Answer
$a.\qquad a(t)=-\displaystyle \frac{1}{4}t^{-3/2}+2 \ \ \mathrm{ft}/\mathrm{sec}^{2}$
$b.\qquad a(4)=\displaystyle \frac{63}{32}\ \ \mathrm{ft}/\mathrm{sec}^{2}$
Work Step by Step
The acceleration of a moving object is the derivative of its velocity,
that is, the second derivative of the position function.
$v=\displaystyle \frac{ds}{dt},\qquad a=\frac{dv}{dt}$
$a.$
$v=\displaystyle \frac{ds}{dt}=\frac{d}{dt}[t^{1/2}+t^{2}]=\frac{1}{2}t^{-1/2}+2t$
$a=\displaystyle \frac{dv}{dt}=\frac{d}{dt}[\frac{1}{2}t^{-1/2}+2t]$
$=\displaystyle \frac{1}{2}(-\frac{1}{2}t^{-3/2})+2$
$=-\displaystyle \frac{1}{4}t^{-3/2}+2 \ \ \mathrm{ft}/\mathrm{sec}^{2}$
$b.$
$a(4)=-\displaystyle \frac{1}{4}(4^{-3/2})+2$
$=-\displaystyle \frac{1}{4}(4^{-1/2})^{3}+2$
$=-\displaystyle \frac{1}{4}(\frac{1}{2^{3}})+2$
$=2-\displaystyle \frac{1}{32}$
$=\displaystyle \frac{63}{32} \ \ \mathrm{ft}/\mathrm{sec}^{2}$
