Answer
$$
a(t)=\frac{15}{2}\sqrt {t}+3e^{-t},\quad \quad [v(0)=-3,\quad s(0)=4]
$$
The distance is given by:
$$
s(t) =(2 t^{5 / 2}+3 e^{-t})+1
$$
Work Step by Step
$$
a(t)=\frac{15}{2}\sqrt {t}+3e^{-t},\quad \quad [v(0)=-3,\quad s(0)=4]
$$
First find $v(t)$ by integrating $a(t)$
$$
\begin{aligned}
v(t) &= \int\left(a(t)\right) d t \\
&=\int\left(\frac{15}{2} \sqrt{t}+3 e^{-t}\right) d t \\
&=\int\left(\frac{15}{2} t^{1 / 2}+3 e^{-t}\right) d t \\
&=\frac{15}{2}\left(\frac{t^{3 / 2}}{\frac{3}{2}}\right)+3\left(\frac{1}{-1} e^{-t}\right)+C \\
&=5 t^{3 / 2}-3 e^{-t}+C
\end{aligned}
$$
for some constant $C$. Find $C$ from the given information that $v(t)=-3$ when $t=0.$
$$
\begin{aligned}
v0) =5 (0)^{3 / 2}-3 e^{-(0)}+C&=-3\\
-3 +C&=-3\\
C&=-3+3=0\\
\end{aligned}
$$
So,
$$
v(t) =5 t^{3 / 2}-3 e^{-t}
$$
Since $v(t)=s^{\prime}(t)$,the function $s(t)$ s an antiderivative of $v(t)$
$$
\begin{aligned} s(t) &=\int v(t) dt \\
&=\int\left(5 t^{3 / 2}-3 e^{-t}\right) d t \\
&=5\left(\frac{t^{5 / 2}}{\frac{5}{2}}\right)-3\left(-\frac{1}{1} e^{-t}\right)+k \\
&=2 t^{5 / 2}+3 e^{-t}+k \end{aligned}
$$
for some constant $k$. Find $k$ from the given information that $s(t)=4$ when $t=0.$
$$
\begin{aligned}
s(0)=2 (0)^{5 / 2}+3 e^{-(0)}+k &=4\\
3+k &=4\\
k &=4-3\\
k&=1\\
\end{aligned}
$$
So,the distance is given by:
$$
s(t) =(2 t^{5 / 2}+3 e^{-t})+1
$$
