Answer
$s(t)=-10\sin{t}-3\cos{t} + \frac{6}{\pi}t+3$
Work Step by Step
First, find the first 2 antiderivatives of $a(t)$.
$$v(t)=-10\cos t+3\sin t +C$$
$$s(t)=-10\sin t -3\cos t + Ct+d$$
using the first initial condition, we find that $0=-3+d \Rightarrow d=3.$ So now we have that $$s(t)=-10\sin t -3\cos t + Ct+3$$
Using the second intial condition we have that $12=-3+2\pi C+3 \Rightarrow \frac{6}{\pi}=C.$
