## Thomas' Calculus 13th Edition

$s(t)= -\cos (\dfrac{3t}{\pi})$
Given: $a=(\dfrac{9}{\pi^2}) \cos (\dfrac{3t}{\pi}), v(0)=0$ and $s(0)=-1$ Since, we know that the derivative of velocity is acceleration. and $v(t)=\dfrac{3}{\pi}\sin \dfrac{3t}{\pi}+C$ Since,we have: $v(0)=0$ So, $0=(\dfrac{3}{\pi})\sin (\dfrac{3(0)}{\pi})+C \implies C=0$ Therefore, $v(t)=\dfrac{3}{\pi}\sin \dfrac{3t}{\pi}$ Since, we know that the derivative of position $s(t)$ is velocity. So, $s(t)=-\cos \dfrac{3t}{\pi}+C'$ Thus, $s(0)=-1$ and $-1=\cos \dfrac{3(0)}{\pi}+C'$ or, $C'=0$ Hence, $s(t)= -\cos (\dfrac{3t}{\pi})$