Answer
$$
a(t)=v^{\prime}(t)= 2t+1 , \quad s(0)=3 , \quad v(0)=-2
$$
The position of the particle is
$$
s(t)=\frac{1}{3} t^{3}+\frac{1}{2}t^{2}-2t+3
$$
Work Step by Step
$$
a(t)=v^{\prime}(t)= 2t+1 , \quad s(0)=3 , \quad v(0)=-2
$$
The general anti-derivative of $
v^{\prime}(t)= 2t+1
$ is
$$
v(t)= t^{2}+t+C
$$
To determine C we use the fact that $v(0)=-2$:
$$
v(0)= (0)^{2}+(0)+C=-2
$$
$ \Rightarrow $
$$
0+C=-2 \quad \Rightarrow \quad C=-2,
$$
so
$$
v(t)= t^{2}+t-2
$$
Since $s^{\prime }(t)=v(t)$, we antidifferentiate again and obtain
$$
s(t)=\frac{1}{3} t^{3}+\frac{1}{2}t^{2}-2t+D
$$
To determine D we use the fact that $s(0)=3$:
$$
s(0)=\frac{1}{3} (0)^{3}+\frac{1}{2}(t)^{2}-2(0)+D=3
$$
$ \Rightarrow $
$$
0+D=3 \quad \Rightarrow \quad D=3,
$$
so the position of the particle is
$$
s(t)=\frac{1}{3} t^{3}+\frac{1}{2}t^{2}-2t+3
$$
