Answer
$$
v(t)=9t^{2}-3\sqrt {t},\quad \quad [s(1)=8]
$$
The distance of the object is given by:
$$
s(t) =(3t^{3}-2t^{\frac{3}{2}})+7.
$$
Work Step by Step
$$
v(t)=9t^{2}-3\sqrt {t},\quad \quad [s(1)=8]
$$
Since $v(t)=s^{\prime}(t)$,the function $s(t)$ s an antiderivative of $v(t)$
$$
\begin{aligned}
s(t) &=\int v(t) dt \\
&=\int (9t^{2}-3\sqrt {t})dt \\
&= (\frac{9}{3}t^{3}-3\frac{2}{3}t^{\frac{3}{2}})+C \\
&= (3t^{3}-2t^{\frac{3}{2}})+C \\
\end{aligned}
$$
for some constant $C$. Find $C$ from the given information that $s(t)=8$ when $t=1.$
$$
\begin{aligned}
s(1)=(3(1)^{3}-2(1)^{\frac{3}{2}})+C&=8\\
(3-2)+C&=8\\
(1)+C&=8\\
C&=8-1=7\\
\end{aligned}
$$
So,
$$
s(t) =(3t^{3}-2t^{\frac{3}{2}})+7.
$$
