Answer
$s(t)= -\cos \dfrac{3t}{\pi}$
Work Step by Step
We are given that $a=\dfrac{9}{\pi^2}\cos \dfrac{3t}{\pi}, v(0)=0$ and $s(0)=-1$
Since the derivative of velocity is acceleration, $v(t)=\dfrac{3}{\pi}\sin \dfrac{3t}{\pi}+C$
As we have, $v(0)=0$ Then $0=\dfrac{3}{\pi}\sin \dfrac{3(0)}{\pi}+C \implies C=0$
Thus, $v(t)=\dfrac{3}{\pi}\sin \dfrac{3t}{\pi}$
Since, the derivative of position is velocity, $s(t)=-\cos \dfrac{3t}{\pi}+C'$
Therefore, $s(0)=-1$ Then $-1=\cos \dfrac{3(0)}{\pi}+C' \implies C'=0$
Hence, $s(t)= -\cos \dfrac{3t}{\pi}$
