Answer
\[
s(t)=1-\cos t-\frac{t^{3}}{3}
\]
Work Step by Step
The velocity function is an integral part of the acceleration function
\[
v(t)=\int a(t) d t=-t^{2}+\sin t+C
\]
Determine $C$ using $v(0)=0$
\[
v(0)=0=C
\]
Conclusion
$-t^{2}+\sin t=v(t)$
The position function is an integral of the velocity function
\[
s(t)=\int v(t) d t=-\frac{t^{3}}{3}-\cos t+C
\]
Determine $C$ using $s(0)=0$
\[
s(0)=C-1=0 \Rightarrow C=1
\]
\[
1-\frac{t^{3}}{3}-\cos t=s(t)
\]
