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Sullivan 04 apcalc4e 45342 ch02 166 233 5pp August 7, 2023 12:54

Section 2.4 • Differentiating the Product and the Quotient of Two Functions; Higher-Order Derivatives 211

Solution
2
(a) Since s = s(t) = −4.9t + 29.4t, then

ds
v = v(t) = = −9.8t + 29.4
dt
v(1) = −9.8 + 29.4 = 19.6
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At t = 1 s, the velocity of the ball is 19.6 m/s.
(b) As the ball gets higher, its velocity decreases until it reaches its maximum height.
Then as the ball drops, its velocity increases. It follows that the ball reaches its maximum
height when v(t) = 0.

v(t) = −9.8t + 29.4 = 0

9.8t = 29.4
t = 3

The ball reaches its maximum height after 3 s.
(c) The maximum height is

2
s = s(3) = −4.9 · 3 + 29.4 · 3 = 44.1
The maximum height of the ball is 44.1 m.
(d) The acceleration of the ball at any time t is

2
NOTE Speed and velocity are not the same. d s dv d 2
Speed measures how fast an object is a = a(t) = 2 = = (−9.8t + 29.4) = −9.8 m/s
moving and is a nonnegative number. dt dt dt
Velocity measures both the speed and the
direction of an object and may be a positive (e) There are two ways to answer the question “How long is the ball in the air?”
number or a negative number or zero. First way: Since it takes 3 s for the ball to reach its maximum height, it follows
that it will take another 3 s to reach the ground, for a total time of 6 s in the air.
Second way: When the ball reaches the ground, s = s(t) = 0. Solve for t:

2
s(t) = −4.9t + 29.4t = 0
t(−4.9t + 29.4) = 0
29.4
v(3) 5 0 m/s t 5 3
t = 0 or t = = 6
4.9
The ball is at ground level at t = 0 and at t = 6, so the ball is in the air for 6 s.
(f) Upon impact with the ground, t = 6 s. So the velocity is
v(1) 5 19.6 m/s t 5 1
44.1 m
v(6) = (−9.8)(6) + 29.4 = −29.4
Upon impact the direction of the ball is downward, and its speed is 29.4 m/s.
(g) The total distance traveled by the ball is
v(0) 5 29.4 m/s t 5 0 t 5 6
Figure 29 s(3) + s(3) = 2 s(3) = 2(44.1) = 88.2 m

See Figure 29 for an illustration.

R
NOW WORK Problem 83 and AP Practice Problem 6.

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