3 kg for the mass.
(b) Repeating the above computation (now to solve for the tension) with a = +2.4 m/s2 will, of course,
leads us right back to T = 89 N. Since the direction of the velocity did not enter our computation,
this is to be expected.
57. The mass of the bundle is m = 449/9.8 = 45.8 kg and we choose +y upward.
(a) Newton’s second law, applied to the bundle, leads to
T −mg = ma =⇒ a = 387− 449
45.8
which yields a = −1.35 m/s2 for the acceleration. The minus sign in the result indicates the
acceleration vector points down. Any downward acceleration of magnitude greater than this is also
acceptable (since that would lead to even smaller values of tension).
(b) We use Eq. 2-16 (with ∆x replaced by ∆y = −6.1 m). We assume v0 = 0.
|v| =
√
2a∆y =
√
2(−1.35)(−6.1) = 4.1 m/s .
For downward accelerations greater than 1.35 m/s2, the speeds at impact will be larger than 4.1 m/s.
126 CHAPTER 5.
58. For convenience, we have labeled the 2.0 kg box m1 and the 3.0 kg box m2 – and their weights w
′ and
w, respectively. The +x axis is “downhill” for m1 and “uphill” for m2 (so they both accelerate with the
same sign).
@
@
@
@
@
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AA
A
AA
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~T
���
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w′y
A
AK ~N2
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~T
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AAUwy
30◦
m2
60◦
We apply Newton’s second law to each box’s x axis:
m1g sin 60
◦ − T = m1a
T −m2g sin 30◦ = m2a
Adding the two equations allows us to solve for the acceleration a = 0.45 m/s2. This value is plugged
back into either of the two equations to yield the tension T = 16 N.
59. (a) There are six legs, and the vertical component of the tension force in each leg is T sin θ where
θ = 40◦. For vertical equilibrium (zero acceleration in the y direction) then Newton’s second law
leads to
6T sin θ = mg =⇒ T = mg
6 sin θ
which (expressed as a multiple of the bug’s weight mg) gives roughly 0.26mg for
the tension.
(b) The angle θ is measured from horizontal, so as the insect “straightens out the legs” θ will increase
(getting closer to 90◦), which causes sin θ to increase (getting closer to 1) and consequently (since
sin θ is in the denominator) causes T to decrease.
60. (a) Choosing the direction of motion as +x, Eq. 2-11 gives
a =
88.5 km/h− 0
6.0 s
= 15 km/h/s .
Converting to SI, this is a = 4.1 m/s2.
(b) With mass m = 2000/9.8 = 204 kg, Newton’s second law gives ~F = m~a = 836 N in the +x
direction.
61. (a) Intuition readily leads to the conclusion (that the heavier block should be the hanging one, for
largest acceleration). The force that “drives” the system into motion is the weight of the hanging
block (gravity acting on the block on the table has no effect on the dynamics, so long as we ignore
friction).
(b) In Sample Problem 5-5 (where it was assumed the m is the hanging block) Eq. 5-21 gave the
acceleration. Now that we have switched m ↔ M (so that now M is the hanging block) our new
version of Eq. 5-21 is
a =
M
m+M
g = 6.5 m/s2 .
(c) Switching m↔M has no effect on Eq. 5-22, which yields
T =
mM
m+M
g = 13 N .
127
62. Making separate free-body diagrams for the helicopter and the truck, one finds there are two forces
on the truck (~T upward, caused by the tension, which we’ll think of as that of a single cable, and m~g
downward, where m = 4500 kg) and three forces on the helicopter (~T downward, ~Flift upward, and M~g
downward, where M = 15000 kg). With +y upward, then a = +1.4 m/s2 for both the helicopter and
the truck.
(a) Newton’s law applied to the helicopter and truck separately gives
Flift − T −Mg = Ma
T −mg = ma
which we add together to obtain
Flift − (M +m)g = (M +m)a .
From this equation, we find Flift = 2.2× 105 N.
(b) From the truck equation T −mg = ma we obtain T = 5.0× 104 N.
63. (a) With SI units understood, the net force is
~Fnet = ~F1 + ~F2 = (3.0 + (−2.0)) ıˆ + (4.0 + (−6.0)) jˆ
which yields ~Fnet = 1.0 ıˆ− 2.0 jˆ in Newtons.
(b) Using magnitude-angle notation (especially convenient on a vector-capable calculator), the answer
to part (a) becomes
~Fnet = (2.2N 6 − 63◦).
(c) Since ~Fnet is equal to ~a multiplied by a positive scalar (which cannot affect the direction of the
vector it multiplies), then the acceleration has the same angle as the net force. The magnitude of
~a comes from dividing the magnitude of ~Fnet by the mass (m = 1.0 kg). Thus, in magnitude-angle
notation, the answer is ~a = (2.2m/s2 6 − 63◦).
64. We take rightwards as the +x direction. Thus, ~F1 = 20 ıˆ in Newtons. In each case, we use Newton’s
second law ~F1 + ~F2 = m~a where m = 2.0 kg.
(a) If ~a = +10 ıˆ in SI units, then the equation above gives ~F2 = 0.
(b) If ~a = +20 ıˆ m/s2, then that equation gives ~F2 = 20 ıˆ N.
(c) If ~a = 0, then the equation gives ~F2 = −20 ıˆ N.
(d) If ~a = −10 ıˆ m/s2, the equation gives ~F2 = −40 ıˆ N.
(e) If ~a = −20 ıˆ m/s2, the equation gives ~F2 = −60 ıˆ N.
65. (a) Since the performer’s weight is (52)(9.8) = 510 N, the rope breaks.
(b) Setting T = 425 N in Newton’s second law (with +y upward) leads to
T −mg = ma =⇒ a = T
m
− g
which yields |a| = 1.6 m/s2.
66. The mass of the pilot is m = 735/9.8 = 75 kg. Denoting the upward force exerted by the spaceship (his
seat, presumably) on the pilot as ~F and choosing upward the +y direction, then Newton’s second law
leads to
F −mgmoon = ma =⇒ F = (75)(1.6 + 1.0) = 195 N .
128 CHAPTER 5.
67. With SI units understood, the net force on the box is
~Fnet = (3.0 + 14 cos 30
◦ − 11) ıˆ + (14 sin 30◦ + 5.0− 17) jˆ
which yields ~Fnet = 4.1 ıˆ− 5.0 jˆ in Newtons.
(a) Newton’s second law applied to the m = 4.0 kg box leads to
~a =
~Fnet
m
= 1.0 ıˆ− 1.3 jˆ m/s2 .
(b) The magnitude of ~a is
√
1.02 + (−1.3)2 = 1.6 m/s2. Its angle is tan−1(−1.3/1.0) = −50◦ (that is,
50◦ measured clockwise from the rightward axis).
68. The net force is in the y direction, so the unknown force must have an x component that cancels the
(8.0N)ˆı value of the known force, and it must also have enough y component to give the 3.0 kg object
an acceleration of (3.0m/s
2
)ˆj . Thus, the magnitude of the unknown force is∣∣∣~F ∣∣∣ =√F 2x + F 2y =√(−8.0)2 + 9.02 = 12 N .
69. We are only concerned with horizontal forces in this problem (gravity plays no direct role). Thus,∑
~F = m~a reduces to ~Favg = m~a, and we see that the magnitude of the force is ma, where m = 0.20 kg
and
a = |~a| =
√
a2x + a
2
y
and the direction of the force is the same as that of ~a. We take east as the +x direction and north as
+y. The acceleration is the average acceleration in the sense of Eq. 4-15.
(a) We find the (average) acceleration to be
~a =
~v − ~v0
∆t
=
(−5.0 ıˆ)− (2.0 ıˆ)
0.50
= −14 ıˆ m/s2 .
Thus, the magnitude of the force is (0.20 kg)(14m/s
2
) = 2.8N and its direction is − ıˆ which means
west in this context.
(b) A computation similar to the one in part (a) yields the (average) acceleration with two components,
which can be expressed various ways:
~a = −4.0 ıˆ− 10.0 jˆ → (−4.0,−10.0) → (10.8 6 − 112◦)
Therefore, the magnitude of the force is (0.20 kg)(10.8 m/s2) = 2.2 N and its direction is 112◦
clockwise from east – which means it is 22◦ west of south, stated more conventionally.
70. The “certain force” is denoted F is assumed to be the net force on the object when it gives m1 an
acceleration a1 = 12 m/s
2 and when it gives m2 an acceleration a2 = 3.3 m/s
2. Thus, we substitute
m1 = F/a1 and m2 = F/a2 in appropriate places during the following manipulations.
(a) Now we seek the acceleration a of an object of mass m2 −m1 when F is the net force on it. Thus,
a =
F
m2 −m1 =
F
(F/a2)− (F/a1) =
a1a2
a1 − a2
which yields a = 4.6 m/s2.
(b) Similarly for an object of mass m2 +m1 :
a =
F
m2 +m1
=
F
(F/a2) + (F/a1)
=
a1a2
a1 + a2
which yields a = 2.6 m/s2.
129
71. We mention that the textbook treats this particular arrangement of blocks and pulleys in extensive detail
in Sample Problem 5-5. Using the usual coordinate system (right = +x and up = +y) for both blocks
has the important consequence that for the 3.0 kg block to have a positive acceleration (a > 0), block
M must have a negative acceleration of the same magnitude (−a). Thus, applying Newton’s second law
to the two blocks, we have
T = (3.0 kg)
(
1.0 m/s
2
)
along x axis
T −Mg = M
(
−1.0 m/s2
)
along y axis .
(a) The first equation yields the tension T = 3.0 N.
(b) The second equation yields the mass M = 3.0/8.8 = 0.34 kg.
72. We take +x uphill for the m = 1.0 kg box and +x rightward for theM = 3.0 kg box (so the accelerations
of the two boxes have the same magnitude and the same sign). The uphill force onm is F and the downhill
forces on it are T and mg sin θ, where θ = 37◦. The only horizontal force on M is the rightward-pointed
tension. Applying Newton’s second law to each box, we find
F − T −mg sin θ = ma
T = Ma
which are added to obtain F −mg sin θ = (m+M)a. This yields the acceleration
a =
12− (1.0)(9.8) sin 37◦
1.0 + 3.0
= 1.53 m/s2 .
Thus, the tension is T =Ma = (3.0)(1.53) = 4.6 N.
73. (a) With v0 = 0, Eq. 2-16 leads to
a =
v2
2∆x
=
(
6.0× 106m/s)2
2(0.015m)
which yields 1.2× 1015 m/s2 for the acceleration. The force responsible for producing this acceler-
ation is
F = ma =
(
9.11× 10−31 kg) (1.2× 1015m/s2) = 1.1× 10−15 N .
(b) The weight is mg = 8.9 × 10−30 N, many orders of magnitude smaller than the result of part (a).
As a result, gravity plays a negligible role in most atomic and subatomic processes.
74. We denote the thrust as T and choose +y upward. Newton’s second law leads to
T −Mg =Ma =⇒ a = 2.6× 10
5
1.3× 104 − 9.8
which yields a = 10 m/s2.
75. (a) The reaction force to ~FMW = 180 N west is, by Newton’s third
law, ~FWM = 180 N east.
(b) Applying ~F = m~a to the woman gives an acceleration a = 180/45 = 4.0m/s2, directed west.
(c) Applying ~F = m~a to the man gives an acceleration a = 180/90 = 2.0m/s
2
, directed east.
76. We note that mg = (15)(9.8) = 147 N.
(a) The penguin’s weight is W = 147 N.
(b) The normal force exerted upward on the penguin by the scale is equal to the gravitational pullW on
the penguin because the penguin is not accelerating. So, by Newton’s second law, N =W = 147 N.
130 CHAPTER 5.
(c) The reading on the scale, by Newton’s third law, is the reaction force to that found in part (b). Its
magnitude is therefore the same: 147 N.
77. Sample Problem 5-8 has a good treatment of the forces in an elevator. We apply Newton’s second law
(with +y up)
N −mg = ma
where m = 100 kg and a must be estimated from the graph (it is the instantaneous slope at the various
moments).
(a) At t = 1.8 s, we estimate the slope to be +1.0m/s2. Thus, Newton’s law yields N ≈ 1100 N (up).
(b) At t = 4.4 s, the slope is zero, so N = 980 N (up).
(c) At t = 6.8 s, we estimate the slope to be -1.7m/s
2
. Thus, Newton’s law yields N = 810 N (up).
78. From the reading when the elevator was at rest, we know the mass of the object is m = 65/9.8 = 6.6 kg.
We choose +y upward and note there are two forces on the object: mg downward and T upward (in the
cord that connects it to the balance; T is the reading on the scale by Newton’s third law).
(a) “Upward at constant speed” means constant velocity, which means no acceleration. Thus, the
situation is just as it was at rest: T = 65 N.
(b) The term “deceleration” is used when the acceleration vector points in the direction opposite to
the velocity vector. We’re told the velocity is upward, so the acceleration vector points downward
(a = −2.4 m/s2). Newton’s second law gives
T −mg = ma =⇒ T = (6.6)(9.8− 2.4) = 49 N .
79. Since (x0, y0) = (0, 0) and ~v0 = 6.0 ıˆ, we have from Eq. 2-15
x = (6.0)t+
1
2
axt
2
y =
1
2
ayt
2 .
These equations express uniform acceleration along each axis; the x axis points east and the y axis
presumably points north (the assumption is that the figure shown in the problem is a view from above).
Lengths are in meters, time is in seconds, and force is in newtons.
Examination of any non-zero (x, y) point will suffice, though it is certainly a good idea to check results by
examining more than one. Here we will look at the t = 4.0 s point, at (8.0, 8.0). The x equation becomes
8.0 = (6.0)(4.0) + 12ax(4.0)
2. Therefore, ax = −2.0 m/s2. The y equation becomes 8.0 = 12ay(4.0)2.
Thus, ay = 1.0 m/s
2
. The force, then, is
~F = m~a = −24 ıˆ + 12 jˆ −→ (27 6 153◦)
where the vector has been expressed in unit-vector and then magnitude-angle notation. Thus, the force
has magnitude 27 N and is directed 63◦ west of north (or, equivalently, 27◦ north of west).
80. We label the 1.0 kg mass m and label the 2.0 kg mass M . We first analyze the forces on m.
The +x direction
HH
HH
HH
HH
HH
HH
HH
HH
HH
�
�
�
�
�
�
HH
H
HHHHHjmg sinβ
�
�
�
��� mg cosβ�
�
�
���
~N
HHj ~T
131
is “downhill”
(parallel to ~T ).
With the acceleration
(5.5m/s
2
) in the positive x direction for m, then Newton’s second law, applied to the x axis, becomes
T +mg sinβ = m(5.5m/s
2
)
But for M , using the more familiar vertical y axis (with up as the positive direction), we have the
acceleration in the negative direction:
F + T −Mg =M(−5.5m/s2)
where the tension comes in as an upward force (the cord can pull, not push).
(a) From the equation for M , with F = 6.0N, we find the tension T = 2.6N.
(b) From the equation for m, using the result from part (a), we obtain the angle β = 17◦.
81. (a) The bottom cord is only supporting a mass of 4.5 kg against gravity, so its tension is (4.5)(9.8) =
44 N.
(b) The top cord is supporting a total mass of 8.0 kg against gravity, so the tension there is (8.0)(9.8) =
78 N.
(c) In the second picture, the lowest cord supports a mass of 5.5 kg against gravity and consequently
has a tension of (5.5)(9.8) = 54 N.
(d) The top cord, we are told, has tension 199 N which supports a total of 199/9.8 = 20.3 kg, 10.3 of
which is accounted for in the figure. Thus, the unknown mass in the middle must be 20.3− 10.3 =
10.0 kg, and the tension in the cord above it must be enough to support 10.0 + 5.5 = 15.5 kg, so
T = (15.5)(9.8) = 152 N. Another way to analyze this is to examine the forces on the 4.8 kg piece;
one of the downward forces on it is this T .
82. The mass of the automobile is 17000/9.8 = 1735 kg, so the net force has magnitude F = (1735)(3.66) =
6.35× 102 N.
83. (First problem in Cluster 1)
(a) Using the coordinate system and force resolution shown in the textbook Figure 5-18(c), we apply
Newton’s second law along the x axis
−mg sin θ = ma
where θ = 30.0◦. Thus, a = −4.9m/s2. The magnitude of the acceleration, then, is 4.9m/s2.
(b) Applying Newton’s second law along the y axis (where there is no acceleration), we have
N −mg cos θ = 0 .
Thus, with m = 10.0 kg, we obtain N = 84.9 N.
84. (Second problem in Cluster 1)
(a) Newton’s second law applied to the x axis yields F −mg sin θ = ma. Thus, with F = 40.0 N, we
find a = −0.90 m/s2. The interpretation is that the magnitude of the acceleration is 0.90 m/s2 and
its direction is downhill.
(b) Substituting F = 60.0 N into F −mg sin θ = ma, we find a = 1.1 m/s2. Thus, the acceleration is
1.1 m/s
2
uphill.
132 CHAPTER 5.
85. (Third problem in Cluster 1)
The coordinate system we wish to use is shown in Figure 5-18(c) in the textbook, so we resolve this
vertical force into appropriate components.
��
��
��
��
��
��
��
��
��
A
A
A A
A
A��
�
6~F
A
A
AAKFy =
F sin 60◦
��*
Fx = F cos 60
◦
θ = 30◦
(a) Assuming the block is not pulled entirely off the incline, Newton’s second law applied to the x axis
yields
Fx −mg sin θ = ma .
This leads to a = −1.9m/s2, which we interpret as a acceleration of 1.9m/s2 directed downhill.
(b) The assumption stated in part (a) implies there is no acceleration in the y direction. Newton’s
second law along the y axis gives
N + Fy −mg cos θ = 0 .
Therefore, N = 32.9 N. We note that a negative value of N would have been a sure sign that our
assumption was incorrect.
(c) The equation in part (a) can be used to solve for the equilibrium (a = 0) value of F :
F cos 60◦ = mg sin 30◦ = 49 N .
Therefore, F = 98 N.
(d) There are three forces acting on the block: ~N, ~F , and m~g. Equilibrium generally suggests that the
“vector triangle” formed by three such vectors closes on itself. In this case, however, two sides of
that “triangle” are vertical! ~F is up and m~g is down! The insight behind this “squashed triangle”
is that ~N (the only vector that is not vertical) has zero magnitude. Thus, the block is not “bearing
down” on the incline surface. In fact, in this circumstance, the incline is not needed at all for
support; the value F = 98.0 N is just what is needed to hold the block (which weighs 98.0 N) aloft.
86. (Fourth problem in Cluster 1)
The coordinate system we wish to use is shown in Fig. 5-18 in the textbook, so we resolve this horizontal
force into appropriate components.
��
��
��
��
��
��
��
��
��
A
A
A A
A
A��
�
-
~FA
AAU
Fy = F sin θ
��
��*
Fx = F cos θ
θ = 30◦
(a) We apply Newton’s second law to the x axis:
Fx −mg sin θ = ma
133
This yields a = −1.44m/s2, which is interpreted as an acceleration of 1.44m/s2 downhill.
(b) Applying Newton’s second law to the y axis (where there is no acceleration), we have
N − Fy −mg cos θ = 0 .
This yields the normal force N = 105 N.
(c) When we set a = 0 in the part (a) equation, we obtain
F cos 30◦ = mg sin 30◦ .
Therefore, F = 56.6
N. Alternatively, we can use a “vector triangle” approach, referred to in the
previous problem solution. We form a closed triangle.
-~F
?
m~g
A
A
A
A
A
A
A
A
AAK
~N
We note that the angle
between the weight vector
and the normal force is θ.
Thus, we see mg tan θ = F ,
which gives F = 56.6 N.
134 CHAPTER 5.
Chapter 6
1. We do not consider the possibility that the bureau might tip, and treat this as a purely horizontal motion
problem (with the person’s push ~F in the +x direction). Applying Newton’s second law to the x and y
axes, we obtain
F − fs,max = ma
N −mg = 0
respectively. The second equation yields the normal force N = mg, whereupon the maximum static
friction is found to be (from Eq. 6-1) fs,max = µsmg. Thus, the first equation becomes
F − µsmg = ma = 0
where we have set a = 0 to be consistent with the fact that the static friction is still (just barely) able
to prevent the bureau from moving.
(a) With µs = 0.45 and m = 45 kg, the equation above leads to F = 198 N. To bring the bureau into
a state of motion, the person should push with any force greater than this value. Rounding to two
significant figures, we can therefore say the minimum required push is F = 2.0× 102 N.
(b) Replacing m = 45 kg with m = 28 kg, the reasoning above leads to roughly F = 1.2× 102 N.
2. An excellent discussion and equation development related to this problem is given in Sample Problem 6-3.
We merely quote (and apply) their main result (Eq. 6-13)
θ = tan−1 µs = tan−1 0.04 ≈ 2◦ .
3. The free-body diagram for the player is shown below. ~N is the normal force of the ground on the player,
m~g is the force of gravity, and ~f is the force of friction. The force of friction is related to the normal
force by f = µkN . We use Newton’s second law applied
to the vertical axis to find the normal
force. The vertical component of the ac-
celeration is zero, so we obtainN−mg =
0; thus, N = mg. Consequently,
µk =
f
N
=
470N
(79 kg)
(
9.8m/s
2
)
= 0.61 .
•......
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.............................................................................
...
...
...
...
..
..
..
..
..
..
............................
~N
m~g
~f
135
136 CHAPTER 6.
4. To maintain the stone’s motion, a horizontal force (in the +x direction) is needed that cancels the
retarding effect due to kinetic friction. Applying Newtons’ second to the x and y axes, we obtain
F − fk = ma
N −mg = 0
respectively. The second equation yields the normal force N = mg, so that (using Eq. 6-2) the kinetic
friction becomes fk = µkmg. Thus, the first equation becomes
F − µkmg = ma = 0
where we have set a = 0 to be consistent with the idea that the horizontal velocity of the stone should
remain constant. With m = 20 kg and µk = 0.80, we find F = 1.6× 102 N.
5. We denote ~F as the horizontal force of the person exerted on the crate (in the +x direction), ~fk is the
force of kinetic friction (in the −x direction), ~N is the vertical normal force exerted by the floor (in the
+y direction), andm~g is the force of gravity. The magnitude of the force of friction is given by fk = µkN
(Eq. 6-2). Applying Newtons’ second to the x and y axes, we obtain
F − fk = ma
N −mg = 0
respectively.
(a) The second equation yields the normal force N = mg, so that the friction is
fk = µkmg = (0.35)(55 kg)
(
9.8m/s
2
)
= 1.9× 102 N .
(b) The first equation becomes
F − µkmg = ma
which (with F = 220 N) we solve to find
a =
F
m
− µkg = 0.56 m/s2 .
6. An excellent discussion and equation development related to this problem is given in Sample Problem
6-3. We merely quote (and apply) their main result (Eq. 6-13)
θ = tan−1 µs = tan−1 0.5 = 27◦
which implies that the angle through which the slope should be reduced is φ = 45◦ − 27◦ ≈ 20◦.
7. The free-body diagram for the puck is shown below. ~N is the normal force of the ice on the puck, ~f is
the force of friction (in the −x direction), and m~g is the force of gravity.
(a) The horizontal component of Newton’s second law gives −f = ma, and constant acceleration
kinematics (Table 2-1) can be used to find the acceleration.
Since the final velocity is zero, v2 =
v20 + 2ax leads to a = −v20/2x. This is
substituted into the Newton’s law equa-
tion to obtain
f =
mv20
2x
=
(0.110 kg)(6.0m/s)2
2(15m)
= 0.13 N .
•......
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.............................................................................
...
...
...
...
..
..
..
..
..
..
............................
~N
m~g
~f
137
(b) The vertical component of Newton’s second law gives N−mg = 0, so N = mg which implies (using
Eq. 6-2) f = µkmg. We solve for the coefficient:
µk =
f
mg
=
0.13N
(0.110 kg)
(
9.8m/s
2
) = 0.12 .
8. (a) The free-body diagram for the person (shown as an L-shaped block) is shown below. The force
that she exerts on the rock slabs is not directly shown (since the diagram should
only show forces
exerted on her), but it is related by Newton’s third law) to the normal forces ~N1 and ~N2 exerted
horizontally by the slabs onto her shoes and back, respectively. We will show in part (b) that
N1 = N2 so that we there is no ambiguity in saying that the magnitude of her push is N2 . The
total upward force due to (maximum) static friction is ~f = ~f1+ ~f2 where (using Eq. 6-1) f1 = µs 1N1
and f2 = µs 2N2 . The problem gives the values µs 1 = 1.2 and µs 2 = 0.8.
?
m~g
ff ~N2
-
~N1
6
~f1
6~f2
(b) We apply Newton’s second law to the x and y axes (with +x rightward and +y upward and there
is no acceleration in either direction).
N1 −N2 = 0
f1 + f2 −mg = 0
The first equation tells us that the normal forces are equal N1 = N2 = N . Consequently, from
Eq. 6-1
f1 = µs 1N
f2 = µs 2N
we conclude that
f1 =
(
µs 1
µs 2
)
f2 .
Therefore, f1 + f2 −mg = 0 leads to (
µs 1
µs 2
+ 1
)
f2 = mg
which (with m = 49 kg) yields f2 = 192 N. From this we find N = f2/µs 2 = 240 N. This is equal
to the magnitude of the push exerted by the rock climber.
(c) From the above calculation, we find f1 = µs 1N = 288 N which amounts to a fraction
f1
W
=
288
(49)(9.8)
= 0.60
or 60% of her weight.
138 CHAPTER 6.
9. (a) The free-body diagram for the block is shown below. ~F is the applied force, ~N is the normal force
of the wall on the block, ~f is the force of friction, and m~g is the force of gravity. To determine if the
block falls, we find the magnitude f of the force of friction required to hold it without accelerating
and also find the normal force of the wall on the block.
We compare f and µsN . If
f < µsN , the block does
not slide on the wall but if
f > µsN , the block does slide.
The horizontal component of
Newton’s second law is F −
N = 0, so N = F = 12N and
µsN = (0.60)(12N) = 7.2N.
The vertical component is f−
mg = 0, so f = mg = 5.0N.
Since f < µsN the block does
not slide.
•.................
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.........................................................................................................................................
....
...
...
...
...
..
..
..
..
..
..
.................................................................................................................
...
...
...
...
..
..
..
..
..
..
............................
~N
m~g
~F
~f
(b) Since the block does not move f = 5.0N and N = 12N. The force of the wall on the block is
~Fw = −N ıˆ + f jˆ = −(12N) ıˆ + (5.0N) jˆ
where the axes are as shown on Fig. 6-21 of the text.
10. In addition to the forces already shown in Fig. 6-22, a free-body diagram would include an upward
normal force ~N exerted by the floor on the block, a downward m~g representing the gravitational pull
exerted by Earth, and an assumed-leftward ~f for the kinetic or static friction. We choose +x rightwards
and +y upwards. We apply Newton’s second law to these axes:
(6.0N)− f = ma
P +N −mg = 0
where m = 2.5 kg is the mass of the block.
(a) In this case, P = 8.0 N leads to N = (2.5)(9.8)− 8.0 so that the normal force is N = 16.5 N. Using
Eq. 6-1, this implies fs,max = µsN = 6.6 N, which is larger than the 6.0 N rightward force – so
the block (which was initially at rest) does not move. Putting a = 0 into the first of our equations
above yields a static friction force of f = P = 6.0 N. Since its value is positive, then our assumption
for the direction of ~f (leftward) is correct.
(b) In this case, P = 10 N leads to N = (2.5)(9.8)− 10 so that the normal force is N = 14.5 N. Using
Eq. 6-1, this implies fs,max = µsN = 5.8 N, which is less than the 6.0 N rightward force – so the
block does move. Hence, we are dealing not with static but with kinetic friction, which Eq. 6-2
reveals to be fk = µkN = 3.6 N. Again, its value is positive, so our assumption for the direction of
~f (leftward) is correct.
(c) In this last case, P = 12 N leads to N = 12.5 N and thus to fs,max = µsN = 5.0 N, which (as
expected) is less than the 6.0 N rightward force – so the block moves. The kinetic friction force,
then, is fk = µkN = 3.1 N. Once again, its value is positive, so our assumption for the direction of
~f (leftward) is correct.
11. A cross section of the cone of sand is shown below. To pile the most sand without extending the radius,
sand is added to make the height h as great as possible. Eventually, however, the sides become so steep
that sand at the surface begins to slide. The goal is to find the greatest height (corresponding to greatest
slope) for which the sand does not slide. A grain of sand is shown on the diagram and the forces on it
are labeled. ~N is the normal force of the surface, m~g is the force of gravity, and ~f is the force of (static)
friction. We take the x axis to be down the plane and the y axis to be in the direction of the normal
139
force. We assume the grain does not slide, so its acceleration is zero. Then the x component of Newton’s
second law is mg sin θ − f = 0 and the y component is N −mg cos θ = 0.
The first equation gives f =
mg sin θ and the second
gives N = mg cos θ. If
the grain does not slide, the
condition f < µsN must
hold. This meansmg sin θ <
µsmg cos θ or tan θ < µs.
The surface of the cone has
the greatest slope (and the
height of the cone is the
greatest) if tan θ = µs.
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