The height of an object t seconds after it is dropped from a height of 250 meters is $s ( t ) = - 4.9 t ^ { 2 } + 250$. Find the time during the first 8 seconds of fall at which the instantaneous velocity equals the average velocity.
A) 32 seconds
B) $19.6$ seconds
C) $6.38$ seconds
D) 4 seconds
E) $2.45$ seconds

Question

Quizplus · Accepted Answer

The instantaneous velocity is the derivative of the height function, $v(t) = s'(t) = -9.8t$. The average velocity over the first 8 seconds is $\frac{s(8) - s(0)}{8}$. Solving $-9.8t = \frac{s(8) - s(0)}{8}$ gives $t = 4$ seconds.

The Height of an Object T Seconds After It Is s(t)=−4.9t2+250s ( t ) = - 4.9 t ^ { 2 } + 250s(t)=−4.9t2+250

The Height of an Object T Seconds After It Is $s ( t ) = - 4.9 t ^ { 2 } + 250$