## University Calculus: Early Transcendentals (3rd Edition)

$s(t)= \sin 2t-3$
Since the derivative of velocity is acceleration, $v(t)=2 \cos 2t+C$ As we have, $v(0)=2$ Then $2=2 \cos 2(0)+C \implies C=0$ Thus, $v(t)=2 \cos 2t$ Since the derivative of position is velocity, $s(t)=\sin 2t+C'$ Therefore, $s(0)=-3$ Then $-3=\sin 2(0)+C' \implies C'=-3$ Hence, $s(t)= \sin 2t-3$