Example 11 from Precalculus, 12.4 Derivatives

Finding the Instantaneous Velocity

A ball is tossed upward from a height of 200 feet with an initial velocity of 36 ft/sec. If the height of the ball in feet after $t$ seconds is given by $s (t) = −16 t^{2} + 36 t + 200,$ find the instantaneous velocity of the ball at $t = 2.$

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

Solution
First, we must find the derivative $s^{'} (t)$ . Then we evaluate the derivative at $t = 2,$ using $s (a + h) = - 16 {(a + h)}^{2} + 36 (a + h) + 200$ and $s (a) = - 16 a^{2} + 36 a + 200.$
$\begin{array}{l} s^{'} (a) = \lim_{h \to 0} \frac{s (a + h) - s (a)}{h} \\ = \lim_{h \to 0} \frac{- 16 {(a + h)}^{2} + 36 (a + h) + 200 - (- 16 a^{2} + 36 a + 200)}{h} \\ = \lim_{h \to 0} \frac{- 16 (a^{2} + 2 a h + h^{2}) + 36 (a + h) + 200 - (- 16 a^{2} + 36 a + 200)}{h} \\ = \lim_{h \to 0} \frac{- 16 a^{2} - 32 a h - 16 h^{2} + 36 a + 36 h + 200 + 16 a^{2} - 36 a - 200}{h} \\ = \lim_{h \to 0} \frac{- 16 a^{2} - 32 a h - 16 h^{2} + 36 a + 36 h + 200 + 16 a^{2} - 36 a - 200}{h} \\ = \lim_{h \to 0} \frac{- 32 a h - 16 h^{2} + 36 h}{h} \\ = \lim_{h \to 0} \frac{h (- 32 a - 16 h + 36)}{h} \\ = \lim_{h \to 0} (- 32 a - 16 h + 36) \\ = - 32 a - 16 \cdot 0 + 36 \\ s^{'} (a) = - 32 a + 36 \\ s^{'} (2) = - 32 (2) + 36 \\ = - 28 \end{array}$

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

A fireworks rocket is shot upward out of a pit 12 ft below the ground at a velocity of 60 ft/sec. Its height in feet after $t$ seconds is given by $s = - 16 t^{2} + 60 t - 12.$ What is its instantaneous velocity after 4 seconds?

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The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Solution

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Solution

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