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CHAPTER 2 MOTION ALONG A STRAIGHT LINE
Select odd-numbered solutions, marked with a dagger (†), appear in the Student SolutionsManual, available for purchase. Answers to all solutions below are underscored.
2-1. Time required, ∆t = distancespeed
= 30 0.3 s100
=
2-2. Avg speed = distancetime
= 100 yd9.0 s
× 1 mi1760 yd
× 3600s 23 mi/h1h
=
†2-3. 1 year = 3.156 × 107 sec, so 20 m/year = 77
20 m 6.34 × 101 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 is7 46.34 × 10 m/s × 8.64 × 10 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 = 33500 × 10 m 1× 81 days.
0.5 m/s 24 hr/day × 3600 s/hrdv
= ≈
2-5. 6 days 12 hrs = 156 hrs.dist. 5068 32.5 km/htime 156
ν = = =
2-6.9
7 15 7
1.4 × 10 ly2.16 × 10 m/s × 9.47 × 10 m/ly × 1/(3.16 × 10 )yr/s
dtν
= =
10= 1.9 × 10 yrt
2-7. Estimated distance (by sea) between Java and England is 20,000 km.20,000 km 600 km/h
32 hdνt
= = ≈
2-8. (a) 4000 nmiAverage speed 8.3 nmi/hr.20 days × 24 hr/day
= = (b) He must cover the remaining 1720
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 thehighest possible speed for the entire 7 days, he should probably conclude that he will not be ableto complete the trip within the 20-day limit.
2-9. 35 kmAverage speed 14 km/hr2.5 hr
dt
= = =
2-10.3110 km 110 × 10 mAverage speed 1.27 m/s.
1 day 24 h × 3600 sec= = = Burst speed = 32 km/hour =
332 × 10 m 8.9 m/s3600 s
=
†2-11. 41 2
1 km 1000 m 2.5 × 10 years.4 cm/year 4 × 10 m/year
dt tv −= ⇒ = = = Moving 1000 km will take
1000 times as long, or t2 = 2.5 × 107 years.
CHAPTER 2
13
2-12. 402 mAverage speed 16.9 m/s23.8 s
dt
= = =
2-13. 500 mAverage speed 12.8 m/s,39.10 s
= = or 12.8 × 10−3 km/m × 3600 s/hr = 46.0 km/hr
2-14. 23.8 m 0.326 s(263 km/hour × 1000 m/km / 3600 s/hr)
dtv
= = =
†2-15. Use the formula: .dtv
= 5280 km900 km/hrairt = = 5.87 hr. 5280 km
35 km/hourshiipt = = 151 hr.
2-16. 1100 m 9.49 m/s.10.54 s
v = = 2200 m 9.37 m/s21.34 s
v = =
2-17. 16 m/s × (s)100
d v t= = =i 0.06 m
2-18. Time taken for the arrow to reach the deer is50 m 10 s 0.77 s.
65 m/s 13dtv
= = = =
In this time the deer traveled from 40 m to 50 m, i.e., 10 m. Thus10 m 13 m/s.
(10/13s)dvt
= = =
†2-19. (a) Take x = 0 to be the cheetah’s starting position. Then the cheetah’s position is given by.c cx v t= The antelope’s starting position is 50 m from the cheetah’s starting position, so the
position of the antelope is given by 50.a ax v t= + When the cheetah catches the antelope, theirpositions are the same, and we get 50.c av t v t= + The speeds are νc = 101 km/h = 28.1 m/s and
va = 88 km/h = 24.4 m/s. Solving the equation for t gives 50 m 50 m28.1 m/s 24.4 m/sc a
tv 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 av t v t x= + but now weset 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 ax 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. 100 mAverage speed 10.1 m/s9.86 s
dt
= = =
†2-21. 3 426 mi × 1.6 × 10 m/mi 385 yd × 0.9144 m/yd 4.195 × 10 md = + =2 hr 24 min 52 s 2 hr × 3600 s/hr 24 min × 60 s / min 52 s 8692 st = = + + =
44.195 × 10 maverage speed8692 s
dt
= = = 4.83 m/s
2-22. 2 × 0.9 cm 0.094 cm/s60 s
secondsecond
second
dvt
π= = =
32 × 0.9 cm 1.6 × 10 cm/s60 min × 60 s/min
minuteminute
minute
dvt
π −= = =
CHAPTER 2
14
52 × 0.5 cm 7.3 × 10 cm/s12 hr × 60 min/hr × 60 s/min
hourhour
hour
dvt
π −= = =
†2-23. 24.0 0.50 .x t t= − To find the maximum value of x, differentiate with respect to t and set the
derivative equal to zero: 4.0 0.dx tdt
= − = 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 is24.0(4.0) 0.50(4.0) 8.0 m.x = − = At t = 8 seconds, x = 24.0(8.0) 0.5(8.0) 0;− = that is when he
comes back to the starting line. The total distance traveled is 16 m. Then average speed =distance 16 m
time 8 s= = 2.0 m/s.
2-24. Distance = 100 km time = 50 km 50 km 1.4660 km/hour 80 km/hr
+ = hr average speed
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 radiusMercury 3.64 × 108 7.61 × 106 47.8 1.68 8.56Venus 6.79 × 108 1.94 × 107 35.0 1.54 8.83Earth 9.42 × 108 3.16 × 107 29.8 1.47 8.97Mars 1.43 × 109 5.93 × 107 24.1 1.38 9.16Jupiter 4.89 × 109 3.76 × 108 13.0 1.11 9.69Saturn 8.98 × 109 9.31 × 108 9.65 0.985 9.95Uranus 1.80 × 1010 2.65 × 109 6.79 0.832 10.26Neptune 2.83 × 1010 5.21 × 109 5.43 0.735 10.45Pluto 3.71 × 1010 7.83 × 109 4.74 0.676 10.57
The slope of the line through the nine points is 1 .2
−
CHAPTER 2
15
This means that log speed 12
= − log radius + log C.
Therefore log speed = log[C(radius)−1/2
]. Thus1/2Speed = C(radius) where Cis some constant.−
2-26. Avg speed = distancetime
= 8 8 6.27 m/s2.55
+ =
Avg velocity = displacement 0 m/stime
=
2-27. Avg speed (for t = 0 to t = 10 s) = 200 20 m/s10
=
Avg speed (for t = 10 to t = 14.3 s) = 270 200 16.3 m/s14.3 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
= = The total change in position is
(12 floors 1 floors) × 4 m/floorx∆ = − = 44 m. The average velocity is44 m 0.88 m/s.50 s
xvt
∆= = =∆
†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
= = The total displacement is ∆x = (12 blocks − 6 blocks +
3 blocks) × 81 m/block = 729 m. The average velocity is 729 m 0.502 m/s.1453 s
xvt
∆= = =∆
2-30. 2 2( 0) 0; ( 8) 4 × 8 0.5 × 8 0; ( 10) 4 × 10 0.5 × 10 10x t x t x t= = = = − = = = − = − m.
Average velocity between t = 0 to t = 8.0 s is: ( 8) ( 0) 08 0
x t x t= − = =−
;
Average velocity between t = 8.0 s to t = 10.0 s is: ( 10) ( 8) 10 m 0 5.0 m/s.10 8 2 s
x t x ts s
= − = − −= = −−
†2-31. Total distance = 3 × 0.25 mile × 1609 m/mile 1207 m.= Displacement = 0 m because the horsereturns to the starting point. Total time = 1 min 40 s = 100 s. Then average speed =distance 1207 m 12.1 m/s.
time 100 s= = The average velocity is 0.xv
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
= = The total displacement is ∆x = 35 m − 22 m = 13 m. Then
average velocity is x 13 m 1.6 m/s.t 8.1 s
v ∆= = =∆
CHAPTER 2
16
†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.time 30 s
= = = The total
displacement is ∆x = 16 m, so the average velocity is 13 m 0.53 m/s.8.1 s
∆xv∆t
= = =
2-34. For 0 2 s,t≤ ≤ displacement = 25 m. displacement 25 mtime 2s
v = = = 12.5 m/s. For
2.0 4.0 s,t≤ ≤ displacement = 40 m − 25 m = 15 m. (This requires an estimate for the position
at 4.0 s. Your value may be slightly different.) displacement 15 m 7.5 m/s.time 2s
v = = =
To find the instantaneous velocity at any time, draw a tangent to the position vs time curve at thattime and determine the slope of the line. Your numbers may be slightly different from the onesgiven 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, or7.5 m/s. Again, your value may be slightly different if you estimate a different position at 4.0 s.
2-35. a = ∆ν/∆t ⇒ ∆ν = a ∆t = 282.6 gee × (9.807 m/s )
gee × 0.04 s32.4 m/sv∆ =
2-36. a = ∆ν/∆t = [(96 − 0) km/h]/2.2 s × 1 h/3600 s × 1000 m/km 212 m/s=
†2-37. 3 22 13
2 1
27 m/s 0 3.4 × 10 m/s .8.0 × 10 s
v vat t −
− −= = =−
2-38. 80 (km/hr) 22.22 m/s,iv = = 0, 2.8 s.fv t= =
20 22.22 m/s 7.94 m/s2.8 s
f iv va
t− −= = = −
2-39. (a) t(s) 2(m/s )a (in gees)a0 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 ∆ν and ∆t.30 0.56 0.057 iii) Convert from km/h to m/s.40 0.49 0.050
(b) t(s) 2(m/s )a (gees)a
0 −0.74 −0.07510 −0.44 −0.04520 −0.44 −0.04530 −0.31 −0.03240 −0.22 −0.022
2-40. 2 32.5 3.1 4.5 ;x t t t= + − 22.5 6.2 13.5 ;dxv t tdt
= = + − 6.2 27.0 .dva tdt
= = −
At t = 0 s, instantaneous velocity = 20| 2.5 6.2 × 0 13.5 × 0 2.5 m/s.t
dxdt = = + − = Instantaneous
acceleration = 6.2 m/s2. At t = 2. 0 s, instantaneous velocity =
CHAPTER 2
17
22| 2.5 6.2 × 2 13.5 × 2t
dxdt = = + − = −39.1 m/s (−39 m/s to two significant figures).
Instantaneous acceleration = 26.2 27.0(2) 48 m/s .− = −
For 0 2 s,t≤ ≤ 2 3( 2 ) ( 0) 2.5(2) 3.1(2) 4.5(2) 9.3 m/s.
2 0 2x t s x tv = − = + −= = = −
−( 2) ( 0) 39.1 m/s 2.5 m/s
2 0 2 sv t v ta = − = − −= = =
− – 21 m/s2.
†2-41. 2 33.6 2.4 ,x t t= − 27.2 7.2 .dxv t tdt
= = − 0v = if
27.2 7.2t t− = 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 1.52.4
= s.
Also, dx/dt = 0 when t = 0 and t = 1s.
2-42. 2.v Bt Ct= − V = 0 if t = 0 or t =2
3
6.0 m/s 3.0 s.2.0 m/s
BC
= = 2 .dva B Ctdt
= = − a = 0 if
1.5 s.2BtC
= = To make a sketch: consider the
above information, plus when t = 1.5 s, ν =2 3 2(6.0 m/s ) (1.5 s) (2.0 m/s ) (1.5 s)− = 4.5 m/s.
†2-43. For 0 5.0 s,t≤ ≤ ( 5 s) ( 0) 5 m/s 05 s 5 s
v t v ta = − = −= ≈ = 1 m/s2. (Your value may be slightly
different depending on how you read the values of ν at 0 and 5 s.) For 5.0 10.0 s,t≤ ≤
2( 10 s) ( 5 s) 9.5 m/s 5.0 m/s 0.9 m/s .5 s 5 s
v t v ta = − = −= ≈ = (Again, your value may be slightly
different depending on how you estimate the values of ν at 5 and 10 s.) To find the instantaneousacceleration at 3 s, draw a tangent to the curve at that time. Your estimate may be slightlydifferent 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 25 m/s 0 1.3 m/s .5 s 1 s
a −= =−
2-44. 0/ 2.5 / 2.50[ ( ) ] .
(2.5 s)ft s t s
f f
v vdv da v v v e edt dt
− −−= = + − = − At t = 0,
00
m/s 200 km/hr 18 km/hr|2.5 s 2.5 s
ft
v vdvadt =
− −= = − = −
2182 km/hr × 1000 m/km 20 m/s .2.5 s × 3600 s/hr
= − = −
CHAPTER 2
18
†2-45. 20 0 02 2 2 2 2
2( ) .(1 ) (1 ) (1 )
v v Av tdv d da Atdt dt At At dt At
⎡ ⎤= = = − = −⎢ ⎥+ + +⎣ ⎦
At t = 0, a = 0.
at t = 2 s, 2
222 2
2(25 m/s)(2 s )(2 s) 2.5 m/s .1 (2 s )(2 s)
a−
−= − = −
⎡ ⎤+⎣ ⎦ As ,t → ∞ 0
2 4
2 , so 0.Av ta aA t
→ − →
2-46. (a) Estimated average velocity (by mid-point method) is about 30 km/h. Because v = ∆x/∆t, we
have ∆x = v ∆t ≈ 30 km/h × 5 s × 5 m/s 42 m.18 km/h
≈
(b) Time interval (s) (km/h)v ∆x(m) 5 m / sby km / h × 5s ×18 km / h
v⎡ ⎤⎢ ⎥⎣ ⎦
5−10 70 10010−15 90 12015−20 110 15020−25 130 18025−30 140 19030−35 150 21035−40 160 22040−45 170 240
(c) Total distance traveled is the sum of the last column plus the 42 m traveled in the first5 s: 1450 m 1.45 km.d = =
†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 1890.6−0.9 611.5 1830.9−1.2 596.0 1791.2−1.5 579.5 1741.5−1.8 564.0 1691.8−2.1 549.5 1652.1−2.4 535.0 1612.4−2.7 521.0 1562.7−3.0 508.0 152
Total distance traveled = 1722 m
CHAPTER 2
19
(c) Counting the number of squares under the ν versus t curve gives the same answer withinabout ± 2 m.
2-48. ν = 655.9 − 61.14t + 3.26t2
acceleration, a = dvdt
= −61.14 + 6.52t
a |t = 0 s 261.14 m/s= −
a |t = 1.5 s = −61.14 + (6.52) (1.5) 251.36 m/s= −
a |t = 3.0 s = −61.14 + (6.52) (3.0) 241.58 m/s= −
†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 1.6 st = and 4.7 s.
ν = dx/dt = −2.0 sin tν (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 ta(π/2) = −2.0 cos(π/2) = 20 m / s
a(3π/2) = −2.0 cos(3π/2) = 20 m / s(c) Maximum distance achieved when cos t = ± 1, i.e., when t = 0, π, 2π, or 0s,3.1s and 6.3s.t =
v = −2.0 sin tv(0) = −2.0 sin(0) = 0 m/ s
v(π) = −2.0 sin(π) = 0 m/ s
v(2π) = −2.0 sin(2π) = 0 m/ s
a = dv/dt = −2.0 cos ta(0) = −2.0 cos(0) = 22.0 m / s−
a(π) = −2.0 cos(π) = 22.0 m/ s
a(2π) = −2.0 cos(2π) = 22.0 m / s−
2-50. x = uext + uex(1/b − t) ln(1 − bt )(a) Instantaneous velocity, v = dx/dtdxdt
= uex + uex ln(1 ) (1/ )1
bbt b tbt
−⎡ ⎤− − + −⎢ ⎥−⎣ ⎦ = uex
[1 − ln(1 − bt) − 1]
ln(1 )exv u bt= − −
(b) a = 2
2
d xdt
= −uex[−bt(1 − bt)] ex
1u b
bt=
−
CHAPTER 2
20
(c) dxdt
= −3.0 × 103 ln(1 − 7.5 × 10−3t) m/s
ν(t = 0) = −3.0 × 103 ln 1 = 0 m/s
ν(t = 120) = −3.0 × 103 ln(1 − 7.5 × 10−3 × 120) m/sν(120) = −3.0 × 103 ln(1 − 0.9) m/s = −3.0 × 103 × (−2.80) m/s
3(120) 6.9 × 10 m/sv =
(d) a = ddtυ =
2
2
d xdt
= 3 3
3
3.0 × 10 × 7.5 × 101 7.5 × 10 t
−
−−m/s2 = 3
22.51 7.5 × 10 t−−
m/s2
a(0) = 22.5 m/s2
a(120) = 3
22.51 7.5 × 10 × 120−−
m/s2 = 22.51 0.9−
m/s2
2(120) 225m/sa =
2-51. Use Equation (25): a(x − x0) = 2 20
1 ( )2
v v−
x − x0 = 2100 m; v = 360 km/h × 5 m/s18 km/h
= 100 m/s; v0 = 0
a = 12
× 2 2
0
0
v vx x
−−
= 12
× 2100
2100m/s2
22.4 m/sa =
2-52. Acceleration, a = 2 2
0
02( )v v
x x−−
= 2(657)
2 × 6.63 = 4 23.26 × 10 m/s
Time to travel the length of the barrel, t = 0v va−
t = 4
6573.26 × 10
= 22.02 × 10 s−
2-53. 4.2 ly = 4.2 ly × 9.46 × 1015 m/ly = 3.97 × 1016 m
Use x − x0 = ν0t + 21 ( )2
at with x0 = ν0 = 0 and x = 1 (4.2ly)2
t 1/2 = 2xa
= 16
2
3.97 × 10 m9.807 m / s
= 76.36 × 10 s
Because the magnitude of the acceleration is the same for both parts of the trip, the time for thesecond half is identical to that of the first half. Thus, the total time for the trip, T, isT = 2t 1/2 = 2 × 6.36 × 107 s = 81.3 × 10 s 4.0 yr≈
Speed at midpoint: Use a(x − x0) = 2 20
1 ( )2
v v−
ν0 = 0; x − x0 = 1.99 × 1016 m; a = 9.807 m/s2
Then22v = 2a(x − x0) = (2 × 9.807 × 1.99 × 1016) m2/s2
ν2 = 3.90 × 1017 m2/s2
86.2 × 10 m / s (This exceeds the speed of light!!)v =
CHAPTER 2
21
2-54. 2 20 0 0 02 ( ) 2 ( )a x x v v v a x x− = − ⇒ = − − From the information given in the problem, x − x0 =
290 m, and a = −10 m/s2. Then 20 2( 10 m/s )(290 m) 76 m/s,v = − − = which corresponds to
about 270 km/h, or 170 mph.
2-55. Use 2 20
1 ( )2
v v− = a(x − x0) with v = 0; ν0 = 80 km/h × 5 m/s18 km/h
= 22.2 m/s; x − x0 = 0.7 m.
Therefore a = 2 2 2
0
20
1 1( ) ( 22.2)2 2 .
0.7m/s
v v
x x
− −=
−2a 350m/s (will probably survive)= −
2-56. 2 20 0 0 02 ( ) 2 ( )a x x v v v a 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 20 2( 5 m/s )(9.6 × 10 m) 3.1 × 10 m/s,v = − − =
which corresponds to about 700 mph. The time to stop is 2
02
0 3.1 × 10 m/s 62 s.5 m/s
v vta− −= = =
−
†2-57. Use 2 20
1 ( )2
v v− = a(x − x0) with x − x0 = 50 m; v = 0;
v0 = 96 km/h = 26.67 m/s2 2 2
0
20
1 1( ) ( 26.67)2 2
50 m/s
v va
x x
− −= =
−27.1m/sa = −
Use 0v v at= + with v = 0 m/s, v0 = 96 km/h = 26.7 m/s, and a = −7.1 m/s2. Then
0 3.8 s.vta
= − =
2-58.2 2 3
0
0
(260 × 10 / 3600)2( ) 2 × 1500v va
x x−= = −−
2 = 1.74 m/s (The minus sign denotes deceleration.)a −
Use 0v v at= + with v = 0 m/s, v0 = 260 km/h = 72.2 m/s, and a = −1.74 m/s2. Then
0 42.5 s.vta
= − =
2-59. 20 1.5 m/s × 20 s 30 m/sv v at= + = =
2 2 20
1 10 × 1.5m/s × (20 s) 300 m.2 2
x v t at= + = + =
2-60. 2 20 0
1 11.0 m 5.0 m 3.0 m/s 4.0s (4.0s)2 2
x x v t at a− = + ⇒ − = +i i
⇒ −4 m = 12 m + 8 a ⇒ 22.0 m/s .a = − The velocity is 0v v at= + =23.0 m/s (2.0 m/s )(4.0s) 5.0 m/s.− = −
2-61. 2 20
1 12 2
x v t at at= + = (since v0 = 0). 2
2 2 × 150 m1.2 m/s
xta
= = = 16 s.
CHAPTER 2
22
2-62. 0550 km/hr × 1000 m/km550 km/hr
3600 s/hrv = = = 153 m/s. 0v v at= + =
2 2153 m/s (0.60 m/s )(90 s) 2.1 × 10 m/s.+ =
†2-63. The sketch should be based on the following:
For 0 6t≤ ≤ s, a = 3.0 m/s2; 0 0 3 ;v v at t= + = + 2
0 0
33 .2
t t
x v dt t dt t= = =∫ ∫ At t = 6s,
3 × 6 18 m/s;v = = 23 6 54 m.2
x = =i For 6 10 s,t≤ ≤ a = 4.5− m/s2, and
0 ( 6) 18 4.5 ( 6)v v a t t= + − = − − m/s = 45 − 4.5t m/s.
2 2 20 0
1 1( 6) ( 6) 54 18( 6) 4.5( 6) 2.25( 6) 18( 6) 542 2
x x v t a t t t t t= + − + − = + − − − = − − + − +i
At t = 10 s, 45 4.5 × 10 0;v = − = 22.25 4 18 4 54 90 m.x = − + + =i i
2-64.
2
2 2 22
0 01 700 m 0.5 × 0.05 m/s × 30 s22 30s
atxx v t at v
t
− −= + ⇒ = = = 22.6 m/s.
20 22.6 m/s 0.05 m/s 30sv v at= + = + =i 24 m/s.
†2-65.
2
2 2 22
0 01 550 m 0.5 × 0.5 m/s × 15 s2 32.92 15s
atxx v t at v
t
− −= + ⇒ = = = m/s.
20 32.9 m/s 0.5 m/s 15 sv v at= + = + =i 40.4 m/s.
2-66. Speed at the end of the 440-yard mark,
v = 250.69 × 1760 × 360 × 60
= 367.7 ft/s
CHAPTER 2
23
(a) Average acceleration, a = vt
∆∆
= 2367.7 65.23ft/s5.637
=
(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 = (5.637)2tv
vt = 2 × 440 × 35.637
= 463 ft/s 319 mi/h.=
2-67. The distance traveled by the elevator is2 21 2 3 3
1 12 2
x at vt vt at⎛ ⎞= + + −⎜ ⎟⎝ ⎠
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 = 12
a(5)2 + (a5)7 + (a5)5 − 12
a52
52.5 m = 60 a20.875m/sa =
The maximum speed of the elevator is vmax = at1
vmax = 0.875 m/s2 × 5 smax 4.4 m/sv =
2-68. (a) Average speed = 40055
= 7.3 m/s
(b) For minimum values of acceleration and deceleration, the elevator should travel half thedistance in half the time. Therefore
21 552002 2
a ⎛ ⎞= ⎜ ⎟⎝ ⎠
20.53m/sa = where a is the acceleration and −a is the deceleration.
Maximum speed is given by vmax = at = 0.53 × 55 14.6 m/s2
=
(This is twice the average speed. What is its significance?)
2-69. v0 (km/h) v0 (m/s) 0v t∆ (m) 20
2va
− (m)Total stoppingdistance (m)
15 4.17 8.3 1.1 9.430 8.33 16.7 4.3 21.045 12.5 25.0 10 35.060 26.7 33.3 18 51.375 20.8 41.7 27 68.790 25.0 50.0 39 89.0
CHAPTER 2
24
2-70. Table on Page 47:
0v t∆ = 20
2va
20 2 2 × 8 m/s × 0.75 sv at⇒ = = = 12 m/s.
2-71. 50 km/h = 50 × 518
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 isv = 02 ( )a x x− = 22(200 m/s )(0.610 m) 15.6 m/s.=
2-72. Consider the position of the car in the reference frame of the truckx0c = −12 − 17 = −29 m
The final position of the car is xc = 17 m, therefore,
xc = x0c + v0c t + 12
at2
17 = −29 + 0 + 12
at2
2146 .2
at= (i)
The final speed of the car relative to the truck is 24 km/h
= 324 × 10
60 × 60 = 6.67 m/s.
at = 6.67 m/s (ii)Divide equation (i) by (ii) to get 13.8t s= and
a = 26.67 0.48 m/s .13.8
=
†2-73. v = −gτ + gτe−t/τ
(a) acceleration, /tdv gedt
τ−=
(b)Lim
t → ∞e −t/τ = 0, therefore
Limt → ∞
v = −gτ + Lim
t → ∞e − t/τ gt= −
(c) v = ( )2 / 20xtd g t gt e g
dtττ τ−− − + +
= −gτ + g 2τ
τe − t/τ
/tg g e ττ τ −= − +
(d) for t << τ, the exponential e−t/τ can be expanded as
e−t/τ = 1 − tτ
+ 2
2
12
tτ
+ . . . .
Therefore, x = −gτ t − gτ2(1 − tτ
+ 2
2
12
tτ
) + gτ2 + x0
= −gτ t − gτ2 + gt τ − 12
gt2 + gτ2 + x0 2
012
gt x= − +
CHAPTER 2
25
2-74. x − x0 = v0t − 12
gt2 with x − x0 = −380 m; v0 = 0; g = 9.8 m/s2
−380 m = − 21 (9.8 m/s )2
t2
t2 = 77.6 s2
8.8st =
For the impact velocity use v = v0 − gtv = 0 − 9.8 m/s2 × 8.8 s 86 m/s= −
†2-75. 130 km/h = 36.1 m/s. The acceleration of the freely falling falcon is g, so use2 2 2
2 2 00 0 0 2
(36.1 m/s) 02 ( ) 66.5 m.2 2(9.81 m/s )
v vv v g x x x xg
− −− = − ⇒ − = = =
2-76. 2 20 2 .v v ax= + Here v0 = 0, a = g. 22 2(9.81m s )(8.7 m) 13 m/s.v gh= = =
2-77. 2 20 2 .v v ax= + Here v = 0, a = −g because the displacement is upward and the acceleration is
downward. 20 2 2( 9.81m s )(1.9 m) 6.1 m/s.v gh= − = − − =
2-78. mgh = 12
mv2, where h is the maximum height reached
h = 21
2vg
⎛ ⎞⎜ ⎟⎝ ⎠
= 2(366) 6834 m
2 × 9.8=
2-79. Neglecting air resistance, we have
x − x0 = v0t − 12
gt2 with v0 = 0; g = 9.8 m/s2; t = 3.0 s
x − x0 = 0 − 21 (9.8 m/s ) (3.0 s)2
2 44 m=
2-80. Use 2 20v v− = −2g(x − x0) with v = 0; g = 1.80 m/s2; x − x0 = 200,000 m
20v = 2 × 1.80 m/s2 × 2 × 105 m = 7.2 × 105 m2/s2
0 849 m/sv =
2-81. v = 45 km/h = 12.5 m/s. 2 2
2
(12.5 m/s) 1 floor8.4 m ×2 2(9.81 m/s ) 2.9 mvhg
= = = = 3 floors.
v = 75 km/h = 20.8 m/s. 2 2
2
(20.8 m/s) 1 floor22.1 m ×2 2(9.81 m/s ) 2.9 mvhg
= = = = 8 floors.
v = 105 km/h = 29.2 m/s. 2 2
2
(29.2 m/s) 1 floor43.4 m ×2 2(9.81 m/s ) 2.9 mvhg
= = = = 15 floors.
2-82. 2
2 2(10 m)9.81 m/s
htg
= = = 1.43 s. This is 1.43 s/(3 half turns) = 0.48 s per half turn.
†2-83. 2
2 2(9.5 m) 1.39 s.9.81 m/sup
htg
= = = Total time = 2(1.39 s) = 2.8 s.
20 (9.81 m/s )(1.4 s) 14 m/s.upv gt= = = This is the initial speed. The initial velocity is 14 m/s
up.
CHAPTER 2
26
2-84. v0 = 1.0 m/s downward, and the ball is initially some distance h above the ground. After fallingthat distance h, the ball will strike the ground with some speed v and will rebound (reverse itsdirection of motion) at the same speed if the collision with the ground is elastic. (This conceptwill be introduced in a later chapter.) Since it starts moving back up with the same speed it hadjust before it hit the ground, the time required to return to its starting point a distance h above thefloor will be the same as the time required for it to reach the floor in the first place, and it willarrive 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 0 ,2tv v g= + where t is the total travel time. Then
the height h can be found using 2 20 2 ,v v gh= + which gives
2 20 .
2v vh
g−= Substituting, we get
2 0.75 s1 m/s (9.81 m/s ) 4.68 m/s.2
v ⎛ ⎞= + =⎜ ⎟⎝ ⎠
Then 2 2
2
(4.68 m/s) (1 m/s)2(9.81 m/s )
h −= = 1.1 m.
Comment: This requires conservation of momentum! The ball will collide with the ground andrebound with a speed equal to its speed just before it hit, a topic that obviously isn’t covered inthis 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
2 2
0 09.2 m ( 9.81 m/s )(2.5 s)2 8.57 m/s,
2 2 2.5 s 2
atdat d atd v t vt t
− − −= + ⇒ = = − = − = or 8.6
m/s (to two significant figures). The height the penny reaches above its launch point is2 2 2 20 0 0
2
(8.57 m/s) 3.7 m.2 2( ) 2 2(9.81 (m/s )v v vha g g
= − = − = = =−
2-86. v0 = 0, a = −g, x = −h ⇒ 20 2 2 .v v ax gh= + = 0
2v vv += for constant acceleration, so
2.
2gh
v =
†2-87. Assume that the collision of the ball with the floor simply reverses the direction of the ball’svelocity. Then the time for the ball to reach the floor is the same as the time for the ball to returnto 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.2downt = = 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. 20
12
h v t gt= +
2 2 2
02 2(1.5 m) (9.81 m/s )(0.45 s)
2 2(0.45 s)h gtv
t− −
⇒ = = = 1.13 m/s, or 1.1 m/s to two significant
figures. The velocity just before hitting the floor is 0v v gt= + =21.13 m/s (9.81 m/s )(0.45 s)+ = 5.5 m/s.
CHAPTER 2
27
2-88. 21 .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 givesthe ratios 1:3:5 and so on.
2-89.2
0 .2
ath v t= + v0 = 0 2
2 .hat
⇒ =
2
2 hat∆∆ =
10 cm4 fingers ×4 fingers × 100% 0.22%.46 cm100 cubits ×
cubit
a ha h
∆ ∆= = =
2-90. Time taken to fall through a distance of 45 m,
y2 = y1 + v1t + 12
at2
0 = 45 + 0 − 1 (9.8)2
t2
t = 3.03 s
Since 12 ,ytg
= to measure y1 within 10%, the time must be known to within 5% or within 0.15 s.
Therefore a stopwatch must be used. An ordinary wristwatch will have an uncertainty of ±1 s.
2-91. Velocity of ball on impact 212 2(9.8 m/s ) (1.5 m)gx= − = − = −5.42 m/s
Velocity of ball after impact = 22gx = 22(9.8m/s ) (1.1m) = 4.64 m/svat
∆=∆
= 4 24
4.64 m / s ( 5.42m/s) 1.6 × 10 m/s6.2 × 10 s−
− − =
2-92. (a) Impact speed, v = 2gh = 2 × 9.8 × 96 = 43.4 m/s
(b) a = 2 22
22( )v v
x x−−
= 2
2(43.4) 254.5m / s2 × 3.7
− = −
(The minus sign denotes deceleration.)†2-93. The muzzle speed is given by 2 3 3
0 2 2(9.81 m/s )(180 × 10 m) 1.88 × 10 m/s.v gh= = = To
find out how long the projectile remains above 100 km, we can use the fact that the time for theprojectile to climb from 100 km to 180 km is the same as the time for it to fall from 180 km backto 100 km. So we can just calculate the time to fall a distance of 80 km from rest and double that
value. The time to fall can be calculated from 2 3
2
2 2(80 × 10 m) 128s,2 9.81 m/s
gt yy tg
= ⇒ = = =
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) 2980.6190 cm/s=
At the pole Θ = 90°, g = 978.0318 cm/s2 × (1 + 53.024 × 10−4) 2983.2177 cm/s=
CHAPTER 2
28
(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
( sin 2 4 sin 2 cos 2 )dg A B Cd
= Θ − Θ ΘΘ
= A (B − 4C cos 2 Θ)sin 2 Θ = 0 gives extrema at Θ = 0, ± 2π
To distinguish between extrema at Θ = 0, ± ,2π evaluate
2
2
d gd θΘ
at Θ = 0, ± 2π
2
02 θ 0(2 cos 2 8 cos 4 sin 4 )d g A B C
d = == −Θ Θ Θ Θ
Θ= 2AB > 0 (Thus at Θ = 0, g has a minimum.)Similarly
2
2θ / 2
2 0d g ABd π=±
= − <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 speedrelative to ground is −700 m/s − 83.3 m/s, or −783.3 m/s.
Use 2 20
1 ( )2
v v− = −g(x − x0) with g = 9.8 m/s2; x − x0 = −1500 m
v2 = 20v − 2g(x − x0) = (783.3)2 − 2 × 9.8(−1500 m) m2/s2
802 m/sv = ±
To find the time, use x − x0 = v0t − 12
gt2, with x − x0 = −1500 m; v0 = −783.3 m/s
−1500 m = 21783.3 9.82
t t⎛ ⎞− −⎜ ⎟⎝ ⎠
m
4.9t 2 + 783.3t − 1500 = 01.9t s=
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 is2
0 ,2p
gty y= − where y0 = 335 m. When the elevator
and penny meet, ye = yp 2
0 ,2
gtvt y⇒ = − which can be rearranged as 202 2 0.gt vt y+ − = The
root is 2 2 2
02
2 (2 ) 8 2(6.17 m/s) 4(6.17 m/s) 8(9.81 m/s )(335 m/s)2 2(9.81 m/s )
v v gyt
g− ± + − ± +
= =
t = 7.66 s (after dropping the negative root).
ye = yp 2
0 2gty= −
29.81 × 7.663352
= − = 47.2 m.
CHAPTER 2
29
†2-97. tup = 1.0 s, h = 10 m, a = g− = 9.81 m/s2.2 2 2
20 0
( 9.81 m/s )(1.0 s)101 2 22 1.0 s
ath mh v t at v
t
−− −= + ⇒ = = = 14.9 m/s. The impact speed is
20 14.9 m/s (9.81 m/s )(1.0 s)v v at= + = − = 5.1 m/s.
2-98. (a) The trajectory of the first stone is
x1 = x0 + v0t − 12
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. Thereforex2 = v0(t − 1) − 4.9(t − 1)2
We want the stones to collide at a height of 11 m. Thereforex1 = 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 0 1.84 or1.22st =
Using these values for t 0, we get for v0:
1.84 s: v0 = 2
02
0
11 4.9 ( 1) 17.2 m/s11 4.9 (0.84)1
0.84 m/s
t
t
+ − =+− =
1.22 s: v0 = 211 4.9(0.22) 51.1m/s
0.22+ =
The 1.84 s corresponds to hitting the stone on its way down, whereas the 1.22 s corresponds tohitting 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 mon the way up, so the collision can take place only on the way down.
v0 = 2
0
0
11 4.9( 1.30)1.30 m/s
tt+ −
− =
211 4.9 (1.84 1.30) 23.0 m/s1.84 1.30 m/s+ − =
−
2-99. (a) Time taken for drops to fall is 2 / .t h g= If there are n drops per second, the number ofdrops in the air at any one time is 1/ 2(2 / ) .nt n h g=
(b) Since the drops are falling at a constant rate, the median height is the distance fallen by a dropin 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 − 12
gt 2 = − 12
g(h/2g) = − 14
h
Thus the position of the median is 14
h below the edge of the spout
or 34
h above the ground.
CHAPTER 2
30
(c) The density of drops is proportional to 1/v, where v is the velocity of a particular drop. Thus
density 1 1 .2 ( )v g h x
=−
∼
The average height calculation must take into account the weighting factor of the density ofdrops. Thus
2 ( ).
2 ( )
x dxg h x
h dxg h x
−=
−
∫
∫Let ξ = h − x, so that
0
0
( )h
h
h d
h d
ξ ξξξξ
−
=∫
∫ = h −
3 / 2
1/ 2
[2 / 3 ]0
[2 ]0
h
h
ξ
ξ = h −
3 / 2
1/ 2
2 / 32
hh
= h − 13
h 2 .3
h=
2-100. 2.dva Ctdt
= = 3 3
2
8.0m/s 0 0
8.0 m/s 8.0 m/s3 3
tv t Ct Ctdv Ct dt v v= ⇒ − = ⇒ = +∫ ∫C = 0.25 m/s4 ⇒ 4 38.0 m/s (0.0833 m/s ) ,v t= + 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
(0.0833 m/s )8.0 m/s (0.0833 m/s) (8.0 m/s4
(8.0 m/s (0.0208 m/s )
t t tx vdt t dt )t
)t t
⎡ ⎤= = + = +⎣ ⎦
= +
∫ ∫
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
0 2(1 )4.0 s
dv ta adt
= = −
1s 2 31 s 2
0 0 02 20 0
1(1 ) ( ) | 20 m/s (1 ) s4.0 s 12.0 s 12
v t tv adt a dt a t= = − = − = − =∫ ∫ 18.3 m/s.
After 2 s, the acceleration becomes zero, so the velocity becomes constant at whatever value ithad at t = 2 s. So to find v after a long time (t >> 2 s), find its value at 2 s:
2 s 2 32 s 2
0 0 02 20 0
8(1 ) ( ) | 20 m/s (2 ) s 26.7 m/s4.0 s 12.0 s 12
v t tv adt a dt a t= = − = − = − =∫ ∫
The distance traveled is2 s 2 s 3 2 4 2 4
2 s 20 0 0 02 2 2
0 0
(2 s) (2 s)( ) ( ) | (20 m/s )12.0 s 2 48 s 2 48 s
t t tx x vdt a t dt a⎡ ⎤
− = = − = − = −⎢ ⎥⎣ ⎦
∫ ∫= 33.3 m.
2-102.2 s 2 3
2 2 s 2 2 2 4 3 30
0 0
1 1( ) | 15 m/s 2 s 25 m/s 2 s2 3 2 3
v At Btv adt At Bt dt= = + = + = +∫ ∫ i i i i
= 96.7 m/s
CHAPTER 2
31
Distance traveled 2 s 2 s 2 3 3 4
2 s1 s
1 s 1 s
( ) |2 3 6 12
At Bt At Btx vdt dt∆ = = + = +∫ ∫
= 3 3 3 4 4 415 m/s (2 1) s 25 m/s (2 1) s
6 12− −+ = 48.75 m
†2-103. dva g Avdt
= = − ⇒ dv dtg Av
=−
0
v
v
dv tg Av
⇒ =−∫
00 0
1 ln ( )At Atg Av g Avt e g Av g Av eA g Av g Av
− −− −⇒ − = ⇒ = ⇒ − = −
− −
0 01 [ ( ) ] (1 )At At Atgv g g Av e e v eA A
− − −⇒ = − − = − +
After a long time, 1/ ,t A>> 0,Ate− → .gvA
→
2-104. 2dva Bvdt
= = − 0
0
2
1 |v
vv
v
dv B t B tv v
⇒ = − ⋅ ∆ ⇒ − = − ∆∫
4 10
4 1
1 1 1 1 1 16.1 × 10 m 90 km/hr 120 km/hr
1 1 1 × 3600 s/hr6.1 × 10 m 1000 m/km 90 km/hr 120 km/hr
tB v v − −
− −
⎛ ⎞ ⎛ ⎞∆ = − = −⎜ ⎟ ⎜ ⎟
⎝ ⎠⎝ ⎠⎛ ⎞
= −⎜ ⎟⎝ ⎠i
= 16.4 seconds
†2-105. Cram’s speed is 1 mile 1 mile .3 min 46.32 s 226.32 sCv = = The time difference for the two runners
getting to the finish line is 3min 46.32sec 3min 44.39sect∆ = − = 1.93 s. So Cram is behind
by a distance of 3C
1 mile 1609 m× 1.93 s 8.53 × 10 mile × 13.7 m.226.32 s mile
v t −∆ = = =
2-106. distance 1500 1500 m 14.43 m/stime 1min 43.95sec 103.95 s
mv = = = =
2-107. Average speed = distance (100 100) m 1.3 m/stime (10 60 80) s
+= =+ +
Average velocity = 0, because the woman returns to her starting point.
2-108. (a) 1135 km/hr × 30 min 35 km/hr × × 30 min 17.5 km;
60 min/hrd = = =
2185 km/hr × 30 min 85 km/hr × × 30 min 42.5 km;
60 min/hrd = = =
(b) Average speed 1 2 (17.5 42.5) kmtime 60 min
d d+ += = = 1.0 km/min, or 60 km/hr
†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 Theseparation between the fish is 20 m, so the time for the sailfish to catch the mackerel is given by109 km/hr × 33 km/hr × 20 mt t− =
CHAPTER 2
32
20 m 20 m 3600 s/hr76 km/hr 76 km/hr 1000 m/km
t⇒ = = ii
= 0.95 s. During this time, the sailfish travels a
distance 109 km/hr × 1000 m/km × 0.95 s109 km/hr × 0.95 s 28.8 m.3600 s/hr
d = = =
2-110. Distance traveled by the first plane = d1 = 720 km/hr ,ti where t is measured beginning at 10:00.Distance traveled by the second plane = d2 = 640 km/hr ( 1 hr),t −i since that plane left one hour
later. 1 2 1286d d+ = km 720 640( 1) 1286 kmt t⇒ + − = (1286 640) km 1.42 hr1360 km/hr
t +⇒ = =
= 1 hr 25 min. The distance traveled by the first plane is 1 (720 km/hr)(1.42 hr) 1022 km,d = =so the planes meet 1022 km north of San Francisco. Since the total elapsed time is 1 hr 25 afterdeparture of the first plane, they meet at 11:25 AM.
†2-111. Distance that my car travels = 80 km/hr ti Distance that the other car travels = 50 km/hr .ti To gofrom 10 m behind the slower car to 10 m ahead of it requires traveling a total relative distance of10 m + 10 m + 4 m, because of the length of the car. Thus 80 km/hr ti − 50 km/hr ti = (10 + 10 +
4) m 24 m 24 m 3600 s/hr30 km/hr 30 km/hr 1000 m/km
t⇒ = = =ii
2.9 s.
2-112.2 2 2 2 2
3 20 (105 0) km /hr 1.72 × 10 km/hr2 2(3.2 km)
v vax
− −= = =
03 2
105 km/hr 0 0.0610 hr,1.72 × 10 km/hr
v vta− −= = = or 219 seconds.
†2-113. (a) 22.0 6.0 3.0 .x t t= + − At t = 0.50 s, 22.0 6.0 × 0.50 3.0 (0.50)x = + − i = 4.3 m.
(b) 6.0 6.0 .dxv tdt
= = − At t = 5.0 s, v = 6.0 6.0 0.50− =i 3.0 m/s
(c) (6.0 6.0 )dv da tdt dt
= = − = −6.0 m/s2 at all times.
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 ssx 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 itsacceleration 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
2 2p
p
atx = = = from the starting position. At this time the position of
the speeder is ,2 ,2 27.8 m/s × 18 s 500 m,s sx 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 upto the speeder. When the cruiser catches up to the speeder, both vehicles have traveled the samedistance from the point where the cruiser reached 120 km/hr. Mathematically this means(33.3 m/s) (27.8 m/s) 332 m,t t= + which gives t = 60 s. So the total time that has elapsed sincethe cruiser began pursuit is 70 s. During this time the cruiser traveled a total distance of
3168 m (33.3 m/s)(60 s) 2.18 × 10 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 0 1 (25 m/s)(0.75 s) 18.8 m.d v t= = = The remaining distance to the cow is d2
= 30 m − 18.8 m = 11.2 m. The car’s acceleration as it travels this distance is a = −8.0 m/s2. Its
CHAPTER 2
33
final speed when it hits the cow is given by2 2 2 2
0 2 0 22 2 (25 m/s) 2( 8.0 m/s)(11.2 m) 21.1 m/s,v v ad v v ad− = ⇒ = + = + − = or76 km/hr.
2-116. Use v2 − v 20 = −2g(x − x0) with v = 0; g = 9.8 m/s2; v0 = 26 m/s
x − x0 = 2
2
(26 m/s) 34m2 × 9.8 m/s
=−
The total time of flight for any particle of water is:
t = 02vg
= 2
2 × 26 m/s9.8 m/s
= 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 minvol 25 l=
2-117. Speed upon impact, v = 2gh = 2 × 9.8 × 56 33.1m/s=
Average deceleration, a = vt
= 233.1 2209 m/s0.015
=
2-118. The final speed of the part with acceleration g is v = 200 km/hr = 55.6 m/s.
The time with acceleration is 01 2
55.6 m/s 5.66 s.9.81 m/s
v vtg−= = = The distance she falls during this
time period 2 2 2 2
01 2
55.6 (m/s)2 2 × 9.81 m/s
v vhg
−= = = 158 m. The rest of the height is h2 = 1000 − 158 m
= 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 22
842 m55.6 m/s
htv
= = = 15.2 s ⇒ total time = (5.66 + 15.2) s = 20.9 s.
†2-119. (a) 1 s.t∆ = The distance the first ball falls in that interval is 21
12
y gt=
2 21 (9.81 m/s )(1 s) 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
1 2
2 2 × 13 m 1.63 s.9.81 m/s
htg
= = = 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 is2 2 2
2 21 1 (9.81 m/s )(0.63 s)2 2
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 = 29.81 m/s 1.63 s 16.0 m/s=iv(2nd) = gt2 = 29.81 m/s 0.63 s 6.18 m/s=i⇒ instantaneous velocity of the first ball relative to the second just before the first hits the groundis: 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.