Written Homework 7 Solutions

160
80
0 0.4 0.8 1.2 1.6 2 2.4 2.8 3.2 3.6 4
-80
-160
y = -16x² + 10x + 200
y = -32x + 10
(c) Use your graph to answer the following questions. During what period of time is the
ball rising? During what period of time is it falling? When does it reach the highest
point of its flight?
Solution: The ball is roughly rising for 0 < t < 0.31 and falling for 0.31 < t < 3.85.
The highest point is reached when v = −32t + 10 = 0 which occurs when t ≈ 0.31.
(d) How high does the ball rise?
Solution: By plugging t = 0.31 into our distance equation, we see that the ball
reaches f(0.31) = −16(0.31) 2 + 10(0.31) + 200 = 201.56 feet, or 1.56 feet above the
top of the building.
20. A steel ball is rolling along a 20-inch long straight track so that its distance from the
midpoint of the track (which is 10 inches from either end) is d = 3 sin(t) inches after t
seconds have passed. (Think of the track as aligned from left to right. Positive distances
mean the ball is to the right of the center; negative distances mean it is to the left.)
(a) Find a formula for the velocity of the ball after t seconds. What is happening when
the velocity is positive; when it is negative; when it equals zero? Write a sentence or
two describing the motion of the ball.
Solution: v(t) = d ′ (t) = 3 cos(t). When the velocity goes from positive to negative
and visa versa it is simply switching directions on the track from right to left and
left to right respectively. When the velocity equals zero, the ball is at its maximum
distance from the center. The ball is moving back and forth along the track reaching
its higher speeds towards the center of the track and pausing briefly when its distance
from the center is maximized.
(b) How far from the midpoint of the track does the ball get? How can you tell?
Solution: The maximum distance occurs when d = 3 sin(t) is maximized which
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