Answer
$a.\quad v_{avg}=-14.7$ m/s
$b.\quad t=1.5$ s
Work Step by Step
(a)
$v_{avg}=\displaystyle \frac{s(3)-s(0)}{3-0}=\frac{255.9-300}{3}=-14.7$ m/s
(b)
The Mean Value Theorem:
If $f$ is continuous on the closed interval $[a, b]$
and differentiable on the open interval $(a, b)$ ,
then there exists a number $c$ in $(a, b)$ such that$ f^{\prime}(c)=\displaystyle \frac{f(b)-f(a)}{b-a}.$
$s(t) =-4.9t^{2}+300$
is continuous on $[0,3]$ and differentiable on $(0,3)$ .
The Mean Value Theorem applies.
$v(t)=s^{\prime}(t)=-9.8t$
$-9.8t=-14.7$
$t=\displaystyle \frac{-14.7}{-9.8}=1.5$
