Answer
$$
a(t)=18t+8,\quad \quad [v(1)=15,\quad s(1)=19]
$$
The distance of the object is given by:
$$
s(t) =(3t^{3}+4t^{2}-2t)+14
$$
Work Step by Step
$$
a(t)=18t+8,\quad \quad [v(1)=15,\quad s(1)=19]
$$
First find $v(t)$ by integrating $a(t)$
$$
\begin{aligned}
V(t) &=\int (18t+8) dt \\
&=\frac{18}{2}t^{2}+8t+C\\
&=9t^{2}+8t+C
\end{aligned}
$$
for some constant $C$. Find $C$ from the given information that $v(t)=15$ when $t=1.$
$$
\begin{aligned}
V(1) =9(1)^{2}+8(1)+C&=15\\
9+8+C&=15\\
C&=15-17=-2\\
\end{aligned}
$$
So,
$$
V(t) =9t^{2}+8t-2
$$
Since $v(t)=s^{\prime}(t)$,the function $s(t)$ s an antiderivative of $v(t)$
$$
\begin{aligned}
s(t) &=\int v(t) dt \\
&=\int (9t^{2}+8t-2)dt \\
&= (\frac{9}{3}t^{3}+\frac{8}{2}t^{2}-2t)+k \\
&= (3t^{3}+4t^{2}-2t)+k \\
\end{aligned}
$$
for some constant $k$. Find $k$ from the given information that $s(t)=19$ when $t=1.$
$$
\begin{aligned}
s(1)=(3(1)^{3}+4(1)^{2}-2(1))+k&=19\\
(3+4-2)+k&=19\\
k&=19-5\\
k&=14\\
\end{aligned}
$$
So, the distance of the object is given by:
$$
s(t) =(3t^{3}+4t^{2}-2t)+14
$$
