Answer
$$
v(t)=s^{\prime}(t)= t^{2}-3\sqrt {t} , \quad s(4)=8
$$
The position of the particle is
$$
s(t)= \frac{1}{3}t^{3}-2t^{\frac{3}{2}}+\frac{8}{3}
$$
Work Step by Step
$$
v(t)=s^{\prime}(t)= t^{2}-3\sqrt {t} , \quad s(4)=8
$$
The general anti-derivative of $
s^{\prime}(t)=t^{2}-3\sqrt {t}
$ is
$$
s(t)= \frac{1}{3}t^{3}-2t^{\frac{3}{2}}+C
$$
To determine C we use the fact that $s(4)=8$:
$$
s(4)= \frac{1}{3}(4)^{3}-2(4)^{\frac{3}{2}}+C=8
$$
$ \Rightarrow $
$$
\frac{16}{3}+C=8 \quad \Rightarrow \quad C=\frac{8}{3},
$$
so the position of the particle is
$$
s(t)= \frac{1}{3}t^{3}-2t^{\frac{3}{2}}+\frac{8}{3}
$$
