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2.1: a) During the later 4.75-s interval, the rocket moves a distance
[pic], and so the magnitude of the average velocity is
b) [pic]

2.2: a) The magnitude of the average velocity on the return flight is


The direction has been defined to be the –x-direction [pic]

b) Because the bird ends up at the starting point, the average velocityfor the round
trip is 0.

2.3: Although the distance could be found, the intermediate calculation can be avoided by considering that the time will be inversely proportional to the speed, and the extra time will be
2.4: The eastward run takes [pic] = 40.0 s and the westward run takes [pic] = 70.0 s. a) (200 m + 280 m)/(40.0 s + 70.0 s) = [pic] to two significantfigures. b) The net displacement is 80 m west, so the average velocity is [pic] = [pic] in the –x-direction [pic]

2.5: In time t the fast runner has traveled 200 m farther than the slow runner:
Fast runner has run [pic]
Slow runner has run [pic]

2.6: The s-waves travel slower, so they arrive 33 s after the p-waves.


2.7: a) The van will travel 480 m for the first 60 s and1200 m for the next 60 s, for a total distance of 1680 m in 120 s and an average speed of [pic] b) The first stage of the journey takes [pic] and the second stage of the journey takes [pic] so the time for the 480-m trip is 42 s, for an average speed of [pic] c) The first case (part (a)); the average speed will be the numerical average only if the time intervals are the same.

2.8: From theexpression for x(t), x(0) = 0, x(2.00 s) = 5.60 m and x(4.00 s) = 20.8 m. a) [pic] b) [pic] c) [pic]

2.9: a) At [pic], so Eq (2.2) gives

b) From Eq. (2.3), the instantaneous velocity as a function of time is


so i) [pic]
ii) [pic]
and iii) [pic]
c) The car is at rest when [pic]. Therefore [pic]. The only time after [pic] when the car isat rest is [pic]

2.10: a) IV: The curve is horizontal; this corresponds to the time when she stops. b) I: This is the time when the curve is most nearly straight and tilted upward (indicating postive velocity). c) V: Here the curve is plainly straight, tilted downward (negative velocity). d) II: The curve has a postive slope that is increasing. e) III: The curve is still tilted upward(positive slope and positive velocity), but becoming less so.

2.11: Time (s) 0 2 4 6 8 10 12 14 16
Acceleration (m/s2) 0 1 2 2 3 1.5 1.5 0

a) The acceleration is not constant, but is approximately constant between the times [pic] and [pic]



At [pic]

2.12: The cruising speed of the car is 60 [pic] = 16.7[pic]. a) [pic] (to two significant figures). b) [pic]c) No change in speed, so the acceleration is zero. d) The final speed is the same as the initial speed, so the average acceleration is zero.

2.13: a) The plot of the velocity seems to be the most curved upward near t = 5 s.
b) The only negative acceleration (downward-sloping part of the plot) is between t = 30 s and t = 40 s. c) At t = 20 s,the plot is level, and in Exercise 2.12 the car is said to be cruising at constant speed, and so the acceleration is zero. d) The plot is very nearly a straight line, and the acceleration is that found in part (b) of Exercise 2.12, [pic]

2.14: (a) The displacement vector is:


The velocity vector is the time derivative of the displacement vector:


andthe acceleration vector is the time derivative of the velocity vector:


At t = 5.0 s:


(b) The velocity in both the x- and the y-directions is constant and nonzero; thus the overall velocity can never be zero.
(c) The object's acceleration is constant, since t does not appear in the acceleration vector....
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