Answer
$$
a(t)=v^{\prime}(t)= 10 \sin t +3 \cos t, \quad s(0)=0 , \quad s(2\pi)=12
$$
The position of the particle is
$$
s(t)=-10 \sin t -3 \cos t +\frac{6}{\pi}t+3
$$
Work Step by Step
$$
a(t)=v^{\prime}(t)= 10 \sin t +3 \cos t, \quad s(0)=0 , \quad s(2\pi)=12
$$
The general anti-derivative of $
v^{\prime}(t)= 10 \sin t +3 \cos t,
$ is
$$
v(t)= -10 \cos t +3 \sin t +C
$$
Since $s^{\prime }(t)=v(t)$, we antidifferentiate again and obtain
$$
s(t)=-10 \sin t -3 \cos t +Ct+D
$$
To determine $C, D$ we use the fact that $s(0)=0 , s(2\pi)=12 $:
$$
s(t)=-10 \sin (0) -3 \cos (0) +C(0)+D =0
$$
$ \Rightarrow $
$$
-3+D=0 \quad \Rightarrow \quad D=3,
$$
and
$$
s(2\pi)=-10 \sin (2\pi) -3 \cos (2\pi) +C(2\pi)+3 =12
$$
$ \Rightarrow $
$$
0+C(2\pi)=12 \quad \Rightarrow \quad C=\frac{6}{\pi},
$$
Thus the position of the particle is
$$
s(t)=-10 \sin t -3 \cos t +\frac{6}{\pi}t+3
$$
