Answer
The position vector:
${\bf{r}}\left( t \right) = \left( {\frac{1}{6}{t^3} + 3t,2{t^2} - 2t} \right)$
The velocity vector:
${\bf{v}}\left( t \right) = \left( {\frac{1}{2}{t^2} + 3,4t - 2} \right)$
Work Step by Step
Find the velocity vector:
${\bf{v}}\left( t \right) = \smallint {\bf{a}}\left( t \right){\rm{d}}t = \smallint \left( {t,4} \right){\rm{d}}t = \left( {\frac{1}{2}{t^2},4t} \right) + {{\bf{c}}_0}$
The initial condition ${\bf{v}}\left( 0 \right) = \left( {3, - 2} \right)$ gives
$\left( {3, - 2} \right) = {{\bf{c}}_0}$
Thus,
${\bf{v}}\left( t \right) = \left( {\frac{1}{2}{t^2},4t} \right) + \left( {3, - 2} \right)$
${\bf{v}}\left( t \right) = \left( {\frac{1}{2}{t^2} + 3,4t - 2} \right)$
Find the position vector:
${\bf{r}}\left( t \right) = \smallint {\bf{v}}\left( t \right){\rm{d}}t = \smallint \left( {\frac{1}{2}{t^2} + 3,4t - 2} \right){\rm{d}}t = \left( {\frac{1}{6}{t^3} + 3t,2{t^2} - 2t} \right) + {{\bf{c}}_1}$
The initial condition ${\bf{r}}\left( 0 \right) = \left( {0,0} \right)$ gives
$\left( {0,0} \right) = {{\bf{c}}_1}$
Thus, ${\bf{r}}\left( t \right) = \left( {\frac{1}{6}{t^3} + 3t,2{t^2} - 2t} \right)$
