## Thomas' Calculus 13th Edition

$s(t)=\sin (\dfrac{2t}{\pi})+1$
Since, the derivative of the function, $v= \dfrac{ds}{dt}=\dfrac{2}{\pi} \cos (\dfrac{2t}{\pi})$ is $s(t)=\sin \dfrac{2t}{\pi}+C$ As we are given that $s=1, t=\pi^2$ This implies that $s=1, t=\pi^2$ and $1=\sin (\dfrac{2 \pi^2}{\pi})+C \implies C=1$ Hence, $s(t)=\sin (\dfrac{2t}{\pi})+1$