Chapter 4: Applications of Derivatives - Section 4.2 - The Mean Value Theorem - Exercises 4.2 - Page 198: 47

$s(t)= \sin (2t)-3$

Since, we know that the derivative of velocity is acceleration, so we have $v(t)=2 \cos 2t+C$ Given:$v(0)=2$ so, $2=2 \cos 2(0)+C $ or, $C=0$ This implies, $v(t)=2 \cos 2t$ Since, we know that the derivative of position $s(t)$ is velocity, so $s(t)=\sin 2t+C'$ we have $s(0)=-3$ and $-3=\sin 2(0)+C' $ or, $C'=-3$ Hence, $s(t)= \sin (2t)-3$