Answer
$s(t)=-(\dfrac{\cos \pi t}{\pi})+(\dfrac{1}{\pi})$
Work Step by Step
Since, the derivative of the function $v= \sin \pi t$ is $\dfrac{d}{dt}(-\dfrac{\cos \pi t}{\pi}) =\sin \pi t$ and ,
General form of the function,we have $s(t)=-(\dfrac{\cos \pi t}{\pi})+C$
As we are given that $s(0)=0$
This implies that $s(0)=0$
or, $0=-\dfrac{\cos \pi (0)}{\pi}+C \implies C=\dfrac{1}{\pi}$
Hence, $s(t)=-(\dfrac{\cos \pi t}{\pi})+(\dfrac{1}{\pi})$
