Chapter 4 - Section 4.2 - The Mean Value Theorem - Exercises - Page 223: 46

$s(t)=\sin \dfrac{2t}{\pi}+1$

Here, $v= \dfrac{ds}{dt}=\dfrac{2}{\pi} \cos \dfrac{2t}{\pi}$ and thus, the general form of the function is: $s(t)=\sin \dfrac{2t}{\pi}+C$ Since we have $s=1, t=\pi^2$ Therefore, $s=1, t=\pi^2 \implies 1=\sin \dfrac{2 \pi^2}{\pi}+C$ Thus, $C=1$ Hence, $s(t)=\sin \dfrac{2t}{\pi}+1$