Answer
$$
a(t)=v^{\prime}(t)= 3 \cos t-2 \sin t, \quad s(0)=0 , \quad v(0)=4
$$
The position of the particle is
$$
s(t)=-3 \cos t+ 2 \sin t +2t +3
$$
Work Step by Step
$$
a(t)=v^{\prime}(t)= 3 \cos t-2 \sin t, \quad s(0)=0 , \quad v(0)=4
$$
The general anti-derivative of $
v^{\prime}(t)= 3 \cos t-2 \sin t
$ is
$$
v(t)= 3 \sin t+ 2 \cos t +C
$$
To determine C we use the fact that $v(0)=4$:
$$
v(0)= 3 \sin (0)+ 2 \cos (0) +C =4
$$
$ \Rightarrow $
$$
2+C=4 \quad \Rightarrow \quad C=2,
$$
so
$$
v(t)= 3 \sin t+ 2 \cos t +2
$$
Since $s^{\prime }(t)=v(t)$, we antidifferentiate again and obtain
$$
s(t)=-3 \cos t+ 2 \sin t +2t +D
$$
To determine D we use the fact that $s(0)=0$:
$$
s(0)=-3 \cos (0)+ 2 \sin (0) +2(0) +D =0
$$
$ \Rightarrow $
$$
-3+D=0 \quad \Rightarrow \quad D=3,
$$
so the position of the particle is
$$
s(t)=-3 \cos t+ 2 \sin t +2t +6
$$
