Answer
$a.\qquad a(t)=\displaystyle \frac{2}{t^{3}}+\frac{6}{t^{4}} \ \ \mathrm{ft}/\mathrm{sec}^{2}$
$b.\qquad a(1)=8\ \ \mathrm{ft}/\mathrm{sec}^{2}$
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.
$v=\displaystyle \frac{ds}{dt},\qquad a=\frac{dv}{dt}$
$a.$
$v=\displaystyle \frac{ds}{dt}=\frac{d}{dt}[t^{-1}+t^{-2}]=-t^{-2}-2t^{-3}$
$a=\displaystyle \frac{dv}{dt}=\frac{d}{dt}[-t^{-2}-2t^{-3}]$
$=-(-2t^{-3})-2(-3t^{-4})$
$=\displaystyle \frac{2}{t^{3}}+\frac{6}{t^{4}} \ \ \mathrm{ft}/\mathrm{sec}^{2}$
$b.$
$a(1)=\displaystyle \frac{2}{1^{3}}+\frac{6}{1^{4}}=8 \ \ \mathrm{ft}/\mathrm{sec}^{2}$
