Answer
$a.\qquad a(t)=-\displaystyle \frac{1}{2}t^{-3/2}+6t \ \ \mathrm{ft}/\mathrm{sec}^{2}$
$b.\qquad a(4)=\displaystyle \frac{11}{2}\ \ \mathrm{ft}/\mathrm{sec}^{2}$
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.
$v=\displaystyle \frac{ds}{dt},\qquad a=\frac{dv}{dt}$
$a.$
$v=\displaystyle \frac{ds}{dt}=\frac{d}{dt}[2t^{1/2}+t^{3}]=2(\frac{1}{2}t^{-1/2})+3t^{2}=t^{-1/2}+3t^{2}$
$a=\displaystyle \frac{dv}{dt}=\frac{d}{dt}[t^{-1/2}+3t^{2}]$
$=-\displaystyle \frac{1}{2}t^{-3/2}+6t \ \ \mathrm{ft}/\mathrm{sec}^{2}$
$b.$
$a(1)=-\displaystyle \frac{1}{2}(1)+6(1)=\frac{11}{2} \ \ \mathrm{ft}/\mathrm{sec}^{2}$
