2-34. A cab driver picks up a customer and delivers her 2.00 km away, driving a straight route. The driver accelerates to the speed limit and, upon reaching it, begins to decelerate immediately. The magnitude of the deceleration is three times the magnitude of the acceleration. Find the lengths of the acceleration and deceleration phases of the trip.
The distance traveled by the cab in accelerating from rest to a speed limit of v, with acceleration a, is found from
v2 = v02 + 2ax1 = 0 + 2ax1 = 2ax1
so x1 = v2/2a
In the second phase, the acceleration is -3a, the initial speed is v and the final speed is zero. The distance traveled is found from
0 = v2 - 6ax2
so x2 = v2/6a
Since x1 is 3x2, the total distance is 4x2, and this is 2.00 km. So x2 = 0.5 km and x1 = 1.5 km.
2-52. Two students, Anne and Joan, are bouncing straight up and down on a trampoline. Anne bounces twice as high as Joan. Assuming both are in free-fall, find the ratio of the time Anne spends between bounces to the time Joan spends.
If an object (in this case a student) moves upwards with an initial velocity v, the time taken to return to earth is t = 2v/g, and the maximum height reached is h = v2/2g. The two numbers are related by h = gt2/8. Notice that h varies as t2, or that t varies as the square root of h. If h is increased by a factor of 2, t increases by 2, or approximately 1.414. So Anne spends a factor of 1.414 more time in the air than Joan.
2-60. A hot-air balloon is ascending straight up at a constant speed of 7.0 m/s. When the balloon is 12.0 m above the ground, a gun fires a pellet straight up from ground level with an initial speed of 30.0 m/s. Along the paths of the balloon and the pellet, there are two places where each of them has the same altitude at the same time. How far above ground level are these places?
Measure height upwards, starting from the instant of time when the pellet is fired. The height of the balloon is already 12.0 m at this point, so at later times its height is
y1 = 12.0 + 7.0 t
The height of the pellet is given by
y2 = 30.0t - 4.9t2
Setting the two heights equal gives
12 + 7t = 30t - 4.9t2
or 4.9t2 - 23t + 12 = 0
The solutions of this quadratic equation are
t = 0.598s and t = 4.096s.
The heights of the objects at these two times can be found from either equation:
y = 16.29 m or 40.67 m
The second case corresponds to the pellet missing the balloon on the way up and hitting it on the way down.
2-64. A person who walks for exercise produces the position-time graph given with this problem. (a) Without doing any calculations, decide which segments of the graph (A, B, C, or D) indicate positive, negative, and zero average velocities. (b) Calculate the average velocity for each segment to verify your answers to part (a).
(a) Positive velocity corresponds to positive slope: A and D
Negative velocity corresponds to negative slope: C
Zero velocity corresponds to horizontal line: B
(b)In segment A, v = (1.00 - 0)/(0.20 - 0) = 5.00 km/h.
In segment B, v = (1.00 - 1.00)/(0.40 - 0.20) = 0
In segment C, v = (0.25 - 1.00)/(0.60 - 0.40) = -3.75 km/h.
In segment D, v = (1.25 - 0.25)/(1.00 - 0.60) = 2.50 km/h.