Activity 3.
For the function given by \(s(t) = 64 - 16(t-1)^2\text{,}\) find the most simplified expression you can for the average velocity of the ball on the interval \([2, 2+h]\text{.}\) Use your result to compute the average velocity on \([1.5,2]\) and to estimate the instantaneous velocity at \(t = 2\text{.}\) Finally, compare your earlier work in Activity 2.
Hint.
Note that
\begin{align*}
s(2+h) \amp = 64 - 16(2+h-1)^2 = 64 - 16(1+h)^2\\
\amp = 64 - (16 + 32h + 16h^2) = 48 - 32h - 16h^2\text{.}
\end{align*}
Answer.
\(AV_{[2, 2+h]} = -32 - 16h\)
Solution.
Observe first that
\begin{align*}
s(2+h) \amp = 64 - 16(2+h-1)^2 = 64 - 16(1+h)^2\\
\amp = 64 - (16 + 32h + 16h^2) = 48 - 32h - 16h^2\text{.}
\end{align*}
Next, recall that \(AV_{[2, 2+h]} = \frac{s(2+h) - s(2)}{h}\text{,}\) so
\begin{equation*}
AV_{[2, 2+h]} = \frac{s(2+h) - s(2)}{h} = \frac{(48 - 32h - 16h^2)-48}{h} = \frac{-32h - 16h^2}{h}\text{.}
\end{equation*}
Now, since we assume \(h \ne 0\text{,}\) we can simplify further to find that \(AV_{[2, 2+h]} = -32 - 16h\text{.}\) Setting \(h = -0.5\text{,}\) it follows \(AV_{[1.5,2]} = -32 + 16(0.5) = -24\) ft/sec, and letting \(h\) approach zero, we see that \(-32 - 16h\) will approach \(-32\text{,}\) so the instantaneous velocity at \(t = 2\) appears to be \(-32\) feet/sec. Both results match our earlier work in Activity 1.
