in 2
Motion One Dimension
CHAPTER OUTLINE
2.1 2.2 2.3 2.4 2.5 2.6 2.7 Position, Velocity, and Speed Instantaneous Velocity and Speed Acceleration Motion Diagrams One-Dimensional Motion with Constant Acceleration Freely Falling Objects Kinematic Equations Derived from Calculus
ANSWERS TO QUESTIONS
Q2.1 If I count 5.0 s between lightning and thunder, the sound has traveled 331 m s 5.0 s = 1.7 km . The transit time for the light is smaller by
b
ga f
3.00 10 8 m s = 9.06 10 5 times, 331 m s so it is negligible in comparison. Q2.2 Q2.3 Q2.4 Yes. Yes, if the particle winds up in the +x region at the end. Zero. Yes. Yes.
Q2.5
No. Consider a sprinter running a straight-line race. His average velocity would simply be the length of the race divided by the time it took for him to complete the race. If he stops along the way to tie his shoe, then his instantaneous velocity at that point would be zero. We assume the object moves along a straight line. If its average x velocity is zero, then the displacement must be zero over the time interval, according to Equation 2.2. The object might be stationary throughout the interval. If it is moving to the right at first, it must later move to the left to return to its starting point. Its velocity must be zero as it turns around. The graph of the motion shown to the right represents such motion, as the initial and final positions are the same. In an x vs. t graph, the instantaneous velocity at any time t is the slope of the curve at that point. At t 0 in the graph, the slope of the curve is zero, and thus the instantaneous velocity at that time is also zero.
Q2.6
t0
t
FIG. Q2.6 Q2.7 Yes. If the velocity of the particle is nonzero, the particle is in motion. If the acceleration is zero, the velocity of the particle is unchanging, or is a constant.
21
22 Q2.8
Motion in One Dimension
Yes. If you drop a doughnut from rest v = 0 , then its acceleration is not zero. A common misconception is that immediately after the doughnut is released, both the velocity and acceleration are zero. If the acceleration were zero, then the velocity would not change, leaving the doughnut floating at rest in mid-air. No: Car A might have greater acceleration than B, but they might both have zero acceleration, or otherwise equal accelerations; or the driver of B might have tramped hard on the gas pedal in the recent past. Yes. Consider throwing a ball straight up. As the ball goes up, its v velocity is upward v > 0 , and its acceleration is directed down v0 a < 0 . A graph of v vs. t for this situation would look like the figure to the right. The acceleration is the slope of a v vs. t graph, and is always negative in this case, even when the velocity is positive.
a
f
Q2.9
Q2.10
a f
a
f
t FIG. Q2.10 Q2.11 (a) (d) (g) (h) Q2.12 Q2.13 Accelerating East Braking West (b) (e) Braking East Accelerating West (c) (f) Cruising East Cruising West
Stopped but starting to move East Stopped but starting to move West
No. Constant acceleration only. Yes. Zero is a constant. The position does depend on the origin of the coordinate system. Assume that the cliff is 20 m tall, and that the stone reaches a maximum height of 10 m above the top of the cliff. If the origin is taken as the top of the cliff, then the maximum height reached by the stone would be 10 m. If the origin is taken as the bottom of the cliff, then the maximum height would be 30 m. The velocity is independent of the origin. Since the change in position is used to calculate the instantaneous velocity in Equation 2.5, the choice of origin is arbitrary. Once the objects leave the hand, both are in free fall, and both experience the same downward acceleration equal to the free-fall acceleration, g. They are the same. After the first ball reaches its apex and falls back downward past the student, it will have a downward velocity equal to vi . This velocity is the same as the velocity of the second ball, so after they fall through equal heights their impact speeds will also be the same. With h = (a) (b) 1 2 gt , 2 1 2 g 0.707t . The time is later than 0.5t. 2 1 2 g 0.5t . The elevation is 0.75h, greater than 0.5h. 2
Q2.14 Q2.15
Q2.16
0.5 h =
a
f
The distance fallen is 0.25 h =
a f
Chapter 2
23
Q2.17
Above. Your ball has zero initial speed and smaller average speed during the time of flight to the passing point.
SOLUTIONS TO PROBLEMS
Section 2.1 P2.1 (a) (b) (c) *P2.2 (a) Position, Velocity, and Speed v = 2.30 m s v= v= v= x 57.5 m - 9.20 m = = 16.1 m s t 3.00 s x 57.5 m - 0 m = = 11.5 m s t 5.00 s x 20 ft 1m = t 1 yr 3.281 ft
FG H
IJ FG 1 yr IJ = 2 10 m s or in particularly windy times K H 3.156 10 s K x 100 ft F 1 m I F = v= GH 3.281 ft JK GH 3.1561 yr s IJK = 1 10 m s . t 1 yr 10
7 -7 7 -6
(b)
The time required must have been t = 3 000 mi 1 609 m x = v 10 mm yr 1 mi
FG H
IJ FG 10 mm IJ = KH 1 m K
3
5 10 8 yr .
P2.3
(a) (b) (c)
v= v= v=
x 10 m = = 5 ms t 2s 5m = 1.2 m s 4s x 2 - x1 5 m - 10 m = = -2.5 m s 4 s-2 s t 2 - t1 x 2 - x 1 -5 m - 5 m = = -3.3 m s 7 s-4 s t 2 - t1 x 2 - x1 0 - 0 = = 0 ms 8-0 t 2 - t1
(d)
v=
(e)
v=
P2.4
x = 10t 2 : For
af xamf
ts
= 2.0 = 40
2.1 44.1
3.0 90
(a) (b)
v= v=
x 50 m = = 50.0 m s t 1.0 s x 4.1 m = = 41.0 m s t 0.1 s
24 P2.5
Motion in One Dimension
(a)
Let d represent the distance between A and B. Let t1 be the time for which the walker has d the higher speed in 5.00 m s = . Let t 2 represent the longer time for the return trip in t1 d d d -3.00 m s = - . Then the times are t1 = and t 2 = . The average speed t2 5.00 m s 3.00 m s is:
b
g
b
g
v=
Total distance = Total time 2 15.0 m 2 s 2 8.00 m s
b
d 5.00 m s
d+d = + 3 .00d m s
g b
b8.00 m sgd g e15.0 m s j
2 2
2d
v= (b)
e
j=
3.75 m s
She starts and finishes at the same point A. With total displacement = 0, average velocity = 0 .
Section 2.2 P2.6 (a)
Instantaneous Velocity and Speed At any time, t, the position is given by x = 3.00 m s 2 t 2 .
i i 2 2
e Thus, at t = 3.00 s: x = e3.00 m s ja3.00 sf = e ja b
j
27.0 m .
(b)
At t f = 3.00 s + t : x f = 3.00 m s 2 3.00 s + t , or x f = 27.0 m + 18.0 m s t + 3.00 m s 2 t
f
2
g e
ja f
2
.
(c)
The instantaneous velocity at t = 3.00 s is: v = lim
t 0
F x - x I = lim e18.0 m s + e3.00 m s jtj = GH t JK
f i 2 t 0
18.0 m s .
P2.7
(a)
at ti = 1.5 s , x i = 8.0 m (Point A) at t f = 4.0 s , x f = 2.0 m (Point B) v= x f - xi t f - ti =
a2.0 - 8.0f m = - 6.0 m = a4 - 1.5f s 2.5 s
-2.4 m s
(b)
The slope of the tangent line is found from points C and D. tC = 1.0 s, x C = 9.5 m and t D = 3.5 s, x D = 0 ,
b
g
b
g
v -3.8 m s . (c) The velocity is zero when x is a minimum. This is at t 4 s .
FIG. P2.7
Chapter 2
25
P2.8
(a)
(b)
At t = 5.0 s, the slope is v
58 m 2.5 s 54 m At t = 4.0 s, the slope is v 3s 49 m At t = 3.0 s, the slope is v 3.4 s 36 m At t = 2.0 s , the slope is v 4.0 s a= v 23 m s 4.6 m s 2 t 5.0 s
23 m s . 18 m s . 14 m s . 9.0 m s .
(c) (d) P2.9 (a)
Initial velocity of the car was zero . v= (5 - 0 ) m (1 - 0) s = 5 ms
(b)
v=
(5 - 10) m (4 - 2) s
= -2.5 m s
(c)
v=
(5 m - 5 m) (5 s - 4 s) 0 - (-5 m) (8 s - 7 s )
= 0
(d) *P2.10
v=
= +5 m s
FIG. P2.9
Once it resumes the race, the hare will run for a time of t= x f - xi vx = 1 000 m - 800 m = 25 s . 8 ms
In this time, the tortoise can crawl a distance x f - xi = 0.2 m s ( 25 s)= 5.00 m .
a
f
26
Motion in One Dimension
Section 2.3 P2.11
Acceleration
Choose the positive direction to be the outward direction, perpendicular to the wall. v f = vi + at : a = v 22.0 m s - -25.0 m s = = 1.3410 4 m s 2 . t 3.50 10-3 s
a
f
P2.12
(a)
Acceleration is constant over the first ten seconds, so at the end, v f = vi + at = 0 + 2.00 m s 2 (10.0 s)= 20.0 m s . Then a = 0 so v is constant from t = 10.0 s to t = 15.0 s. And over the last five seconds the velocity changes to v f = vi + at = 20.0 m s + 3.00 m s 2 (5.00 s)= 5.00 m s .
c
h
c
h
(b)
In the first ten seconds, x f = x i + vi t + 1 2 1 2 at = 0 + 0 + 2.00 m s 2 (10.0 s) = 100 m . 2 2
c
h
Over the next five seconds the position changes to x f = xi + vi t + And at t = 20.0 s , x f = x i + vi t + *P2.13 (a) 1 2 1 2 at = 200 m + 20.0 m s (5.00 s)+ -3.00 m s 2 (5.00 s) = 262 m . 2 2 1 2 at = 100 m + 20.0 m s (5.00 s)+ 0 = 200 m . 2
a
f
a
f
c
h
The average speed during a time interval t is v =
distance traveled . During the first t quarter mile segment, Secretariat's average speed was v1 = 0.250 mi 1 320 ft = = 52.4 ft s 25.2 s 25.2 s
b35.6 mi hg .
During the second quarter mile segment, v2 = 1 320 ft = 55.0 ft s 24.0 s
b37.4 mi hg . b37.7 mi hg , b39.0 mi hg .
For the third quarter mile of the race, v3 = 1 320 ft = 55.5 ft s 23.8 s
and during the final quarter mile, v4 = continued on next page 1 320 ft = 57.4 ft s 23.0 s
Chapter 2
27
(b)
Assuming that v f = v 4 and recognizing that vi = 0 , the average acceleration during the race was a= v f - vi total elapsed time = 57.4 ft s - 0 = 0.598 ft s 2 . ( 25. 2 + 24.0 + 23.8 + 23.0) s
a (m/s2) 2.0 1.6 1.0
P2.14
(a)
Acceleration is the slope of the graph of v vs t. For 0 < t < 5.00 s, a = 0 . For 15.0 s < t < 20.0 s , a = 0 . For 5.0 s < t < 15.0 s , a = v f - vi t f - ti .
a=
8.00 - (-8.00) 15.0 - 5.00
0.0
t (s) 0 5 10 15 20
= 1.60 m s 2
We can plot a(t ) as shown. (b) a= (i) v f - vi t f - ti For 5.00 s < t < 15.0 s , ti = 5.00 s , vi = -8.00 m s , t f = 15.0 s v f = 8.00 m s a= (ii) v f - vi t f - ti =
FIG. P2.14
8.00 - -8.00 = 1.60 m s 2 . 15.0 - 5.00
a
f
ti = 0 , vi = -8.00 m s , t f = 20.0 s , v f = 8.00 m s a= v f - vi t f - ti = 8.00 - (-8.00) 20.0 - 0 = 0.800 m s 2
P2.15
x = 2.00 + 3.00t - t 2 , v = At t = 3.00 s : (a) (b) (c)
dx dv = 3.00 - 2.00t , a = = -2.00 dt dt
x = ( 2.00 + 9.00 - 9.00) m = 2.00 m v = (3.00 - 6.00) m s = -3.00 m s
a = -2.00 m s 2
28 P2.16
Motion in One Dimension
(a)
At t = 2.00 s , x = 3.00( 2.00) - 2.00( 2.00)+ 3.00 m = 11.0 m. At t = 3.00 s , x = 3.00 9.00 so v= x 24.0 m - 11.0 m = 13.0 m s . = 3.00 s - 2.00 s t
2
a f
2
- 2.00 3.00 + 3.00 m = 24.0 m
a f
(b)
At all times the instantaneous velocity is v= d 3.00t 2 - 2.00t + 3.00 = (6.00t - 2.00) m s dt
c
h
At t = 2.00 s , v = 6.00( 2.00)- 2.00 m s = 10.0 m s . At t = 3.00 s , v = 6.00(3.00)- 2.00 m s = 16.0 m s . (c) (d) P2.17 (a) (b) (c) (d) a= v 16.0 m s - 10.0 m s = = 6.00 m s 2 t 3.00 s - 2.00 s d (6.00 - 2.00)= 6.00 m s 2 . (This includes both t = 2.00 s and t = 3.00 s ). dt
At all times a = a=
v 8.00 m s = = 1.3 m s 2 t 6.00 s
Maximum positive acceleration is at t = 3 s, and is approximately 2 m s 2 . a = 0 , at t = 6 s , and also for t > 10 s . Maximum negative acceleration is at t = 8 s, and is approximately -1.5 m s 2 .
Section 2.4 P2.18 (a) (b) (c) (d) (e)
Motion Diagrams
continued on next page
Chapter 2
29
(f)
One way of phrasing the answer: The spacing of the successive positions would change with less regularity. Another way: The object would move with some combination of the kinds of motion shown in (a) through (e). Within one drawing, the accelerations vectors would vary in magnitude and direction.
Section 2.5 P2.19
One-Dimensional Motion with Constant Acceleration
From v 2 = vi2 + 2 ax , we have 10.97 10 3 m s f which is a = 2.79 10 4 times g .
c
h
2
= 0 + 2 a( 220 m) , so that a = 2.7410 5 m s 2
P2.20
(a) (b)
x f - xi = a=
1 1 vi + v f t becomes 40 m = vi + 2.80 m s (8.50 s) which yields vi = 6.61 m s . 2 2
c
h
a
f
v f - vi t
=
2.80 m s - 6.61 m s = -0.448 m s 2 8.50 s
P2.21
Given vi = 12.0 cm s when x i = 3.00 cm(t = 0) , and at t = 2.00 s , x f = -5.00 cm , x f - x i = vi t + 1 1 2 2 at : -5.00 - 3.00 = 12.0( 2.00)+ a( 2.00) 2 2 32.0 a =- = -16.0 cm s 2 . -8.00 = 24.0 + 2 a 2
*P2.22
(a)
Let i be the state of moving at 60 mi h and f be at rest
2 2 v xf = v xi + 2 a x x f - xi 2
d 0 = b60 mi hg
2
ax
mi fFGH 5 1280 ft IJK -3 600 mi F 5 280 ft I F 1 h I = G JG J = -21.8 mi h s 242 h H 1 mi K H 3 600 s K F 1 609 m IJ FG 1 h IJ = -9.75 m s = -21.8 mi h s G H 1 mi K H 3 600 s K + 2 a x 121 ft - 0
i
a
2
.
(b)
Similarly,
0 = 80 mi h ax = - (c)
b
6 400 5 280 422 3 600
b
b
g
2
+ 2 a x 211 ft - 0
g
g
a
f
mi h s = -22.2 mi h s = -9.94 m s 2 .
Let i be moving at 80 mi h and f be moving at 60 mi h .
d i b60 mi hg = b80 mi hg + 2a a211 ft - 121 ftf 2 800b5 280 g a =- mi h s = -22.8 mi h s = -10.2 m s 2a90fb3 600g
2 2 v xf = v xi + 2 a x x f - x i 2 2 x x
2
.
30 *P2.23
Motion in One Dimension
(a)
Choose the initial point where the pilot reduces the throttle and the final point where the boat passes the buoy: x i = 0 , x f = 100 m , v xi = 30 m s , v xf = ?, a x = -3.5 m s 2 , t = ? x f = xi + v xi t + 1 axt 2 : 2 100 m = 0 + 30 m s t +
2 2
a
f
1 -3.5 m s 2 t 2 2
c
h
c1.75 m s ht - a30 m sft + 100 m = 0 .
We use the quadratic formula: t= -b b 2 - 4ac 2a
t=
30 m s 900 m 2 s 2 - 4 1.75 m s 2 (100 m) 2 1.75 m s
c
c
2
h
j
h
=
30 m s 14.1 m s 3.5 m s 2
= 12.6 s or 4.53 s .
The smaller value is the physical answer. If the boat kept moving with the same acceleration, it would stop and move backward, then gain speed, and pass the buoy again at 12.6 s. (b) P2.24 (a) v xf = v xi + a x t = 30 m s - 3.5 m s 2 4.53 s = 14.1 m s Total displacement = area under the v , t curve from t = 0 to 50 s. 1 50 m s 15 s + 50 m s 40 - 15 s 2 1 + 50 m s 10 s 2 x = 1 875 m x = (b) From t = 10 s to t = 40 s , displacement is 1 50 m s + 33 m s 5 s + 50 m s 25 s = 1 457 m . 2 v (50 - 0) m s 0 t 15 s : a1 = = = 3.3 m s 2 t 15 s - 0 15 s < t < 40 s : a 2 = 0 x =
e
a f ga
b b
ga f b ga f
f
b
ga f b
ga f
FIG. P2.24
(c)
40 s t 50 s : a 3 =
continued on next page
v (0 - 50) m s = = -5.0 m s 2 t 50 s - 40 s
Chapter 2
31
(d)
(i) (ii) (iii)
x1 = 0 +
1 1 a1 t 2 = 3.3 m s 2 t 2 or x1 = 1.67 m s 2 t 2 2 2
c
h
c
h
1 x 2 = (15 s) 50 m s - 0 + 50 m s (t - 15 s) or x 2 = 50 m s t - 375 m 2 For 40 s t 50 s , x3 = or x 3 = 375 m + 1 250 m + which reduces to x 3 = 250 m s t - 2.5 m s 2 t 2 - 4 375 m . 1 -5.0 m s 2 t - 40 s 2
a
f
a
f
FG area under v vs t IJ + 1 a (t - 40 s) + a50 m sf(t - 40 s) H from t = 0 to 40 sK 2
3 2
e
ja
f + b50 m sgat - 40 sf
2
b
g e
j
(e) P2.25 (a)
v=
total displacement 1 875 m = = 37.5 m s total elapsed time 50 s
Compare the position equation x = 2.00 + 3.00t - 4.00t 2 to the general form x f = xi + vi t + 1 2 at 2
to recognize that x i = 2.00 m, vi = 3.00 m s, and a = -8.00 m s 2 . The velocity equation, v f = vi + at , is then v f = 3.00 m s - 8.00 m s 2 t . The particle changes direction when v f = 0 , which occurs at t = time is: x = 2.00 m + 3.00 m s 3 s . The position at this 8
c
h
a
fFGH 3 sIJK - c4.00 m s hFGH 3 sIJK 8 8
2
2
= 2.56 m .
(b)
From x f = xi + vi t +
2v 1 2 at , observe that when x f = xi , the time is given by t = - i . Thus, a 2 when the particle returns to its initial position, the time is t= -2 3.00 m s -8.00 m s
2
a
f=3 s
4 -3.00 m s .
and the velocity is v f = 3.00 m s - 8.00 m s 2
c
hFGH 3 sIJK = 4
32 *P2.26
Motion in One Dimension
The time for the Ford to slow down we find from x f = xi + 1 v xi + v xf t 2 2 250 m 2 x t= = = 6.99 s . v xi + v xf 71.5 m s + 0
d
a
i
f
Its time to speed up is similarly t= 2(350 m) 0 + 71.5 m s = 9.79 s .
The whole time it is moving at less than maximum speed is 6.99 s + 5.00 s + 9.79 s = 21.8 s . The Mercedes travels x f = xi + 1 1 v xi + v xf t = 0 + 71.5 + 71.5 m s 21.8 s 2 2 = 1 558 m
d
i
a
fb ga
f
while the Ford travels 250 + 350 m = 600 m, to fall behind by 1 558 m - 600 m = 958 m . P2.27 (a) vi = 100 m s , a = -5.00 m s 2 , v f = vi + at so 0 = 100 - 5t , v 2 = vi2 + 2 a x f - xi so f 0 = (100 ) - 2(5.00) x f - 0 . Thus x f = 1 000 m and t = 20.0 s . (b) P2.28 (a) At this acceleration the plane would overshoot the runway: No . Take ti = 0 at the bottom of the hill where x i = 0 , vi = 30.0 m s, a = -2.00 m s 2 . Use these values in the general equation x f = xi + vi t + to find 1 2 at 2
2 2 2
c
h
c
h
x f = 0 + 30.0t m s + when t is in seconds
a
f 1 c-2.00 m s ht 2 h
x f = 30.0t - t 2 m . To find an equation for the velocity, use v f = vi + at = 30.0 m s + -2.00 m s 2 t , v f = (30.0 - 2.00t ) m s . (b) The distance of travel x f becomes a maximum, x max , when v f = 0 (turning point in the motion). Use the expressions found in part (a) for v f to find the value of t when x f has its maximum value: From v f = (3.00 - 2.00t ) m s , v f = 0 when t = 15.0 s. Then x max = 30.0t - t 2 m = (30.0)(15.0)-(15.0) = 225 m .
c
e
j
c
h
2
Chapter 2
33
P2.29
In the simultaneous equations:
R | Sx | T
v xf = v xi + a x t
f
- xi =
1 v xi + v xf 2
c
U | | we have Rv = v - c5.60 m s h(4.20 s)U . S 62.4 m = 1 v + v (4.20 s) | V V t| hW | | c h 2 T W
xf xi 2 xi xf
So substituting for v xi gives 62.4 m =
1 v xf + 56.0 m s 2 ( 4.20 s)+ v xf ( 4.20 s) 2
c
h
14.9 m s = v xf + Thus
1 5.60 m s 2 ( 4.20 s). 2
c
h
v xf = 3.10 m s .
P2.30
Take any two of the standard four equations, such as substitute into the other: v xi = v xf - a x t x f - xi = Thus
R | Sx | T
v xf = v xi + a x t
f
- xi =
1 v xi + v xf 2
c
U |. Solve one for v ht V | W
xi ,
and
1 v xf - a x t + v xf t . 2
c
h
1 x f - xi = v xf t - a x t 2 . 2 Back in problem 29, 62.4 m = v xf ( 4.20 s)- v xf = v f - vi t 632 1 2 -5.60 m s 2 ( 4. 20 s) 2
c
h
62.4 m - 49.4 m = 3.10 m s . 4.20 s
P2.31
(a)
a=
=
1.40
e j=
5 280 3 600
-662 ft s 2 = -202 m s 2
(b)
x f = vi t +
5 280 1 2 1 1.40 - 662 1.40 at = 632 2 3 600 2
a
fFGH
I a f a fa f JK
2
= 649 ft = 198 m
34 P2.32
Motion in One Dimension
(a)
The time it takes the truck to reach 20.0 m s is found from v f = vi + at . Solving for t yields t= The total time is thus 10.0 s + 20.0 s + 5.00 s = 35.0 s . v f - vi a = 20.0 m s - 0 m s 2.00 m s 2 = 10.0 s .
(b)
The average velocity is the total distance traveled divided by the total time taken. The distance traveled during the first 10.0 s is x 1 = vt =
FG 0 + 20.0 IJ(10.0)= 100 m . H 2 K
With a being 0 for this interval, the distance traveled during the next 20.0 s is x 2 = vi t + 1 2 at = ( 20.0)( 20.0)+ 0 = 400 m. 2
The distance traveled in the last 5.00 s is x 3 = vt =
FG 20.0 + 0 IJ(5.00)= 50.0 m. H 2 K
The total distance x = x1 + x 2 + x 3 = 100 + 400 + 50 = 550 m , and the average velocity is x 550 = 15.7 m s . given by v = = t 35.0 P2.33 We have vi = 2.00 10 4 m s, v f = 6.00 10 6 m s , x f - xi = 1.50 10-2 m . 2 1.50 10-2 m 2 x f - xi 1 = = 4.98 10-9 s vi + v f t : t = 4 6 2 vi + v f 2.00 10 m s + 6.00 10 m s
(a)
x f - xi =
c
h
c
h
c
h
(b)
v 2 = vi2 + 2 a x x f - xi : f v 2 - vi2 f 2( x f - xi )
d
i
ax =
e6.00 10 =
6
ms
j - e2.00 10
2
4
ms
j
2
2(1.50 10 -2 m)
= 1.20 10 15 m s 2
Chapter 2
35
*P2.34
(a)
2 2 v xf = v xi + 2 a x x f - x i : 0.01 3 10 8 m s
c
h
c
h
2
= 0 + 2 a x ( 40 m)
ax (b)
c310 = i
6
ms
h
2
80 m
= 1.12 10 11 m s 2
We must find separately the time t1 for speeding up and the time t 2 for coasting: x f - xi = 1 1 v xf + v xi t1 : 40 m = 3 10 6 m s + 0 t1 2 2 t1 = 2.67 10 -5 s
d
e
j
x f - xi =
1 1 v xf + v xi t 2 : 60 m = 3 10 6 m s + 3 10 6 m s t 2 2 2 t 2 = 2.00 10 -5 s
d
i
e
j
total time = 4.67 10-5 s . *P2.35 (a) Along the time axis of the graph shown, let i = 0 and f = t m . Then v xf = v xi + a x t gives v c = 0 + am tm am = (b) The displacement between 0 and t m is x f - xi = v xi t + The displacement between t m and t 0 is x f - xi = v xi t + The total displacement is x = (c) 1 1 v c t m + v c t 0 - v c t m = v c t 0 - tm 2 2 1 a x t 2 = v c t0 - tm + 0 . 2 1 vc 2 1 1 axt 2 = 0 + t m = v c tm . 2 tm 2 2 vc . tm
a
f
FG H
IJ K
.
For constant v c and t 0 , x is minimized by maximizing t m to t m = t 0 . Then v t 1 x min = v c t 0 - t 0 = c 0 . 2 2
FG H
IJ K
(e) (d) (e)
This is realized by having the servo motor on all the time. We maximize x by letting t m approach zero. In the limit x = v c t 0 - 0 = v c t 0 . This cannot be attained because the acceleration must be finite.
a
f
36 *P2.36
Motion in One Dimension
Let the glider enter the photogate with velocity vi and move with constant acceleration a. For its motion from entry to exit, x f = xi + v xi t + 1 axt 2 2 1 2 = 0 + vi t d + at d = v d t d 2 1 v d = vi + at d 2
(a)
The speed halfway through the photogate in space is given by
2 v hs = vi2 + 2 a
FG IJ = v H 2K
2 i
+ av d t d .
v hs = vi2 + av d t d and this is not equal to v d unless a = 0 . (b) The speed halfway through the photogate in time is given by v ht = vi + a equal to v d as determined above. P2.37 (a) Take initial and final points at top and bottom of the incline. If the ball starts from rest, vi = 0 , a = 0.500 m s 2 , x f - xi = 9.00 m . Then v 2 = vi2 + 2 a x f - xi = 0 2 + 2 0.500 m s 2 9.00 m f v f = 3.00 m s . (b) x f - x i = vi t + 1 2 at 2
FG t IJ and this is H 2K
d
d
i
e
ja
f
9.00 = 0 +
1 0.500 m s 2 t 2 2 t = 6.00 s
e
j
(c)
Take initial and final points at the bottom of the planes and the top of the second plane, respectively: vi = 3.00 m s, v f = 0 , x f - xi = 15.00 m. v 2 = vi2 + 2 a x f - xi gives f a= v 2 - vi2 f 2 x f - xi
c
h
c
h
=
0 - 3.00 m s 2(15.0 m)
a
f
2
= -0.300 m s 2 .
(d)
Take the initial point at the bottom of the planes and the final point 8.00 m along the second: vi = 3.00 m s, x f - xi = 8.00 m , a = -0.300 m s 2 v 2 = vi2 + 2 a x f - xi = 3.00 m s f v f = 2.05 m s .
d
i b
g + 2e-0.300 m s ja8.00 mf = 4.20 m
2 2
2
s2
Chapter 2
37
P2.38
Take the original point to be when Sue notices the van. Choose the origin of the x-axis at Sue's car. For her we have x is = 0 , vis = 30.0 m s , a s = -2.00 m s 2 so her position is given by x s (t )= x is + vis t + 1 1 a s t 2 = 30.0 m s t + -2.00 m s 2 t 2 . 2 2
a
f
c
h
For the van, x iv = 155 m, viv = 5.00 m s , a v = 0 and x v (t xiv )= + viv t + 1 a v t 2 = 155 + 5.00 m s t + 0 . 2
a
f
To test for a collision, we look for an instant t c when both are at the same place:
2 30.0t c - t c = 155 + 5.00t c 2 0 = t c - 25.0t c + 155 .
From the quadratic formula 25.0 ( 25.0) - 4(155) 2
2
tc =
= 13.6 s or 11.4 s .
The smaller value is the collision time. (The larger value tells when the van would pull ahead again if the vehicles could move through each other). The wreck happens at position 155 m + 5.00 m s (11.4 s)= 212 m . *P2.39 As in the algebraic solution to Example 2.8, we let t represent the time the trooper has been moving. We graph x car = 45 + 45t and x (km) 1.5 car 1 police officer t (s)
a
f
x trooper = 1.5t .
They intersect at t = 31 s .
2
0.5
10
20
30
40
FIG. P2.39
38
Motion in One Dimension
Section 2.6 P2.40
Freely Falling Objects
Choose the origin y = 0 , t = 0 at the starting point of the ball and take upward as positive. Then yi = 0 , vi = 0 , and a = -g = -9.80 m s 2 . The position and the velocity at time t become: y f - yi = vi t + and v f = vi + at : v f = -gt = - 9.80 m s 2 t . (a) at t = 1.00 s : y f = - 1 2 9.80 m s 2 (1.00 s) = -4.90 m 2 1 2 at t = 2.00 s : y f = - 9.80 m s 2 ( 2.00 s) = -19.6 m 2 1 2 at t = 3.00 s : y f = - 9.80 m s 2 (3.00 s) = -44.1 m 2 1 1 1 2 at : y f = - gt 2 = - 9.80 m s 2 t 2 2 2 2
a
f
e
j
c
h
c c c
h h h
(b)
at t = 1.00 s : v f = - 9.80 m s 2 (1.00 s)= -9.80 m s at t = 2.00 s : v f at t = 3.00 s : v f
2 2
c h = -c9.80 m s h( 2.00 s)= -19.6 m s = -c9.80 m s h(3.00 s)= -29.4 m s
P2.41
Assume that air resistance may be neglected. Then, the acceleration at all times during the flight is that due to gravity, a = -g = -9.80 m s 2 . During the flight, Goff went 1 mile (1 609 m) up and then 1 mile back down. Determine his speed just after launch by considering his upward flight: v 2 = vi2 + 2 a y f - yi : f
d
i
0 = vi2 - 2 9.80 m s 2 1 609 m vi = 178 m s .
e
jb
g
His time in the air may be found by considering his motion from just after launch to just before impact: y f - yi = vi t + 1 1 2 at : 0 = 178 m s t - -9.80 m s 2 t 2 . 2 2
a
f
c
h
The root t = 0 describes launch; the other root, t = 36.2 s , describes his flight time. His rate of pay may then be found from pay rate = $1.00 = 0.027 6 $ s 3 600 s h = $99.3 h . 36.2 s
b
gb
g
We have assumed that the workman's flight time, "a mile", and "a dollar", were measured to threedigit precision. We have interpreted "up in the sky" as referring to the free fall time, not to the launch and landing times. Both the takeoff and landing times must be several seconds away from the job, in order for Goff to survive to resume work.
Chapter 2
39
P2.42
1 We have y f = - gt 2 + vi t + yi 2 0 = - 4.90 m s 2 t 2 - 8.00 m s t + 30.0 m . Solving for t, t= 8.00 64.0 + 588 . -9.80
c
h a
f
Using only the positive value for t, we find that t = 1.79 s . P2.43 (a) (b) y f - yi = vi t + 1 2 2 at : 4.00 = (1.50)vi -(4.90)(1.50) and vi = 10.0 m s upward . 2
v f = vi + at = 10.0 -(9.80)(1.50) = -4.68 m s v f = 4.68 m s downward
P2.44
The bill starts from rest vi = 0 and falls with a downward acceleration of 9.80 m s 2 (due to gravity). Thus, in 0.20 s it will fall a distance of y = vi t - 1 2 2 gt = 0 - 4.90 m s 2 (0. 20 s) = -0.20 m . 2
c
h
This distance is about twice the distance between the center of the bill and its top edge 8 cm . Thus, David will be unsuccessful . *P2.45 (a) From y = vi t + 1 2 at with vi = 0 , we have 2 t= 2 y a
a
f
a f=
h
2(-23 m) -9.80 m s 2
= 2.17 s .
(b) (c)
The final velocity is v f = 0 + -9.80 m s 2 ( 2.17 s)= -21.2 m s . The time take for the sound of the impact to reach the spectator is t sound = y v sound = 23 m = 6.76 10-2 s , 340 m s
c
so the total elapsed time is t total = 2.17 s + 6.76 10 -2 s 2.23 s .
40 P2.46
Motion in One Dimension
At any time t, the position of the ball released from rest is given by y1 = h - position of the ball thrown vertically upward is described by y 2 = vi t -
1 2 gt . The time at which the 2 h 1 h first ball has a position of y1 = is found from the first equation as = h - gt 2 , which yields 2 2 2 h h . To require that the second ball have a position of y 2 = at this time, use the second t= g 2 equation to obtain ball as vi = gh . P2.47 (a) v f = vi - gt : v f = 0 when t = 3.00 s , g = 9.80 m s 2 . Therefore, vi = gt = 9.80 m s 2 (3.00 s)= 29.4 m s . (b) y f - yi = 1 v f + vi t 2 h h 1 h = vi - g . This gives the required initial upward velocity of the second 2 g 2 g
1 2 gt . At time t, the 2
F I GH JK
c
h
c
h
y f - yi = *P2.48 (a)
1 29.4 m s 3.00 s = 44.1 m 2
b
ga
f
Consider the upward flight of the arrow.
2 2 v yf = v yi + 2 a y y f - yi 2
d i 0 = b100 m sg + 2e -9.8 m s jy
2
y = (b)
10 000 m 2 s 2 19.6 m s 2
= 510 m
Consider the whole flight of the arrow. y f = yi + v yi t + 1 ayt 2 2
0 = 0 + 100 m s t +
b
g
1 -9.8 m s 2 t 2 2
e
j
The root t = 0 refers to the starting point. The time of flight is given by t= 100 m s 4.9 m s 2 = 20.4 s . 1 9.80 m s 2 t 2 , t = 0.782 s. 2
P2.49
Time to fall 3.00 m is found from Eq. 2.12 with vi = 0 , 3.00 m = (a) (b)
c
h
With the horse galloping at 10.0 m s, the horizontal distance is vt = 7.82 m . t = 0.782 s
Chapter 2
41
P2.50
Take downward as the positive y direction. (a) While the woman was in free fall, y = 144 ft , vi = 0 , and a = g = 32.0 ft s 2 . Thus, y = vi t + before impact is: 1 2 at 144 ft = 0 + 16.0 ft s 2 t 2 giving t fall = 3.00 s . Her velocity just 2
c
h
v f = vi + gt = 0 + 32.0 ft s 2 (3.00 s)= 96.0 ft s . (b) While crushing the box, vi = 96.0 ft s , v f = 0 , and y = 18.0 in. = 1.50 ft . Therefore, a= v 2 - vi2 f 2 y
c
h
a f
=
0 - 96.0 ft s 2(1.50 ft )
a
f
2
= -3.07 10 3 ft s 2 , or a = 3.07 10 3 ft s 2 upward .
(c) P2.51
Time to crush box: t =
2(1.50 ft) y y = v +v = or t = 3.13 10-2 s . f i v 0 + 96.0 ft s
2 3
y = 3.00t 3 : At t = 2.00 s , y = 3.00 2.00
a f
= 24.0 m and dy = 9.00t 2 = 36.0 m s . dt 1 1 2 gt = 24.0 + 36.0t - (9.80)t 2 . 2 2
vy =
A
If the helicopter releases a small mailbag at this time, the equation of motion of the mailbag is y b = y bi + vi t - Setting y b = 0 , 0 = 24.0 + 36.0t - 4.90t 2 . Solving for t, (only positive values of t count), t = 7.96 s . *P2.52 Consider the last 30 m of fall. We find its speed 30 m above the ground: y f = yi + v yi t + 1 ayt 2 2
0 = 30 m + v yi 1.5 s + v yi =
-30 m + 11.0 m = -12.6 m s . 1.5 s
a f 1 e-9.8 m s ja1.5 sf 2
2
2
Now consider the portion of its fall above the 30 m point. We assume it starts from rest
i b-12.6 m sg = 0 + 2e-9.8 m s jy
2 2 v yf = v yi + 2 a y y f - yi 2 2
d
y =
160 m 2 s 2 -19.6 m s 2
= -8.16 m .
Its original height was then 30 m + -8.16 m = 38.2 m .
42
Motion in One Dimension
Section 2.7 P2.53 (a)
Kinematic Equations Derived from Calculus J= da = constant dt da = Jdt
a = J dt = Jt + c 1
but a = ai when t = 0 so c 1 = ai . Therefore, a = Jt + ai dv dt dv = adt a=
z
v = adt =
z zb
v=
Jt + ai dt =
g
1 2 Jt + ai t + c 2 2
but v = vi when t = 0, so c 2 = vi and v = dx dt dx = vdt
1 2 Jt + ai t + vi 2
x = vdt = x=
z z FGH
1 2 Jt + ai t + vi dt 2
IJ K
1 3 1 2 Jt + ai t + vi t + c 3 6 2 x = xi when t = 0, so c 3 = xi . Therefore, x = (b) 1 3 1 2 Jt + ai t + vi t + xi . 6 2
a 2 = Jt + ai
a
a 2 = ai2 + J 2 t 2 + 2 Jai t
c
f
2
= J 2 t 2 + ai2 + 2 Jai t
1 2 a 2 = ai2 + 2 J Jt + ai t 2
FG H
h
IJ K
1 2 1 Jt + ai t + vi . So v - vi = Jt 2 + ai t . Therefore, 2 2
Recall the expression for v: v =
a
f
a 2 = ai2 + 2 J v - vi
a
f
.
Chapter 2
43
P2.54
(a)
See the graphs at the right. Choose x = 0 at t = 0. At t = 3 s, x = 1 8 m s (3 s)= 12 m . 2
a
f
At t = 5 s, x = 12 m + 8 m s ( 2 s)= 28 m . At t = 7 s, x = 28 m + (b) For 0 < t < 3 s, a = 1 8 m s ( 2 s)= 36 m . 2
a
f
a
f
8 ms = 2.67 m s 2 . 3s For 3 < t < 5 s, a = 0 . 16 m s = -4 m s 2 . 4s
(c) (d) (e)
For 5 s < t < 9 s , a = -
At t = 6 s, x = 28 m + 6 m s (1 s)= 34 m . At t = 9 s, x = 36 m + 1 -8 m s ( 2 s)= 28 m . 2
a
f
a
f
FIG. P2.54
P2.55
(a)
a=
dv d = -5.00 10 7 t 2 + 3.00 10 5 t dt dt a = - 10.0 10 7 m s 3 t + 3.00 10 5 m s 2
c
h
Take x i = 0 at t = 0. Then v =
t
dx dt
t
x - 0 = vdt =
0
z ze
0
-5.00 10 7 t 2 + 3.00 10 5 t dt
j
x = -5.00 10 7
t3 t2 + 3.00 10 5 3 2 3 3 7 x = - 1.67 10 m s t + 1.50 10 5 m s 2 t 2 .
e
j e
j
(b)
The bullet escapes when a = 0 , at - 10.0 10 7 m s 3 t + 3.00 10 5 m s 2 = 0 t= 3.00 10 5 s = 3.00 10-3 s . 10.0 10 7
2 5 -3
c
h
(c)
New v = -5.00 10 7 3.00 10-3
c
hc
h + c3.0010 hc3.0010 h
5 -3 2
v = -450 m s + 900 m s = 450 m s . (d) x = - 1.67 10 7 3.00 10-3
c
hc
h + c1.5010 hc3.0010 h
3
x = -0.450 m + 1.35 m = 0.900 m
44 P2.56
Motion in One Dimension
a=
dv = -3.00 v 2 , vi = 1.50 m s dt dv = -3.00 v 2 dt
Solving for v,
v = vi
z
v
v -2 dv = -3.00 dt
t =0
z
t
- When v =
1 1 1 1 + = -3.00t or 3.00t = - . v vi v vi
vi 1 , t= = 0.222 s . 2 3.00 vi
Additional Problems *P2.57 The distance the car travels at constant velocity, v 0 , during the reaction time is x 1 = v 0 t r . The time for the car to come to rest, from initial velocity v 0 , after the brakes are applied is t2 = v f - vi a = 0 - v0 v =- 0 a a
a f
and the distance traveled during this braking period is
axf
2
= vt 2 =
Fv GH
f
+ vi 2
I t = FG 0 + v IJ FG - v IJ = - v . JK H 2 K H a K 2 a
2 0 0 2 0 2
Thus, the total distance traveled before coming to a stop is sstop = x
a f + a x f
1
= v 0 t r -
2 v0 . 2a
*P2.58
(a)
If a car is a distance sstop = v 0 t r -
2 v0 (See the solution to Problem 2.57) from the 2a intersection of length s i when the light turns yellow, the distance the car must travel before the light turns red is v2 x = sstop + si = v 0 t r - 0 + si . 2a
Assume the driver does not accelerate in an attempt to "beat the light" (an extremely dangerous practice!). The time the light should remain yellow is then the time required for the car to travel distance x at constant velocity v 0 . This is t light (b) v s x v 0 t r - 2 0 + si a = = = t r - 0 + i . v0 v0 2 a v0
v2
With si = 16 m, v = 60 km h , a = -2.0 m s 2 , and t r = 1.1 s , t light = 1.1 s -
F 0.278 m s I + 16 m F 1 km h I = G J G J 2e -2.0 m s j H 1 km h K 60 km h H 0.278 m s K
60 km h
2
6. 23 s .
Chapter 2
45
*P2.59
(a)
As we see from the graph, from about -50 s to 50 s Acela is cruising at a constant positive velocity in the +x direction. From 50 s to 200 s, Acela accelerates in the +x direction reaching a top speed of about 170 mi/h. Around 200 s, the engineer applies the brakes, and the train, still traveling in the +x direction, slows down and then stops at 350 s. Just after 350 s, Acela reverses direction (v becomes negative) and steadily gains speed in the -x direction.
200 100 0 50 0 100
v t 100 200 300 400
t (s)
FIG. P2.59(a)
(b)
The peak acceleration between 45 and 170 mi/h is given by the slope of the steepest tangent to the v versus t curve in this interval. From the tangent line shown, we find a = slope = v (155 - 45) mi h = = 2. 2 mi h s = 0.98 m s 2 . (100 - 50) s t
a
f
(c)
Let us use the fact that the area under the v versus t curve equals the displacement. The train's displacement between 0 and 200 s is equal to the area of the gray shaded region, which we have approximated with a series of triangles and rectangles. x 0 200 s = area 1 + area 2 + area 3 + area 4 + area 5
200 100 0 3
5 4 1 2 0
100 200 300 400
t (s)
b ga f b ga f + b160 mi hga100 sf 1 + a50 sfb100 mi hg 2 1 + a100 sfb170 mi h - 160 mi hg 2 = 24 000bmi hgasf
50 mi h 50 s + 50 mi h 50 s
FIG. P2.59(c)
Now, at the end of our calculation, we can find the displacement in miles by converting hours to seconds. As 1 h = 3 600 s , x 0 200 s
F 24 000 mi I asf = GH 3 600 s JK
6.7 mi .
46 *P2.60
Motion in One Dimension
Average speed of every point on the train as the first car passes Liz: x 8.60 m = = 5.73 m s. 1.50 s t The train has this as its instantaneous speed halfway through the 1.50 s time. Similarly, halfway 8.60 m through the next 1.10 s, the speed of the train is = 7.82 m s . The time required for the speed 1.10 s to change from 5.73 m/s to 7.82 m/s is 1 1 (1.50 s)+ (1.10 s)= 1.30 s 2 2 so the acceleration is: a x = v x 7.82 m s - 5.73 m s = = 1.60 m s 2 . t 1.30 s
P2.61
The rate of hair growth is a velocity and the rate of its increase is an acceleration. Then mm d v xi = 1.04 mm d and a x = 0.132 . The increase in the length of the hair (i.e., displacement) w during a time of t = 5.00 w = 35.0 d is
FG H
IJ K
x = v xi t +
x = 1.04 mm d 35.0 d + or x = 48.0 mm . P2.62
b
1 axt 2 2
ga
f 1 b0.132 mm d wga35.0 dfa5.00 wf 2
Let point 0 be at ground level and point 1 be at the end of the engine burn. Let point 2 be the highest point the rocket reaches and point 3 be just before impact. The data in the table are found for each phase of the rocket's motion. (0 to 1) v 2 - 80.0 f
a f
2
2
= 2 4.00 1 000
a fb
g
h
so giving
v f = 120 m s t = 10.0 s x f - xi = 735 m t = 12.2 s FIG. P2.62
120 = 80.0 +( 4.00)t (1 to 2) 0 -(120) = 2(-9.80) x f - xi
c
giving
0 - 120 = -9.80t giving This is the time of maximum height of the rocket. (2 to 3)
v 2 - 0 = 2 -9.80 -1 735 f
v f = -184 = (-9.80)t
a
fb
g
giving t = 18.8 s
(a) (b)
t total = 10 + 12.2 + 18.8 = 41.0 s
cx
f
- xi
h
continued on next page
total
= 1.73 km
Chapter 2
47
(c)
v final = -184 m s t 0.0 10.0 22.2 41.0 x 0 1 000 1 735 0 v 80 120 0 184 a +4.00 +4.00 9.80 9.80
0 #1 #2 #3 P2.63
Launch End Thrust Rise Upwards Fall to Earth
Distance traveled by motorist = 15.0 m s t 1 Distance traveled by policeman = 2.00 m s 2 t 2 2
a
c
f
h
(a) (b) (c) P2.64
intercept occurs when 15.0t = t 2 , or t = 15.0 s v(officer)= 2.00 m s 2 t = 30.0 m s x(officer )= 1 2.00 m s 2 t 2 = 225 m 2
c
h
c
h
Area A1 is a rectangle. Thus, A1 = hw = v xi t . 1 1 Area A 2 is triangular. Therefore A 2 = bh = t v x - v xi . 2 2 The total area under the curve is
b
g
vx vx vxi A1 0 FIG. P2.64 t t A2
A = A1 + A 2 = v xi t + and since v x - v xi = a x t A = v xi t +
bv
x
- v xi t 2
g
1 axt 2 . 2 1 a x t 2 , the 2
The displacement given by the equation is: x = v xi t + same result as above for the total area.
48 P2.65
Motion in One Dimension
(a)
Let x be the distance traveled at acceleration a until maximum speed v is reached. If this is achieved in time t1 we can use the following three equations: x= The first two give 100 = 10.2 - 1 v + vi t1 , 100 - x = v 10.2 - t1 and v = vi + at1 . 2
a
f
a
f
FG 1 t IJ v = FG10.2 - 1 t IJ at H 2 K H 2 K 200 a= b20.4 - t gt .
1 1 1 1
1
For Maggie: a =
a18.4fa2.00f = 200 For Judy: a = a17.4fa3.00f = a fa f Judy: v = a3.83fa3.00f =
x=
200
5.43 m s 2 3.83 m s 2
(b)
v = a1 t Maggie: v = 5.43 2.00 = 10.9 m s 11.5 m s
(c)
At the six-second mark 1 2 at1 + v 6.00 - t1 2
2
a
f
Maggie: x =
1 5.43 2.00 2 1 Judy: x = 3.83 3.00 2
a fa f + a10.9fa4.00f = 54.3 m a fa f + a11.5fa3.00f = 51.7 m
2
Maggie is ahead by 2.62 m . P2.66 a1 = 0.100 m s 2 1 1 2 2 x = 1 000 m = a1 t1 + v1 t 2 + a 2 t 2 2 2 at at 1 1 2 1 000 = a1 t1 + a1 t1 - 1 1 + a 2 1 1 2 a2 2 a2 a 2 = -0.500 m s 2 t = t1 + t 2 and v1 = a1 t1 = -a 2 t 2
2
FG H
IJ K
FG IJ H K
1 000 = t1 =
a 1 2 a1 1 - 1 t1 2 a2
FG H
IJ K
t2 =
a1 t1 12.9 = 26 s -a 2 0.500
20 000 = 129 s 1.20
Total time = t = 155 s
Chapter 2
49
P2.67
Let the ball fall 1.50 m. It strikes at speed given by
2 2 v xf = v xi + 2 a x f - x i : 2 v xf = 0 + 2 -9.80 m s 2 (-1.50 m)
c
h
c
h
v xf = -5.42 m s and its stopping is described by
2 2 v xf = v xi + 2 a x x f - x i
0 = -5.42 m s ax = -2.00 10
b
d
g
i e j
2
+ 2 a x -10 -2 m
-29.4 m 2 s 2
-2
m
= +1.47 10 3 m s 2 .
Its maximum acceleration will be larger than the average acceleration we estimate by imagining constant acceleration, but will still be of order of magnitude ~ 10 3 m s 2 . *P2.68 (a) x f = xi + v xi t + 1 a x t 2 . We assume the package starts from rest. 2 -145 m = 0 + 0 + 1 -9.80 m s 2 t 2 2
c
h
t=
2(-145 m) -9.80 m s 2
= 5. 44 s
(b)
x f = xi + v xi t +
1 1 2 a x t 2 = 0 + 0 + -9.80 m s 2 (5.18 s) = -131 m 2 2
c
h
distance fallen = x f = 131 m (c) (d) speed = v xf = v xi + a x t = 0 + -9.8 m s 2 5.18 s = 50.8 m s The remaining distance is 145 m - 131.5 m = 13.5 m . During deceleration, v xi = -50.8 m s, v xf = 0, x f - xi = -13.5 m
2 2 v xf = v xi + 2 a x x f - x i :
e
j
c
h
0 = -50.8 m s ax =
a
f
2
+ 2 a x (-13.5 m)
-2 580 m 2 s 2 = +95.3 m s 2 = 95.3 m s 2 upward . 2 -13.5 m
a
f
50 P2.69
Motion in One Dimension
(a)
1 1 2 at = 50.0 = 2.00t + (9.80)t 2 , 2 2 4.90t 2 + 2.00t - 50.0 = 0 y f = v i1 t + t= -2.00 + 2.00 2 - 4( 4.90)(-50.0) 2( 4.90)
Only the positive root is physically meaningful: t = 3.00 s after the first stone is thrown. (b) 1 2 at and t = 3.00 - 1.00 = 2.00 s 2 1 2 substitute 50.0 = vi 2 ( 2.00)+ (9.80)( 2.00) : 2 y f = vi 2 t + vi2 = 15.3 m s downward (c) v1 f = vi1 + at = 2.00 +(9.80)(3.00)= 31.4 m s downward v 2 f = vi 2 + at = 15.3 +(9.80)( 2.00)= 34.8 m s downward P2.70 (a) 1 2 d = (9.80)t1 2 t1 + t 2 = 2.40
2 4.90t 2
d = 336 t 2
336t 2 = 4.90 2.40 - t 2
t2 = so
a
f
2
- 359.5t 2 + 28.22 = 0
359.5 359.5 2 - 4( 4.90)( 28.22) 9.80
359.5 358.75 = 0.076 5 s t2 = 9.80 (b) P2.71 (a)
d = 336 t 2 = 26.4 m
1 2 Ignoring the sound travel time, d = (9.80)( 2.40) = 28.2 m , an error of 6.82% . 2 In walking a distance x , in a time t , the length of rope is only increased by x sin . x sin . The pack lifts at a rate t v= x x sin = v boy = v boy t x x + h2
2
(b)
d 1 dv v boy dx = + v boy x dt dt dt v boy v boy x d x d = v = v boy a = v boy - 2 , but dt dt 2 2 2 v boy v boy h 2 h 2 v boy x2 a= = 1- 2 = 2 3 2 x 2 + h2 a=
FG IJ HK
F GH
I JK
c
h
(c) (d)
2 v boy
h
,0
FIG. P2.71
v boy , 0
Chapter 2
51
P2.72
h = 6.00 m, v boy = 2.00 m s v = However, x = v boy t : v = (a) ts 0 1 1.5 2 2.5 3 3.5 4 4.5 5 (b)
v boy x x x sin = v boy = . 12 t x 2 + h2
c
h
2 v boy t
c
2 v boy t 2
+h
2 12
h
=
c 4t
4t
2
+ 36
h
12
.
a f vb m s g
0 0.32 0.63 0.89 1.11 1.28 1.41 1.52 1.60 1.66 1.71
2 h 2 v boy 2 h 2 v boy
0.5
FIG. P2.72(a) 144
2
From problem 2.71 above, a = ts 0 1 1.5 2 2.5 3. 3.5 4. 4.5 5
a f aem s j
2
cx
2
+h
2 3 2
h
=
c
2 v boy t 2
+h
2 3 2
h
=
c4t
+ 36
h
32
.
0.67 0.64 0.57 0.48 0.38 0.30 0.24 0.18 0.14 0.11 0.09
0.5
FIG. P2.72(b)
P2.73
(a)
We require x s = x k when t s = t k + 1.00 1 3.50 m s 2 t k + 1.00 2 t k + 1.00 = 1.183t k xs =
e
jb
g
2
=
1 4.90 m s 2 t k 2
e
jb g
2
= xk
t k = 5.46 s . (b) (c) xk = 1 4.90 m s 2 5.46 s 2
e
ja
f
2
= 73.0 m
v k = 4.90 m s 2 5.46 s = 26.7 m s vs
2
e ja f = e3.50 m s ja6.46 sf =
22.6 m s
52 P2.74
Motion in One Dimension
Time t (s) 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 2.00 2.25 2.50 2.75 3.00 3.25 3.50 3.75 4.00 4.25 4.50 4.75 5.00
Height h (m) 5.00 5.75
h (m) 0.75 0.65
t (s) 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25 0.25
v (m/s) 3.00 2.60 2.16 1.76 1.36 0.96 0.56 0.12 0.24 0.68 1.12 1.48 1.92 2.28 2.72 3.16 3.52 3.96 4.36 4.76
midpt time t (s) 0.13 0.38 0.63 0.88 1.13 1.38 1.63 1.88 2.13 2.38 2.63 2.88 3.13 3.38 3.63 3.88 4.13 4.38 4.63 4.88 FIG. P2.74
6.40 0.54 6.94 0.44 7.38 0.34 7.72 0.24 7.96 0.14 8.10 0.03 8.13 0.06 8.07 0.17 7.90 0.28 7.62 0.37 7.25 0.48 6.77 0.57 6.20 0.68 5.52 0.79 4.73 0.88 3.85 0.99 2.86 1.09 1.77 1.19 0.58 TABLE P2.74
acceleration = slope of line is constant.
a =-1.63 m s 2 = 1.63 m s 2 downward
Chapter 2
53
P2.75
The distance x and y are always related by x 2 + y 2 = L2 . Differentiating this equation with respect to time, we have dy dx 2x + 2y =0 dt dt Now dy dx = -v . is v B , the unknown velocity of B; and dt dt From the equation resulting from differentiation, we have dy x dx x = - (-v). =- dt y dt y But
y
y B x L v A
O x
FG IJ H K
FIG. P2.75
y v v 3 1 = = 0.577 v . = tan so v B = v . When = 60.0 , v B = tan 60.0 3 x tan
FG H
IJ K
ANSWERS TO EVEN PROBLEMS
P2.2 (a) 2 10 -7 m s ; 1 10 -6 m s ; (b) 5 10 yr P2.4 P2.6 (a) 50.0 m s ; (b) 41.0 m s (a) 27.0 m ; 2 (b) 27.0 m + 18.0 m s t + 3.00 m s 2 t ;
8
P2.24
(a) 1.88 km; (b) 1.46 km; (c) see the solution; (d) (i) x 1 = 1.67 m s 2 t 2 ;
(ii) x 2 = 50 m s t - 375 m ;
b
g e
ja f
(iii) x 3
b g = b 250 m sgt - e 2.5 m s jt
2
e
j
2
- 4 375 m ;
(e) 37.5 m s P2.26 P2.28 958 m (a) x f = 30.0t - t 2 m; v f = 30.0 - 2t m s ; (b) 225 m P2.30 P2.32 P2.34 P2.36 P2.38 P2.40 x f - xi = v xf t - 1 a x t 2 ; 3.10 m s 2
(c) 18.0 m s P2.8 P2.10 P2.12 P2.14 (a), (b), (c) see the solution; 4.6 m s 2 ; (d) 0 5.00 m (a) 20.0 m s ; 5.00 m s ; (b) 262 m (a) see the solution; (b) 1.60 m s 2 ; 0.800 m s 2 (a) 13.0 m s; (b) 10.0 m s; 16.0 m s; (c) 6.00 m s 2 ; (d) 6.00 m s 2 P2.18 P2.20 P2.22 see the solution (a) 6.61 m s; (b) -0. 448 m s 2 (a) -21.8 mi h s = -9.75 m s 2 ; (b) -22.2 mi h s = -9.94 m s 2 ; (c) -22.8 mi h s = -10.2 m s 2
e
j
a
f
(a) 35.0 s; (b) 15.7 m s (a) 1.12 10 11 m s 2 ; (b) 4.67 10 -5 s (a) False unless the acceleration is zero; see the solution; (b) True Yes; 212 m; 11.4 s (a) -4.90 m ; -19.6 m; -44.1 m; (b) -9.80 m s; -19.6 m s; -29.4 m s 1.79 s
P2.16
P2.42
54 P2.44 P2.46
Motion in One Dimension
No; see the solution The second ball is thrown at speed vi = gh (a) 510 m; (b) 20.4 s (a) 96.0 ft s ; (b) a = 3.07 10 ft s upward ; (c) t = 3.13 10 -2 s
3 2
P2.60 P2.62 P2.64 P2.66 P2.68
1.60 m s 2 (a) 41.0 s; (b) 1.73 km; (c) -184 m s v xi t + 1 a x t 2 ; displacements agree 2
P2.48 P2.50
155 s; 129 s (a) 5.44 s; (b) 131 m; (c) 50.8 m s ; (d) 95.3 m s 2 upward
P2.52 P2.54
38.2 m (a) and (b) see the solution; (c) -4 m s ; (d) 34 m; (e) 28 m 0.222 s (a) see the solution; (b) 6.23 s
2
P2.70 P2.72 P2.74
(a) 26.4 m; (b) 6.82% see the solution see the solution; a x = -1.63 m s 2
P2.56 P2.58