Answer
(a) $t-\cos t-2=s(t)$
(b) $\frac{t^{4}}{12}-\frac{t^{3}}{2}+\frac{t^{2}}{2}=s(t)$
Work Step by Step
(a)You have $\sin t+1=v(t)$
$$
\begin{aligned}
\therefore \int(\sin t+1) d t=s(t) & \\
&=t-\cos t+C \\
s(0) &=0-1+C \\
C-1=-3 & \\
\therefore -2=C & \\
t-\cos t-2=s(t) &
\end{aligned}
$$
(b) You have $t^{2}-3 t+1=a(t)$
$$
\begin{aligned}
\therefore \int\left(t^{2}-3 t+1\right) d t =v(t) &\\
&=\frac{t^{3}}{3}-\frac{3 t^{2}}{2}+t+C \\
0-0+0+C=v(0) &
\end{aligned}
$$
$$
\begin{array}{l}
\therefore 0=C \\
\qquad v(t)=\frac{t^{3}}{3}-\frac{3 t^{2}}{2}+t
\end{array}
$$
We will find the position $s(t)$
$$
\begin{array}{l}
\begin{aligned}
\int\left(\frac{t^{3}}{3}-\frac{3 t^{2}}{2}+t\right) d t=s(t) & \\
&=\frac{t^{4}}{12}-\frac{t^{3}}{2}+\frac{t^{2}}{2}+C
\end{aligned} \\
0-0+0+C=s(0) \\
0=C \\
\therefore \frac{t^{4}}{12}-\frac{t^{3}}{2}+\frac{t^{2}}{2}=s(t)
\end{array}
$$
