1. CHAPTER 2 MOTION ALONG A STRAIGHT LINE
Select odd-numbered solutions, marked with a dagger (†), appear in the Student Solutions
Manual, available for purchase. Answers to all solutions below are underscored.
= = (b) He must cover the remaining 1720
= = = Burst speed = 32 km/hour =
1 km 1000 m 2.5 × 10 years.
4 cm/year 4 × 10 m/year
v − = ⇒ = = = Moving 1000 km will take
1000 times as long, or t2 = 2.5 × 107 years.
12
2-1. Time required, Δt = distance
speed
= 30 =
0.3 s
100
2-2. Avg speed = distance
time
= 100 yd
9.0 s
× 1 mi
1760 yd
× 3600s 23 mi/h
1h
=
20 m 6.34 × 10
†2-3. 1 year = 3.156 × 107 sec, so 20 m/year = 7
7
1 year × 3.156 × 10 s/year
= − m/s (6.3 ×
10−7 m/s to two significant figures). 1 day = 24 hr = 86,400 s. In cm/day the rate is
6.34 × 10−7 m/s × 8.64 × 104 s/day × 100 cm/m = 5.4 cm/day.
2-4. Assume the butterfly’s speed is 0.5 m/sec. Then the travel time is
t =
3500 × 103 m 1 × 81 days.
0.5 m/s 24 hr/day × 3600 s/hr
d
v
= ≈
2-5. 6 days 12 hrs = 156 hrs.
dist. 5068 32.5 km/h
time 156
ν= = =
2-6.
9
1.4 × 10 ly
7 15 7
= =
t = 1.9 × 1010 yr
2.16 × 10 m/s × 9.47 × 10 m/ly × 1/(3.16 × 10 )yr/s
t d
ν
2-7. Estimated distance (by sea) between Java and England is 20,000 km.
20,000 km 600 km/h
32 h
ν d
= = ≈
t
2-8. (a) Average speed 4000 nmi 8.3 nmi/hr.
20 days × 24 hr/day
nmi in 7 days, which requires an average speed of 1720 nmi 10.2 nmi/hr.
7 days × 24 hr/day
= This is
about the same as his maximum possible speed. Since it’s unlikely that he can maintain the
highest possible speed for the entire 7 days, he should probably conclude that he will not be able
to complete the trip within the 20-day limit.
2-9. Average speed 35 km 14 km/hr
2.5 hr
d
t
= = =
2-10.
110 km 110 × 103 m Average speed 1.27 m/s.
1 day 24 h × 3600 sec
32 × 103 m =
8.9 m/s
3600 s
t d t
†2-11. 4
1 2

2. CHAPTER 2
= = or 12.8 × 10−3 km/m × 3600 s/hr = 46.0 km/hr
= = =
= = = =
In this time the deer traveled from 40 m to 50 m, i.e., 10 m. Thus
= = π = −
13
2-12. Average speed 402 m 16.9 m/s
23.8 s
d
t
= = =
2-13. Average speed 500 m 12.8 m/s,
39.10 s
2-14. 23.8 m 0.326 s
(263 km/hour × 1000 m/km/ 3600 s/hr)
t d
v
†2-15. Use the formula: t d .
= 5280 km
v
900 km/hr air t = =5.87 hr. 5280 km
35 km/hour shiip t = =151 hr.
2-16. 1
100 m 9.49 m/s.
10.54 s
v= = 2
200 m 9.37 m/s
21.34 s
v= =
2-17. 6 m/s × 1 (s)
d = v i t = = 0.06 m
100
2-18. Time taken for the arrow to reach the deer is
50 m 10s 0.77s.
65 m/s 13
t d
v
10m 13 m/s.
(10/13s)
v d
= = =
t
†2-19. (a) Take x = 0 to be the cheetah’s starting position. Then the cheetah’s position is given by
x = v t . The antelope’s starting position is 50 m from the cheetah’s starting position, so the
c c position of the antelope is given by x = v t + 50. When the cheetah catches the antelope, their
a a positions are the same, and we get v t = v t + 50. The speeds are νc = 101 km/h = 28.1 m/s and
c a va = 88 km/h = 24.4 m/s. Solving the equation for t gives = 50 m =
50 m
28.1 m/s 24.4 m/s c a
t
v v
− −
=
13.8 s, or 14 s to two significant figures. During this time, the cheetah travels (28.1 m/s)(13.8 s)
= 380 m.
(b) The cheetah must catch the antelope within 20 s. Call the antelope’s initial position x0.
We use the same equation that says the cheetah catches the antelope, 0 , c a v t = v t + x but now we
set t = 20 s and calculate what head start x0 the antelope needs. We get
0 ( ) (28.1 m/s 24.4 m/s) (20 s) c a x = v − v t = − = 72 m. If the antelope is farther away than 72 m,
the cheetah will not be able to catch it.
2-20. Average speed 100 m 10.1 m/s
9.86 s
d
t
= = =
†2-21. d = 26 mi × 1.6 × 103m/mi + 385 yd × 0.9144 m/yd = 4.195 × 104 m
t = 2 hr 24 min 52 s = 2 hr × 3600 s/hr + 24min × 60 s /min + 52 s = 8692 s
4.195 × 104m average speed
8692 s
d
t
= = = 4.83 m/s
v d
= = π =
2-22. 2 ×0.9 cm 0.094 cm/s
60 s
second
second
second
t
2 × 0.9 cm 1.6 × 10 3 cm/s
60 min × 60 s/min
minute
v d
minute
minute
t

3. CHAPTER 2
= = π = −
2 × 0.5 cm 7.3 × 10 5 cm/s
12 hr × 60 min/hr × 60 s/min
+ = hr average speed
14
hour
v d
hour
hour
t
†2-23. x = 4.0t − 0.50t2. To find the maximum value of x, differentiate with respect to t and set the
derivative equal to zero: dx 4.0 t 0.
dt
= − = The result is t = 4.0 s. (This is the point at which the
runner turns around and moves back toward the starting line.) The distance traveled at this time is
x = 4.0(4.0) − 0.50(4.0)2 = 8.0 m. At t = 8 seconds, x = 4.0(8.0) − 0.5(8.0)2 = 0; that is when he
comes back to the starting line. The total distance traveled is 16 m. Then average speed =
distance = 16 m
= 2.0 m/s.
time 8 s
2-24. Distance = 100 km time = 50 km 50 km 1.46
60 km/hour 80 km/hr
distance 100 km 69 km/hr.
time 1.46 hour
= = = The average speed is not exactly 70 km/hr because the car
moves at 80 km/hr for a shorter period of time than it does at 60 km/hr.
†2-25. Planet Orbit circumference (km) Period (s) Speed (km/s) log speed log radius
Mercury 3.64 × 108 7.61 × 106 47.8 1.68 8.56
Venus 6.79 × 108 1.94 × 107 35.0 1.54 8.83
Earth 9.42 × 108 3.16 × 107 29.8 1.47 8.97
Mars 1.43 × 109 5.93 × 107 24.1 1.38 9.16
Jupiter 4.89 × 109 3.76 × 108 13.0 1.11 9.69
Saturn 8.98 × 109 9.31 × 108 9.65 0.985 9.95
Uranus 1.80 × 1010 2.65 × 109 6.79 0.832 10.26
Neptune 2.83 × 1010 5.21 × 109 5.43 0.735 10.45
Pluto 3.71 × 1010 7.83 × 109 4.74 0.676 10.57
The slope of the line through the nine points is 1 .
2
−

4. CHAPTER 2
= − log radius + log C.
− =
−
= = The total change in position is
= = The total displacement is Δx = (12 blocks − 6 blocks +
= = The total displacement is Δx = 35 m − 22 m = 13 m. Then
15
This means that log speed 1
2
Therefore log speed = log[C(radius)−1/2]. Thus
Speed = C(radius)−1/2 whereCis some constant.
2-26. Avg speed = distance
time
+ =
= 8 8 6.27 m/s
2.55
Avg velocity = displacement 0m/s
time
=
2-27. Avg speed (for t = 0 to t = 10 s) = 200 20 m/s
10
=
Avg speed (for t = 10 to t = 14.3 s) = 270 200 16.3 m/s
14.3 10
2-28. Distance = (8 floors + 4 floors + 7 floors) × 4 m/floor = 76 m.
Average speed = distance 76 m 1.5 m/s.
time 50 s
Δx = (12 floors − 1 floors) × 4 m/floor = 44 m. The average velocity is
= Δ = =
Δ
44 m 0.88 m/s.
50 s
v x
t
†2-29. Distance = (12 blocks + 6 blocks + 3 blocks) × 81 m/block = 1701 m.
The elapsed time = 14 min 5 s + 6 min 28 s + 3 min 40 s = 23 min 73 s = 1453 s. Then average
speed = distance 1701 m 1.17 m/s.
time 1453 s
= Δ = =
Δ
3 blocks) × 81 m/block = 729 m. The average velocity is 729 m 0.502 m/s.
1453 s
v x
t
2-30. x(t = 0) = 0; x(t = 8) = 4 × 8 − 0.5 × 82 = 0; x(t = 10) = 4 × 10 − 0.5 × 102 = −10m.
x t = − x t = =
Average velocity between t = 0 to t = 8.0 s is: ( 8) ( 0) 0
−
8 0
;
x t x t
= − = = − − = −
Average velocity between t = 8.0 s to t = 10.0 s is: ( 10) ( 8) 10m 0 5.0 m/s.
s −
s
10 8 2 s
†2-31. Total distance = 3 × 0.25 mile × 1609 m/mile = 1207 m. Displacement = 0 m because the horse
returns to the starting point. Total time = 1 min 40 s = 100 s. Then average speed =
distance = 1207 m = 12.1 m/s.
The average velocity is v = Δ x =
0.
time 100 s
Δ
t
2-32. Total distance = 35 m + 22 m = 57 m. The total time is = 4.5 s + 3.6 s = 8.1 s. The average speed
is distance 57 m 7.0 m/s.
time 8.1 s
v = Δ = =
average velocity is x 13m 1.6 m/s.
Δ
t 8.1s

5. CHAPTER 2
†2-33. From the graph, the total distance traveled by the squirrel is 6 m + 6 m + 2 m + 6 m = 20 m. The
total elapsed time is 30 s. Then average speed distance 20 m 0.67 m/s.
v= = = 12.5 m/s. For
= = + − a dv 6.2 27.0t.
dx
dt = = + − = Instantaneous
16
= = = The total
time 30 s
displacement is Δx = 16 m, so the average velocity is = = 13 m =
0.53 m/s.
8.1 s
v Δx
Δt
2-34. For 0 ≤ t ≤ 2 s, displacement = 25 m. displacement 25 m
time 2s
2.0 ≤ t ≤ 4.0 s, displacement = 40 m − 25 m = 15 m. (This requires an estimate for the position
at 4.0 s. Your value may be slightly different.) v= displacement = 15 m =
7.5 m/s.
time 2s
To find the instantaneous velocity at any time, draw a tangent to the position vs time curve at that
time and determine the slope of the line. Your numbers may be slightly different from the ones
given here. At 1.0 s, we get a tangent passing through points with coordinates (0.3 s, 0 m) and
(2.5 s, 25 m). This gives a slope of 11 m/s. At 3.0 s, the position vs time graph is a straight line,
so the instantaneous velocity will be the same as the average velocity between 2.0 s and 4.0 s, or
7.5 m/s. Again, your value may be slightly different if you estimate a different position at 4.0 s.
2-35. a = Δν/Δt ⇒ Δν = a Δt =
82.6 gee × (9.807 m/s2 )
gee × 0.04 s
Δv = 32.4 m/s
2-36. a = Δν/Δt = [(96 − 0) km/h]/2.2 s × 1 h/3600 s × 1000 m/km= 12 m/s2
†2-37. = − − 2 1 = =
3 2
3
2 1
27 m/s 0 3.4 × 10 m/s .
8.0 × 10 s
a v v
t −
t −
2-38. 80 (km/hr) 22.22 m/s, i v= = 0, 2.8 s. fv = t =
− − = = =−
0 22.22 m/s 7.94 m/s2
2.8 s
f i v v
a
t
2-39. (a) t(s) a(m/s2 ) a(in gees)
0 6.1 0.62 Method:
10 1.4 0.14 i) Draw tangent to curve.
20 0.83 0.085 ii) Get slope of line by counting squares to find Δν and Δt.
30 0.56 0.057 iii) Convert from km/h to m/s.
40 0.49 0.050
(b) t(s) a(m/s2 ) a(gees)
0 −0.74 −0.075
10 −0.44 −0.045
20 −0.44 −0.045
30 −0.31 −0.032
40 −0.22 −0.022
2-40. x = 2.5t + 3.1t2 − 4.5t3; v dx 2.5 6.2t 13.5t2 ;
dt
= = −
dt
At t = 0 s, instantaneous velocity = 2
0 | 2.5 6.2 × 0 13.5 × 0 2.5 m/s. t
acceleration = 6.2 m/s2. At t = 2. 0 s, instantaneous velocity =

6. CHAPTER 2
dx
dt = = + − = −39.1 m/s (−39 m/s to two significant figures).
Instantaneous acceleration = 6.2 − 27.0(2) = −48 m/s2 .
For 0 ≤ t ≤ 2 s,
v = x t = s − x t = = + − = −
a = v t = − v t = = − − =
= = − v = 0 if
= = − a = 0 if
= = To make a sketch: consider the
above information, plus when t = 1.5 s, ν =
(6.0 m/s2 ) (1.5 s) − (2.0 m/s3 ) (1.5 s)2 = 4.5 m/s.
a = v t = − v t = ≈ − = 1 m/s2. (Your value may be slightly
a = v t = − v t = ≈ − = (Again, your value may be slightly
different depending on how you estimate the values of ν at 5 and 10 s.) To find the instantaneous
acceleration at 3 s, draw a tangent to the curve at that time. Your estimate may be slightly
different from ours. We get a tangent line that passes through the points (1 s, 0 m/s) and (5 s, 5
m/s). The slope of this line is the instantaneous acceleration 5 m/s 0 1.3 m/s2.
dv d v v a v v v e e
t s f t s
dv v v a
− − = =− =−
182 km/hr × 1000 m/km 20 m/s2 .
m/s 200 km/hr 18 km/hr |
17
2
2 | 2.5 6.2×2 13.5×2 t
( 2 ) ( 0) 2.5(2) 3.1(2)2 4.5(2)3 9.3 m/s.
−
2 0 2
( 2) ( 0) 39.1 m/s 2.5 m/s
−
2 0 2s
– 21 m/s2.
†2-41. x = 3.6t2 − 2.4t3 , v dx 7.2t 7.2t2 .
dt
7.2t − 7.2t2 = 0 ⇒ t = 0 s or t = 1.0 s. At t = 0, x = 0.
At t = 1.0 s, x = 3.6 − 2.4 = 1.2 m. To make a sketch,
consider that x = 0 when t = 0 and t = 3.6 1.5
2.4
= s.
Also, dx/dt = 0 when t = 0 and t = 1s.
2-42. v = Bt − Ct2 . V = 0 if t = 0 or t =
2
3
6.0 m/s 3.0 s.
2.0 m/s
B
C
= = a dv B 2Ct.
dt
1.5 s.
t B
C
2
†2-43. For 0 ≤ t ≤ 5.0 s, ( 5 s) ( 0) 5 m/s 0
5 s 5 s
different depending on how you read the values of ν at 0 and 5 s.) For 5.0 ≤ t ≤ 10.0 s,
( 10 s) ( 5 s) 9.5 m/s 5.0 m/s 0.9 m/s2 .
5 s 5 s
a = − =
−
5 s 1 s
2-44. / 2.5 0 / 2.5
0 [ ( ) ] .
(2.5 s)
f f
dt dt
− − −
= = + − =− At t = 0,
0
0
f
2.5 s 2.5 s
t
dt =
= − = −
2.5 s × 3600 s/hr

7. CHAPTER 2
a dv d v v d At Av t
( ) 2 .
†2-45. 0 0 2 0
2 22 2 2
dt dt At At dt At
(1 ) (1 ) (1 )
2(25 m/s)(2 s )(2 s) 2.5 m/s .
1 (2s )(2s)
18
⎡ ⎤
= = ⎢ ⎥ = − = − ⎣ + ⎦ + +
At t = 0, a = 0.
at t = 2 s,
2
2
= − = −
2 2 2
a
−
⎡⎣ + −
⎤⎦
a 2Av t , so a 0.
As t → ∞, 0
→ − →
2 4
A t
2-46. (a) Estimated average velocity (by mid-point method) is about 30 km/h. Because v = Δx/Δt, we
have Δx = v Δt ≈ 30 km/h × 5 s × 5 m/s 42 m.
18 km/h
≈
⎡ ⎤
⎢ ⎥
⎣ ⎦
(b) Time interval (s) v (km/h) Δx(m) by km/ h × 5s × 5 m/ s
18 km / h
v
5−10 70 100
10−15 90 120
15−20 110 150
20−25 130 180
25−30 140 190
30−35 150 210
35−40 160 220
40−45 170 240
(c) Total distance traveled is the sum of the last column plus the 42 m traveled in the first
5 s: d = 1450m = 1.45 km.
†2-47. (a)
(b) Time interval (s) Avg speed (m/s) Distance traveled (m)
0−0.3 647.5 194
0.3−0.6 628.5 189
0.6−0.9 611.5 183
0.9−1.2 596.0 179
1.2−1.5 579.5 174
1.5−1.8 564.0 169
1.8−2.1 549.5 165
2.1−2.4 535.0 161
2.4−2.7 521.0 156
2.7−3.0 508.0 152
Total distance traveled = 1722 m

8. CHAPTER 2
(c) Counting the number of squares under the ν versus t curve gives the same answer within
about ± 2 m.
bt b t b
⎡− − + − − ⎤ ⎢⎣ − ⎥⎦
19
2-48. ν = 655.9 − 61.14t + 3.26t2
acceleration, a = dv
dt
= −61.14 + 6.52t
a |t = 0 s = −61.14m/s2
a |t = 1.5 s = −61.14 + (6.52) (1.5) = −51.36m/s2
a |t = 3.0 s = −61.14 + (6.52) (3.0) = −41.58 m/s2
†2-49. (a)
(b) Calculus method:
If x = 0, then cos t = 0, so t = π/2 or 3π/2 s. Particle crosses x = 0 at t = 1.6 s and 4.7 s.
ν = dx/dt = −2.0 sin t
ν (π/2 s) = −2.0 sin(π/2) = −2.0 m/s
ν (3π/2 s) = −2.0 sin(3π/s) = 2.0 m/s
a = dv/dt = −2.0 cos t
a(π/2) = −2.0 cos(π/2) = 0m/ s2
a(3π/2) = −2.0 cos(3π/2) = 0m/ s2
(c) Maximum distance achieved when cos t = ± 1, i.e., when t = 0, π, 2π, or t = 0s,3.1s and 6.3s.
v = −2.0 sin t
v(0) = −2.0 sin(0) = 0m/ s
v(π) = −2.0 sin(π) = 0m/ s
v(2π) = −2.0 sin(2π) = 0m/ s
a = dv/dt = −2.0 cos t
a(0) = −2.0 cos(0) = −2.0m/ s2
a(π) = −2.0 cos(π) = 2.0 m/ s2
a(2π) = −2.0 cos(2π) = −2.0m/ s2
2-50. x = uext + uex(1/b − t) ln(1 − bt )
(a) Instantaneous velocity, v = dx/dt
dx
dt
= uex + uex ln(1 ) (1/ )
1
bt
= uex
[1 − ln(1 − bt) − 1]
ln(1 ) ex v = −u − bt
(b) a =
2
2
d x
dt
u b
bt
= −uex[−bt(1 − bt)] ex
1
=
−

9. CHAPTER 2
−
1( v − v
)
2
20
(c) dx
dt
= −3.0 × 103 ln(1 − 7.5 × 10−3t) m/s
ν(t = 0) = −3.0 × 103 ln 1 = 0m/s
ν(t = 120) = −3.0 × 103 ln(1 − 7.5 × 10−3 × 120) m/s
ν(120) = −3.0 × 103 ln(1 − 0.9) m/s = −3.0 × 103 × (−2.80) m/s
v(120) = 6.9 × 103m/s
(d) a = d
υ =
dt
2
2
d x
dt
=
3 3
3.0 × 10 × 7.5 × 10
3
1 7.5×10 t
−
− −
22.5
m/s2 = 3
1 − 7.5 × 10− t
m/s2
a(0) = 22.5 m/s2
a(120) = 22.5
3
1 − 7.5 × 10− × 120
m/s2 = 22.5
1 − 0.9
m/s2
a(120) = 225m/s2
1( )
2
2-51. Use Equation (25): a(x − x0) = v 2 − v
2
0
x − x0 = 2100 m; v = 360 km/h × 5 m/s
18 km/h
= 100 m/s; v0 = 0
a = 1
2
×
2 2
v v
x x
0
0
−
−
= 1
2
×
1002
2100
m/s2
a = 2.4m/s2
2-52. Acceleration, a =
2 2
v −
v
x −
x
0
0 2( )
=
(657)2
2 × 6.63
= 3.26 × 104m/s2
Time to travel the length of the barrel, t = 0 v v
a
657
t = 4
3.26 × 10
= 2.02 × 10−2 s
2-53. 4.2 ly = 4.2 ly × 9.46 × 1015 m/ly = 3.97 × 1016 m
Use x − x0 = ν0t + 1 ( 2 )
2
at with x0 = ν0 = 0 and x = 1 (4.2ly)
2
t 1/2 = 2x
a
=
16
3.97 × 10 m
9.807m/ s
2
= 6.36 × 107 s
Because the magnitude of the acceleration is the same for both parts of the trip, the time for the
second half is identical to that of the first half. Thus, the total time for the trip, T, is
T = 2t 1/2 = 2 × 6.36 × 107 s = 1.3 × 108 s ≈ 4.0 yr
Speed at midpoint: Use a(x − x0) = 2 2
0
ν0 = 0; x − x0 = 1.99 × 1016 m; a = 9.807 m/s2
Then
2
2 v = 2a(x − x0) = (2 × 9.807 × 1.99 × 1016) m2/s2
ν2 = 3.90 × 1017 m2/s2
v = 6.2 × 108m/ s (This exceeds the speed of light!!)

10. CHAPTER 2
2-54. 2 2
2a(x − x0 ) = v − v0 ⇒ v0 = −2a(x − x0 ) From the information given in the problem, x − x0 =
290 m, and a = −10 m/s2. Then 2
0 v = −2(−10 m/s )(290 m) = 76 m/s, which corresponds to
v − v = a(x − x0) with v = 0; ν0 = 80 km/h × 5 m/s
v − v = a(x − x0) with x − x0 = 50 m; v = 0;
= = = 16 s.
21
about 270 km/h, or 170 mph.
1( )
2
2-55. Use 2 2
0
18 km/h
= 22.2 m/s; x − x0 = 0.7 m.
Therefore a =
1 ( 2 2 ) 1 ( 22.2)
2
2 0
2 .
2
0
0.7m/s
v v
x x
− −
=
−
a = −350m/s2 (will probably survive)
2-56. 2 2
0 0 0 0 2a(x − x ) = v − v ⇒ v = −2a(x − x ) From the information given in the problem, x − x0 =
9.6 km = 9.6 × 103 m, and a = −5 m/s2. Then 2 3 2
0 v = −2(−5 m/s )(9.6 × 10 m) = 3.1 × 10 m/s,
which corresponds to about 700 mph. The time to stop is
2
= − = − 0
=
0 3.1×10 m/s 62 s.
2
5 m/s
t v v
a
−
1( )
2
†2-57. Use 2 2
0
v0 = 96 km/h = 26.67 m/s
1 ( ) 1 ( 26.67)
2 2
2 2 2
− −
0
2
= =
0
50m/s
v v
a
x −
x
a = −7.1m/s2
Use v = v + at with v = 0 m/s, v0 = 96 km/h = 26.7 m/s, and a = −7.1 m/s2. Then
0 t = − v0 =
3.8 s.
a
2-58.
2 2 3
a v v
= − = −
0
0
(260 × 10 / 3600)
x −
x
2( ) 2 × 1500
a = − 1.74m/s2 (The minus sign denotes deceleration.)
Use v = v + at with v = 0 m/s, v0 = 260 km/h = 72.2 m/s, and a = −1.74 m/s2. Then
0 t = − v0 =
42.5 s.
a
2-59. 2
0 v = v + at = 1.5 m/s × 20 s = 30 m/s
2 2 2
0
1 0 1 × 1.5m/s × (20 s) 300 m.
2 2
x = v t + at = + =
2-60. 2 2
0 0
1 1.0m 5.0m 3.0 m/s 4.0s 1 (4.0s)
2 2
x − x = v t + at ⇒ − = i + a i
⇒ −4 m = 12 m + 8 a ⇒ a = −2.0 m/s2 . The velocity is 0 v = v + at =
3.0m/s − (2.0m/s2 )(4.0s) = − 5.0m/s.
2-61. 2 2
0
1 1
2 2
2 2×150m
x = v t + at = at (since v0 = 0). 2
1.2 m/s
t x
a

11. CHAPTER 2
v= = = 153 m/s. 0 v = v + at =
153 m/s + (0.60 m/s2 )(90 s) = 2.1 × 102 m/s.
x = i = For 6 ≤ t ≤ 10 s, a = −4.5m/s2, and
x = x + v t − + a t − = + t − − i t − = − t − + t − +
− − = + ⇒ = = = 22.6 m/s.
1 2 700m 0.5 × 0.05m/s × 30 s
2 30s
− − = + ⇒ = = = m/s.
1 2 550m 0.5 × 0.5m/s × 15 s 32.9
2 15s
22
2-62. 0
550 km/hr 550 km/hr × 1000 m/km
3600 s/hr
†2-63. The sketch should be based on the following:
For 0 ≤ t ≤ 6 s, a = 3.0 m/s2; 0 v = v + at = 0 + 3t; 2
0 0
3 3 .
2
t t
x = ∫ v dt = ∫ t dt = t At t = 6s,
v = 3 × 6 = 18 m/s; 3 62 54 m.
2
0 v = v + a(t − 6) = 18 − 4.5 (t − 6) m/s = 45 − 4.5t m/s.
2 2 2
0 0
( 6) 1 ( 6) 54 18( 6) 1 4.5( 6) 2.25( 6) 18( 6) 54
2 2
At t = 10 s, v = 45 − 4.5 × 10 = 0; x = −2.25i 42 + 18i 4 + 54 = 90 m.
2-64.
2
2 2 2
2
0 0
x at
x vt at v
t
2
0 v = v + at = 22.6m/s + 0.05m/s i30s = 24 m/s.
†2-65.
2
2 2 2
2
0 0
x at
x vt at v
t
2
0 v = v + at = 32.9 m/s + 0.5 m/s i15 s = 40.4 m/s.
2-66. Speed at the end of the 440-yard mark,
v = 250.69 × 1760 × 3
60 × 60
= 367.7 ft/s

12. CHAPTER 2
23
(a) Average acceleration, a = v
Δ
Δ
t
= 367.7 =
65.23ft/s2
5.637
(b) For a constant acceleration, the distance traveled would be = avg speed × time =
367.7 (5.637)
⎛ ⎞
⎜ ⎝ 2
⎟
⎠
= 1036.36 ft = 345.45 yd.
Therefore the acceleration was not constant.
(c) Assuming constant acceleration, distance = 440 × 3 = v
(5.637)
2 t
vt = 2 × 440 × 3
5.637
= 463 ft/s = 319mi/h.
2-67. The distance traveled by the elevator is
1 1
2 2
x = at 2 + vt + ⎛⎜ vt − at 2
⎞⎟
1 2 3 3
⎝ ⎠
where v = constant speed reached at at1; t1, t2 and t3 are the times spent in the following,
respectively: accelerating, traveling at constant speed, and decelerating.
21 × 2.5 m = 1
2
a(5)2 + (a5)7 + (a5)5 − 1
2
a52
52.5 m = 60 a
a = 0.875m/s2
The maximum speed of the elevator is vmax = at1
vmax = 0.875 m/s2 × 5 s
max v = 4.4m/s
2-68. (a) Average speed = 400
55
= 7.3 m/s
(b) For minimum values of acceleration and deceleration, the elevator should travel half the
distance in half the time. Therefore
2 200 1 55
= a ⎛⎜ ⎞⎟
2 2
⎝ ⎠
a = 0.53m/s2 where a is the acceleration and −a is the deceleration.
Maximum speed is given by vmax = at = 0.53 × 55 14.6m/s
2
=
(This is twice the average speed. What is its significance?)
2-69. v0 (km/h) v0 (m/s) 0 v t Δ (m) 2
v
a
− (m)
0
2
Total stopping
distance (m)
15 4.17 8.3 1.1 9.4
30 8.33 16.7 4.3 21.0
45 12.5 25.0 10 35.0
60 26.7 33.3 18 51.3
75 20.8 41.7 27 68.7
90 25.0 50.0 39 89.0

13. CHAPTER 2
= at (i)
24
2-70. Table on Page 47:
0 v t Δ =
2
0
2
v
a
2
⇒ v0 = 2at = 2 × 8 m/s × 0.75 s = 12 m/s.
2-71. 50 km/h = 50 × 5
18
m/s = 13.9 m/s; 2.0 ft = (2 × 0.3048) m = 0.610 m. In the inertial frame that is
traveling at constant velocity with the car, all velocities are zero. In this frame, a = 200 m/s2 also.
Therefore the speed with which the dashboard hits the passenger is
v = 0 2a(x − x ) = 2(200 m/s2 )(0.610m) = 15.6m/s.
2-72. Consider the position of the car in the reference frame of the truck
x0c = −12 − 17 = −29 m
The final position of the car is xc = 17 m, therefore,
xc = x0c + v0c t + 1
2
at2
17 = −29 + 0 + 1
2
at2
46 1 2 .
2
The final speed of the car relative to the truck is 24 km/h
=
24 × 103
60 × 60
= 6.67 m/s.
at = 6.67 m/s (ii)
Divide equation (i) by (ii) to get t = 13.8s and
a = 6.67 0.48m/s2 .
13.8
=
†2-73. v = −gτ + gτe−t/τ
(a) acceleration, dv ge t /
dt
= − τ
(b)
Lim
t → ∞
e −t/τ = 0, therefore
Lim
t → ∞
v = −gτ +
Lim
t → ∞
e − t/τ = −gt
0 d g t gt e t g x
dt
(c) v = ( 2 / 2 )
− τ − − τ + τ +
= −gτ + g
τ 2
τ
e − t/τ
= −gτ + gτ e−t /τ
(d) for t << τ, the exponential e−t/τ can be expanded as
e−t/τ = 1 − t
τ
+
2
2
1
2
t
τ
+ . . . .
Therefore, x = −gτ t − gτ2(1 − t
τ
+
2
2
1
2
t
τ
) + gτ2 + x0
= −gτ t − gτ2 + gt τ − 1
2
gt2 + gτ2 + x0
2
0
1
2
= − gt + x

14. CHAPTER 2
v v g x x x x v v
− = − ⇒ − = − = − =
2 ( ) (36.1 m/s) 0 66.5 m.
0 v − v = −2g(x − x0) with v = 0; g = 1.80 m/s2; x − x0 = 200,000 m
(12.5 m/s) 8.4 m × 1 floor
= = = = 3 floors.
(20.8 m/s) 22.1 m × 1 floor
= = = = 8 floors.
(29.2 m/s) 43.4 m × 1 floor
= = = = 15 floors.
= = = 1.43 s. This is 1.43 s/(3 half turns) = 0.48 s per half turn.
= = = Total time = 2(1.39 s) = 2.8 s.
25
2-74. x − x0 = v0t − 1
2
gt2 with x − x0 = −380 m; v0 = 0; g = 9.8 m/s2
−380 m = − 1 (9.8 m/s2 )
2
t2
t2 = 77.6 s2
t = 8.8s
For the impact velocity use v = v0 − gt
v = 0 − 9.8 m/s2 × 8.8 s = −86m/s
†2-75. 130 km/h = 36.1 m/s. The acceleration of the freely falling falcon is g, so use
2 2 2
2 2 0
0 0 0 2
g
2 2(9.81 m/s)
2-76. 2 2
0 v = v + 2ax. Here v0 = 0, a = g. v = 2gh = 2(9.81m s2 )(8.7 m) = 13 m/s.
2-77. 2 2
0 v = v + 2ax. Here v = 0, a = −g because the displacement is upward and the acceleration is
downward. 2
0 v = −2gh = −2(−9.81m s )(1.9m) = 6.1 m/s.
2-78. mgh = 1
2
mv2, where h is the maximum height reached
h =
⎛ ⎞
⎜ ⎟
⎝ ⎠
1 v
2
2
g
=
(366)2 =
6834 m
2 × 9.8
2-79. Neglecting air resistance, we have
x − x0 = v0t − 1
2
gt2 with v0 = 0; g = 9.8 m/s2; t = 3.0 s
x − x0 = 0 − 1 (9.8m/s2 ) (3.0 s)
2
2 = 44m
2-80. Use 2 2
2
0 v = 2 × 1.80 m/s2 × 2 × 105 m = 7.2 × 105 m2/s2
0 v = 849m/s
2-81. v = 45 km/h = 12.5 m/s.
2 2
2
h v
g
2 2(9.81 m/s ) 2.9 m
v = 75 km/h = 20.8 m/s.
2 2
2
h v
g
2 2(9.81 m/s ) 2.9 m
v = 105 km/h = 29.2 m/s.
2 2
2
h v
g
2 2(9.81 m/s ) 2.9 m
2 2(10m)
2-82. 2
9.81 m/s
t h
g
t h
2 2(9.5m) 1.39 s.
†2-83. 2
up
g
9.81 m/s 2
0 (9.81 m/s )(1.4 s) 14 m/s. up v = gt = = This is the initial speed. The initial velocity is 14 m/s
up.

15. CHAPTER 2
2-84. v0 = 1.0 m/s downward, and the ball is initially some distance h above the ground. After falling
that distance h, the ball will strike the ground with some speed v and will rebound (reverse its
direction of motion) at the same speed if the collision with the ground is elastic. (This concept
will be introduced in a later chapter.) Since it starts moving back up with the same speed it had
just before it hit the ground, the time required to return to its starting point a distance h above the
floor will be the same as the time required for it to reach the floor in the first place, and it will
arrive at the distance h above the ground with same speed with which it was initially thrown.
Thus the downward and upward travel times are each equal to half the total travel time. The speed
of the ball just before striking the ground is v = v + g t ,
where t is the total travel time. Then
0 − − − = + ⇒ = = − = − = or 8.6
m/s (to two significant figures). The height the penny reaches above its launch point is
(8.57 m/s) 3.7 m.
= − = − = = =
⇒ = − = − = 1.13 m/s, or 1.1 m/s to two significant
figures. The velocity just before hitting the floor is 0 v = v + gt =
1.13 m/s + (9.81 m/s2 )(0.45 s) = 5.5 m/s.
26
2
the height h can be found using 2 2
0 v = v + 2gh, which gives
2 2
h v v
= − 0 .
Substituting, we get
2
g
1 m/s (9.81 m/s2 ) 0.75 s 4.68 m/s.
v = + ⎛⎜ ⎞⎟ =
2
⎝ ⎠
Then
2 2
h = (4.68 m/s) − (1 m/s)
= 1.1 m.
2
2(9.81 m/s )
Comment: This requires conservation of momentum! The ball will collide with the ground and
rebound with a speed equal to its speed just before it hit, a topic that obviously isn’t covered in
this chapter.
†2-85. Take the coordinate axis to point up. Then the final displacement is −9.2 m, a = −g, t = 2.5 s.
2
d at d v t at v d at
2 2
0 0
2 9.2 m ( 9.81 m/s )(2.5 s) 8.57 m/s,
t t
2 2 2.5 s 2
2 2 2 2
0 0 0
2
h v v v
a −
g g
2 2( ) 2 2(9.81(m/s)
2-86. v0 = 0, a = −g, x = −h ⇒ 2
v = v + v for constant acceleration, so
0 v = v + 2ax = 2gh. 0
2
2
.
gh
2
v =
†2-87. Assume that the collision of the ball with the floor simply reverses the direction of the ball’s
velocity. Then the time for the ball to reach the floor is the same as the time for the ball to return
to its starting point, and its speed upon returning will be the same as the speed with which it
started. Thus 0.90 s 0.45 s.
2 down t = = Since the ball begins by moving down, choose the
direction of the axis for the motion to point down. Then the acceleration and initial velocity are
represented by positive numbers. 2
0
1
2
h = v t + gt
2 2 2
v h gt
0
2 2(1.5 m) (9.81 m/s )(0.45 s)
2 t
2(0.45 s)

16. CHAPTER 2
x = gt At t = 0,1,2,3 s the distance fallen is 0, 4.91 m, 19.6 m, 44.1 m, 78.5 m. The distance
traveled from 0 − 1 s is 4.91 m; from 1 s to 2 s the distance traveled is 19.6 m − 4.91 m = 14.7 m.
Likewise the distance traveled from 2 s to 3 s is 24.5 m. Dividing each of these by 4.91 m gives
the ratios 1:3:5 and so on.
= to measure y1 within 10%, the time must be known to within 5% or within 0.15 s.
− − =
4.64m/ s ( 5.42m/s) 1.6 × 10 m/s
27
2-88. 1 2 .
2
2-89.
2
h = v t + at .
v0 = 0 0 2
2
a 2h .
t
⇒ =
a 2 h
Δ = Δ
2
t
4 fingers × 10 cm
4 fingers × 100% 0.22%. 100 cubits × 46 cm
cubit
a h
a h
Δ = Δ = =
2-90. Time taken to fall through a distance of 45 m,
y2 = y1 + v1t + 1
2
at2
0 = 45 + 0 − 1 (9.8)
2
t2
t = 3.03 s
Since 1 t 2y ,
g
Therefore a stopwatchmust be used.An ordinary wristwatch will have an uncertainty of ±1 s.
2-91. Velocity of ball on impact 2
1 = − 2gx = − 2(9.8m/s ) (1.5m) = −5.42 m/s
Velocity of ball after impact = 2 2gx = 2(9.8m/s2 ) (1.1m) = 4.64 m/s
a v
= Δ
Δ
t
= 4 2
6.2 × 10− 4
s
2-92. (a) Impact speed, v = 2gh = 2 × 9.8 × 96 = 43.4 m/s
(b) a =
v 2 −
v
2
2
2( x −
x
)
2 =
2
− = −
(43.4) 254.5m/ s2
2 × 3.7
(The minus sign denotes deceleration.)
†2-93. The muzzle speed is given by 2 3 3
0 v = 2gh = 2(9.81 m/s )(180 × 10 m) = 1.88 × 10 m/s. To
find out how long the projectile remains above 100 km, we can use the fact that the time for the
projectile to climb from 100 km to 180 km is the same as the time for it to fall from 180 km back
to 100 km. So we can just calculate the time to fall a distance of 80 km from rest and double that
2 3
value. The time to fall can be calculated from
y gt t y
2 2(80×10 m) 128s,
= ⇒ = = =
2
g
2 9.81 m/s
so the total time above 100 km is 2t = 256 s.
2-94. Given,
(a) g = 978.0318 cm/s2 × (1 + 53.024 × 10−4 sin2 Θ − 5.9 × 10−6 sin2 2 Θ) for Θ = 45°
g = 978.0318 cm/s2 × (1 + (53.024 × 10−4)/2 − 5.9 × 10−6) = 980.6190 cm/s2
At the pole Θ = 90°, g = 978.0318 cm/s2 × (1 + 53.024 × 10−4) = 983.2177 cm/s2

17. CHAPTER 2
(b) Let g = A(1 + B sin2 Θ − C sin2 2 Θ)
where A = 978.0318 cm/s2, B = 53.024 × 10−4, C = 5.9 × 10−6. The condition
dg A(Bsin 2 4C sin 2 cos 2 )
d
= Θ − Θ Θ
Θ
= A (B − 4C cos 2 Θ)sin 2 Θ = 0 gives extrema at Θ = 0, ±
π evaluate
v − v = −g(x − x0) with g = 9.8 m/s2; x − x0 = −1500 m
y = y − gt where y0 = 335 m. When the elevator
⇒ vt = y − gt which can be rearranged as 2
v v gy
− ± + − ± +
= =
2 (2) 8 2(6.17 m/s) 4(6.17 m/s) 8(9.81 m/s )(335 m/s)
= − =47.2 m.
28
π
2
To distinguish between extrema at Θ = 0, ± ,
2
2
d g
d Θθ
2
at Θ = 0, ±
π
2
2
2 0 θ 0 d g A(2B cos 2 8C cos 4 sin 4 )
d = = = − Θ Θ Θ Θ
Θ
= 2AB > 0 (Thus at Θ = 0, g has a minimum.)
Similarly
2
2
d g 2AB 0
d =±π
θ /2
= − <
Θ
shows g has maxima at ± π/2 (poles).
At the equator, Θ = 0, so g = 978.0318 cm/s2.
2-95. 300 km/h = 83.3 m/s. If down is negative, then v0 = −83.3 m/s for bomber. Projectile speed
relative to ground is −700 m/s − 83.3 m/s, or −783.3 m/s.
Use 2 2
1( )
2
0
v2 = 2
0 v − 2g(x − x0) = (783.3)2 − 2 × 9.8(−1500 m) m2/s2
v = ±802m/s
To find the time, use x − x0 = v0t − 1
2
gt2, with x − x0 = −1500 m; v0 = −783.3 m/s
−1500 m = 783.3 1 9.8 2
⎛⎜ − t − t ⎞⎟
⎝ 2
⎠
m
4.9t 2 + 783.3t − 1500 = 0
t = 1.9s
2-96. For the elevator, v = 370 m/min = 6.17 m/s. Its distance above the ground is given by . ey = vt
The height of the penny above the ground is
2
0 ,
2 p
and penny meet, ye = yp
2
0 ,
2
0 gt + 2vt − 2y = 0. The
root is
2 2 2
0
2
2 2(9.81 m/s )
t
g
t = 7.66 s (after dropping the negative root).
2
ye = yp
= y − gt
0 2
9.81 × 7.662 335
2

18. CHAPTER 2
− − −
10 ( 9.81 m/s )(1.0 s) 1 2 2
2 1.0 s
= + ⇒ = = =14.9 m/s. The impact speed is
+ − =
29
†2-97. tup = 1.0 s, h = 10 m, a = −g = 9.81 m/s2.
2 2 2
2
0 0
h at m
h vt at v
t
2
0 v = v + at = 14.9 m/s − (9.81 m/s )(1.0 s) = 5.1 m/s.
2-98. (a) The trajectory of the first stone is
x1 = x0 + v0t − 1
2
gt2 = 15t − 4.9t2. The second stone is thrown 1.00 s later,
so x2 = v0t′, where t′ is the time after the second stone is thrown,
so t′ = t − 1.00 s. Therefore
x2 = v0(t − 1) − 4.9(t − 1)2
We want the stones to collide at a height of 11 m. Therefore
x1 = 15t 0 − 4.9t 2
0 = 11 m = v0(t 0 − 1) − 4.9(t 0 − 1)2 = x 2
Solving for t 0 gives t = 1.84 or1.22s
0 Using these values for t 0, we get for v0:
2
1.84 s: v0 =
0
2
0
t
11 4.9 ( 1) 17.2m/s
1 11 4.9 (0.84)
0.84m/s
t
+ − =
− = +
1.22 s: v0 =
11 + 4.9(0.22)2 =
51.1m/s
0.22
The 1.84 s corresponds to hitting the stone on its way down, whereas the 1.22 s corresponds to
hitting it on the way up.
(b) If the second stone is thrown 1.30 s after the first one, the first stone has already passed 11 m
on the way up, so the collision can take place only on the way down.
2
v0 =
t
+ −
−
11 4.9( 1.30)
0
0
1.30 m/s
t
=
11 4.9(1.84 1.30)2 23.0m/s
−
1.84 1.30m/s
2-99. (a) Time taken for drops to fall is t = 2h / g. If there are n drops per second, the number of
drops in the air at any one time is nt = n(2h / g)1/ 2 .
(b) Since the drops are falling at a constant rate, the median height is the distance fallen by a drop
in half the time needed for it to hit the ground. Calling this time τ, we have
τ = 1 (2 / )
2
h g 1/2 = (h/2g)1/2.
Then x = v0t − 1
2
gt 2 = − 1
2
g(h/2g) = − 1
4
h
Thus the position of the median is 1
4
h below the edge of the spout
or 3
4
h above the ground.

19. CHAPTER 2
(c) The density of drops is proportional to 1/v, where v is the velocity of a particular drop. Thus
density 1 1 .
v t t ∫ dv = ∫Ct dt ⇒ v − 8.0 m/s = Ct ⇒ v = 8.0 m/s
+
Ct t t x vdt 8.0 m/s (0.0833 m/s) t dt (8.0 m/s )t
(0.0833 m/s )
t ∫ ∫
= = ⎡⎣ + ⎤⎦ = +
= +
v t t v = ∫ adt = ∫ a − dt = a t − = − = 18.3 m/s.
After 2 s, the acceleration becomes zero, so the velocity becomes constant at whatever value it
had at t = 2 s. So to find v after a long time (t >> 2 s), find its value at 2 s:
v t t v = ∫ adt = ∫ a − dt = a t − = − =
The distance traveled is
x x vdt a t t dt a t t
⎣ ⎦ ∫ ∫
− = = − = − = ⎢ − ⎥
v At Bt v = ∫ adt = ∫ At + Bt dt = + = i i + i i
= 96.7 m/s
30
=
v 2g(h −
x)
∼
The average height calculation must take into account the weighting factor of the density of
drops. Thus
x dx
g h −
x
2 ( )
.
h dx
g h x
2 ( )
=
−
∫
∫
Let ξ = h − x, so that
h d
h ( )
0
h h
d
0
ξ ξ
ξ
ξ
ξ
−
=
∫
∫
= h −
3 / 2
[2 / 3 ]
1/ 2
0
[2 ]
0
h
h
ξ
ξ
= h −
3 / 2
1/ 2
h
h
2 / 3
2
= h − 1
3
h = 2 h
.
3
2-100. a = dv =
Ct2 .
dt
3 3
2
3 3
8.0m/s 0 0
C = 0.25 m/s4 ⇒ v = 8.0 m/s + (0.0833 m/s4 )t3 , At t = 3.0 s, v = 10.25 m/s. Final result = 10
m/s. To find x, integrate v:
4 4
3
0 0
4 4
4
)t t
(8.0 m/s (0.0208 m/s )
At t = 1.0 s, x = 8.0 m. (This is also the change in position, since x = 0 at t = 0.)
†2-101. At t = 1 s,
2
a = dv = a −
t
0 2 (1 )
4.0 s
dt
1s 2 3
1 s 2
0 2 0 2 0
0 0
(1 ) ( ) | 20 m/s (1 1 ) s
4.0 s 12.0 s 12
2 s 2 3
2 s 2
0 2 0 2 0
0 0
(1 ) ( ) | 20 m/s (2 8 ) s 26.7 m/s
4.0 s 12.0 s 12
2 s 2 s 3 2 4 2 4
2 s 2
⎡ ⎤
0 0 2 0 2 0 2
0 0
( ) ( ) | (20 m/s ) (2 s) (2 s)
12.0 s 2 48 s 2 48 s
= 33.3 m.
2-102.
2 s 2 3
2 2s 2 2 2 4 3 3
0
0 0
( ) | 1 15 m/s 2 s 1 25 m/s 2 s
2 3 2 3

20. CHAPTER 2
Δx = ∫ vdt = ∫ At + Bt dt = At + Bt
− + − = 48.75 m
− ∫
1 ln g Av t g Av e At g Av (g Av )e At
A g Av g Av
⇒ − − = ⇒ − = − ⇒ − = − −
v 1 [g (g Av )e At ] g (1 e At ) v e At
⇒ = − − − = − − + −
= =− 0
⎛ ⎞ ⎛ ⎞
1 1 1 1 1 1
Δ = ⎜ − ⎟ = ⎜ − ⎟
⎝ ⎠ ⎝ ⎠
v = = m = =
= + =
d= = =
d= = =
= d + d = + = 1.0 km/min, or 60 km/hr
31
Distance traveled
2 s 2 s 2 3 3 4
2 s
1 s
1 s 1 s
( ) |
2 3 6 12
=
15 m/s3 (23 1) s3 25 m/s4 (24 1) s4
6 12
†2-103. a dv g Av
= = − ⇒ dv dt
dt
g −
Av =
v
v
0
dv t
g Av
⇒ =
0
− −
0 0
0 0
A A
After a long time, t >> 1/ A, e− At → 0, v g .
A
→
2-104. a dv Bv2
dt
0
2
1 |
v
vv
v
dv B t Bt
v v
⇒ ∫ = − ⋅ Δ ⇒ − = − Δ
4 1
0
4 1
6.1 × 10 m 90 km/hr 120 km/hr
1 1 1 × 3600 s/hr
6.1 × 10 m 1000 m/km 90 km/hr 120 km/hr
t
B v v − −
− −
⎛ ⎞
= ⎜ − ⎟
i ⎝ ⎠
= 16.4 seconds
†2-105. Cram’s speed is 1 mile 1 mile .
3 min 46.32 s 226.32 s C v= = The time difference for the two runners
getting to the finish line is Δt = 3min 46.32sec − 3min 44.39sec = 1.93 s. So Cram is behind
by a distance of 3
C
1 mile × 1.93 s 8.53 × 10 mile × 1609 m 13.7 m.
226.32 s mile
v Δt = = − =
2-106. distance 1500 1500 m 14.43 m/s
time 1min 43.95sec 103.95 s
2-107. Average speed = distance (100 100) m 1.3 m/s
+ +
time (10 60 80) s
Average velocity = 0, because the woman returns to her starting point.
2-108. (a) 1
35 km/hr × 30min 35 km/hr × 1 × 30min 17.5 km;
60 min/hr
2
85 km/hr × 30min 85 km/hr × 1 × 30min 42.5 km;
60 min/hr
(b) Average speed 1 2 (17.5 42.5) km
time 60 min
†2-109. Define: t = time from when the sailfish spots the mackerel to when it catches the mackerel. Then:
distance for the sailfish = 109 km/hr × t, distance for the mackerel = 33 km/hr × t. The
separation between the fish is 20 m, so the time for the sailfish to catch the mackerel is given by
109 km/hr × t − 33 km/hr × t = 20 m

21. CHAPTER 2
d= = =
⇒ t = = i =
= − = − = or 219 seconds.
105 km/hr 0 0.0610 hr,
1.72 × 10 km/hr
= = − At t = 5.0 s, v = 6.0 − 6.0i 0.50 = 3.0 m/s
= = − = −6.0 m/s2 at all times.
d = v t = 25 m/s)(0.75 s) = 18.8 m. The remaining distance to the cow is d2
32
20 m 20 m 3600 s/hr
76 km/hr 76 km/hr 1000 m/km
⇒ t = = i
i
= 0.95 s. During this time, the sailfish travels a
distance 109 km/hr × 0.95 s 109 km/hr × 1000 m/km × 0.95 s 28.8 m.
3600 s/hr
2-110. Distance traveled by the first plane = d1 = 720 km/hr i t, where t is measured beginning at 10:00.
Distance traveled by the second plane = d2 = 640 km/hr i (t − 1 hr), since that plane left one hour
later. d + d = 1286 km ⇒ 720t + 640(t − 1) = 1286 km ⇒ t = (1286 + 640) km =
1.42 hr
1 2 1360 km/hr
= 1 hr 25 min. The distance traveled by the first plane is 1 d = (720 km/hr)(1.42 hr) = 1022 km,
so the planes meet 1022 km north of San Francisco. Since the total elapsed time is 1 hr 25 after
departure of the first plane, they meet at 11:25 AM.
†2-111. Distance that my car travels = 80 km/hr i t Distance that the other car travels = 50 km/hr i t. To go
from 10 m behind the slower car to 10 m ahead of it requires traveling a total relative distance of
10 m + 10 m + 4 m, because of the length of the car. Thus 80 km/hr i t − 50 km/hr i t = (10 + 10 +
4) m 24 m 24 m 3600 s/hr
30 km/hr 30 km/hr 1000 m/km
i
2.9 s.
2-112.
2 2 2 2 2
a v v
= − = − =
0 3 2 (105 0) km /hr 1.72 × 10 km/hr
x
2 2(3.2 km)
0
3 2
t v v
a
†2-113. (a) x = 2.0 + 6.0t − 3.0t2 . At t = 0.50 s, x = 2.0 + 6.0 × 0.50 − 3.0 i (0.50)2 = 4.3 m.
(b) v dx 6.0 6.0t.
dt
(c) a dv d (6.0 6.0t)
dt dt
2-114. (a) 100 km/hr = 27.8 m/s. The position of the speeder after 8.0 s is
,1 1 27.8 m/s × 8 s sx = vt = = 222 m from the starting point.
(b) The speed of the police cruiser goes from 0 to 120 km/hr (33.3 m/s) in 10 s, so its
acceleration is 3.33 m/s2. The position of the police cruiser after reaching its final speed is
2 2 2
,1
,1
(3.33 m/s )(10 s) 168 m
p
2 2
p
at
x = = = from the starting position. At this time the position of
the speeder is ,2 ,2 27.8 m/s × 18 s 500 m, s s x = vt = = so the speeder is 332 m ahead of the
police cruiser.
(c) Let t be the time from when the cruiser reaches its final speed of 120 km/hr until it catches up
to the speeder. When the cruiser catches up to the speeder, both vehicles have traveled the same
distance from the point where the cruiser reached 120 km/hr. Mathematically this means
(33.3 m/s)t = (27.8 m/s)t + 332 m, which gives t = 60 s. So the total time that has elapsed since
the cruiser began pursuit is 70 s. During this time the cruiser traveled a total distance of
168 m + (33.3 m/s)(60 s) = 2.18 × 103 m, or 2.18 km.
†2-115. The initial speed of the car is v0 = 90 km/hr = 25.0 m/s. The distance traveled during the reaction
time t1 = 0.75 s is 1 01 (
= 30 m − 18.8 m = 11.2 m. The car’s acceleration as it travels this distance is a = −8.0 m/s2. Its

22. CHAPTER 2
final speed when it hits the cow is given by
2 2 2 2
0 2 0 2 2 2 (v − v = ad ⇒ v = v + ad = 25 m/s) + 2(−8.0 m/s)(11.2 m) = 21.1 m/s, or
76 km/hr.
= − = = The distance she falls during this
55.6 m/s 5.66 s.
9.81 m/s
= − = = 158 m. The rest of the height is h2 = 1000 − 158 m
= = = 15.2 s ⇒ total time = (5.66 + 15.2) s = 20.9 s.
= = = When the first ball hits the ground, the second ball has been
falling for t2 = 0.63 s. The distance second ball has fallen during this time is
y = gt = = 1.95 m, so the second ball is 13 m − 1.95 m = 11.1 m
above the ground when the first ball lands.
(b) v (1st) = gt1 = 9.81 m/s2 i1.63 s = 16.0 m/s
v(2nd) = gt2 = 9.81 m/s2 i 0.63 s = 6.18 m/s
⇒ instantaneous velocity of the first ball relative to the second just before the first hits the ground
is: 16.0 m/s − 6.18 m/s = 9.8 m/s down.
(c) Both balls have same acceleration, (9.81 m/s2 down,) so the relative acceleration is zero.
33
2-116. Use v2 − v 2
0 = −2g(x − x0) with v = 0; g = 9.8 m/s2; v0 = 26 m/s
x − x0 =
2
(26m/s) =
34m
2 × 9.8m/s
2
−
The total time of flight for any particle of water is:
t = 0 2v
g
2 × 26m/s
9.8m/s
= 2
= 5.3 s = 8.84 × 10−2 min
The discharge rate is 280 l/min so the total amount of water in the air after 5.3 s is vol = 280 l/min
× 8.84 × 10−2 min
vol = 25l
2-117. Speed upon impact, v = 2gh = 2 × 9.8 × 56 = 33.1m/s
Average deceleration, a = v
t
= 33.1 =
2209 m/s2
0.015
2-118. The final speed of the part with acceleration g is v = 200 km/hr = 55.6 m/s.
t v v
The time with acceleration is 0
1 2
g
time period
2 2 2 2
h v v
0
55.6 (m/s)
1 2
g
2 2×9.81m/s
= 842 m. She falls the distance h2 with a constant speed of 55.6 m/s. The time for this part of the
fall is 2
2
842 m
55.6 m/s
t h
v
†2-119. (a) Δt = 1 s. The distance the first ball falls in that interval is 2
1
1
2
y = gt
1 (9.81 m/s2 )(1 s)2 4.90 m,
2
= = so the first ball is 13 m − 4.9 m = 8.1 m above the ground when
the second ball is released. The time for the first ball to fall the total distance of 13 m is
2 2×13m 1.63 s.
1 2
9.81 m/s
t h
g
1 1(9.81 m/s )(0.63 s)
2 2
2 2 2
2 2