(III) Mary and Sally are in a fool race (Fig. 2-43). When Mary is 22 m from the finish line, she has a speed of 4.0 m/s and is 5.0 m behind Sally, who has a speed of 5.0 m/s. Sally thinks she has an easy win and so. during the remaining portion of the race, decelerates at a constant rate of 0.50 m/s 2 to the finish line. What constant acceleration does Mary now need during the remaining portion of the race. If she wishes to cross the finish line side-by-side with Sally? FIGURE 2-43 Problem 47. 47. For the runners to cross the finish line side-by-side means they must both reach the finish line in the same amount of time from their current positions. Take Mary's current location as the origin. Use Eq. 2-12b. For Sally: 22 = 5 + 5t + 1 2 (−.5) → t 2 − 20 t + 68 = 0 → t = 20 ± 20 2 − 4 ( 68 ) 2 = 4.343 s . 15.66 s The first time is the time she first crosses the finish line, and so is the time to be used for the problem. Now find Mary's acceleration so that she crosses the finish line in that same amount of time. For Mary: 22 = 0 + 4 t + 1 2 a t 2 → a = 22 − 4 t 1 2 t 2 = 22 − 4 ( 4.343 ) 1 2 ( 4.343 ) 2 = 0.49 m / s 2
(III) Mary and Sally are in a fool race (Fig. 2-43). When Mary is 22 m from the finish line, she has a speed of 4.0 m/s and is 5.0 m behind Sally, who has a speed of 5.0 m/s. Sally thinks she has an easy win and so. during the remaining portion of the race, decelerates at a constant rate of 0.50 m/s 2 to the finish line. What constant acceleration does Mary now need during the remaining portion of the race. If she wishes to cross the finish line side-by-side with Sally? FIGURE 2-43 Problem 47. 47. For the runners to cross the finish line side-by-side means they must both reach the finish line in the same amount of time from their current positions. Take Mary's current location as the origin. Use Eq. 2-12b. For Sally: 22 = 5 + 5t + 1 2 (−.5) → t 2 − 20 t + 68 = 0 → t = 20 ± 20 2 − 4 ( 68 ) 2 = 4.343 s . 15.66 s The first time is the time she first crosses the finish line, and so is the time to be used for the problem. Now find Mary's acceleration so that she crosses the finish line in that same amount of time. For Mary: 22 = 0 + 4 t + 1 2 a t 2 → a = 22 − 4 t 1 2 t 2 = 22 − 4 ( 4.343 ) 1 2 ( 4.343 ) 2 = 0.49 m / s 2
(III) Mary and Sally are in a fool race (Fig. 2-43). When Mary is 22 m from the finish line, she has a speed of 4.0 m/s and is 5.0 m behind Sally, who has a speed of 5.0 m/s. Sally thinks she has an easy win and so. during the remaining portion of the race, decelerates at a constant rate of 0.50 m/s2 to the finish line. What constant acceleration does Mary now need during the remaining portion of the race. If she wishes to cross the finish line side-by-side with Sally?
FIGURE 2-43 Problem 47.
47. For the runners to cross the finish line side-by-side means they must both reach the finish line in the same amount of time from their current positions. Take Mary's current location as the origin. Use Eq. 2-12b.
The first time is the time she first crosses the finish line, and so is the time to be used for the problem. Now find Mary's acceleration so that she crosses the finish line in that same amount of time.
For Mary:
22
=
0
+
4
t
+
1
2
a
t
2
→
a
=
22
−
4
t
1
2
t
2
=
22
−
4
(
4.343
)
1
2
(
4.343
)
2
=
0.49
m
/
s
2
(II) A baseball pitcher throws a baseball with a speed of
43 m/s. Estimate the average acceleration of the ball
during the throwing
motion. In throwing
the baseball, the pitcher
accelerates it through
a displacement of about
3.5 m, from behind
3.5 m
the body to the point
where it is released
(Fig. 2–37).
FIGURE 2-37 Problem 25.
(6) Water striders are insect that propel themselves on the surface of ponds by creating vortices in the water shed by their driving legs. The velocity-versus-time graph of 2 a 17 mm long water strider that moved in a straight line was created from a video. See the figure attached. The insect started from rest, sped up by taking two strides, and then slowed down until it stopped. Estimate
(a) maximum speed (in m/s),
(b) the maximum acceleration (in m/s2 )
(c) the total displacement (in m) of the water strider. Note that the velocity on the graph in the textbook is given in units of length of water strider body per second.
5) A rugby ball on Mars, where the acceleration due to gravity is 0.379g and air resistance is negligible, is hit directly
upward and returns to the same level 9.25 sec later. (a) How fast was it moving just after being hit in ft/sec? (b) How
high above its original point did the ball go just before it fell back in feet?
Chapter 2 Solutions
Physics for Scientists and Engineers with Modern Physics
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