Answer
$a.\qquad a(t)=-32 \ \ \mathrm{ft}/\mathrm{sec}^{2}$
$b.\qquad a(2)=-32\ \ \mathrm{ft}/\mathrm{sec}^{2}$
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.
$v=\displaystyle \frac{ds}{dt},\qquad a=\frac{dv}{dt}$
$a.$
$v=\displaystyle \frac{ds}{dt}=\frac{d}{dt}[12+3t-16t^{2}]=3-32t$
$a=\displaystyle \frac{dv}{dt}=\frac{d}{dt}[3-32t]=-32 \ \ \mathrm{ft}/\mathrm{sec}^{2}$
$b.$
$a(2)=-32\ \ \mathrm{ft}/\mathrm{sec}^{2}$
