Solve the problem.
-Assume that a watermelon dropped from a tall building falls $\mathrm { y } = 16 \mathrm { t } ^ { 2 } \mathrm { ft }$ in $\mathrm { t }$ sec. Find the watermelon's average speed during the first $4 \mathrm { sec }$ of fall and the speed at the instant $t = 4 \mathrm { sec }$.
A) $32 \mathrm { ft } / \mathrm { sec } ; 64 \mathrm { ft } / \mathrm { sec }$
B) $128 \mathrm { ft } / \mathrm { sec } ; 65 \mathrm { ft } / \mathrm { sec }$
C) $64 \mathrm { ft } / \mathrm { sec } ; 128 \mathrm { ft } / \mathrm { sec }$
D) $65 \mathrm { ft } / \mathrm { sec } ; 130 \mathrm { ft } / \mathrm { sec }$

Question

Quizplus · Accepted Answer

The average speed is the total distance divided by time: $y = 16t^2$, so at $t = 4$, $y = 16(4)^2 = 256$ ft. Average speed = $256 \text{ ft} / 4 \text{ sec} = 64 \text{ ft/sec}$. The instantaneous speed is the derivative of $y$, $v = \frac{dy}{dt} = 32t$. At $t = 4$, $v = 32(4) = 128 \text{ ft/sec}$.

Solve the Problem y=16t2ft\mathrm { y } = 16 \mathrm { t } ^ { 2 } \mathrm { ft }y=16t2ft

Solve the Problem $\mathrm { y } = 16 \mathrm { t } ^ { 2 } \mathrm { ft }$