Answer
$$
a(t)=v^{\prime}(t)= t^{2} -4 t+6 , \quad s(0)=0 , \quad s(1)=20
$$
The position of the particle is
$$
s(t)=\frac{1}{12}t^{4} - \frac{2}{3} t^{3}+3t^{2}+\frac{235}{12}t
$$
Work Step by Step
$$
a(t)=v^{\prime}(t)= t^{2} -4 t+6 , \quad s(0)=0 , \quad s(1)=20
$$
The general anti-derivative of $
v^{\prime}(t)= t^{2} - 4 t+6,
$ is
$$
v(t)=\frac{1}{3}t^{3} - 2 t^{2}+6t+C
$$
Since $s^{\prime }(t)=v(t)$, we antidifferentiate again and obtain
$$
s(t)=\frac{1}{12}t^{4} - \frac{2}{3} t^{3}+3t^{2}+Ct+D
$$
To determine $C, D$ we use the fact that $s(0)=0 , s(2\pi)=12 $:
$$
s(0)=\frac{1}{12}(0)^{4} - \frac{2}{3} (0)^{3}+3(0)^{2}+C(0)+D=0
$$
$ \Rightarrow $
$$
0+D=0 \quad \Rightarrow \quad D=0,
$$
and
$$
s(1)=\frac{1}{12}(1)^{4} - \frac{2}{3} (1)^{3}+3(1)^{2}+C(1)+0=20
$$
$ \Rightarrow $
$$
\frac{5}{12}+C=20 \quad \Rightarrow \quad C=\frac{235}{12},
$$
Thus the position of the particle is
$$
s(t)=\frac{1}{12}t^{4} - \frac{2}{3} t^{3}+3t^{2}+\frac{235}{12}t
$$
