Answer
The position of the mass at $t=2$ is
${\bf{r}}\left( 2 \right) = \left( {20,4} \right)$
Work Step by Step
We have $m=2$ and ${\bf{F}} = \left( {12t + 4,8 - 24t} \right)$. By Newton's Second Law of Motion, the acceleration is
${\bf{a}} = {\bf{F}}/m = \left( {6t + 2,4 - 12t} \right)$
Step 1. Find the velocity vector
We have
${\bf{v}}\left( t \right) = \smallint {\bf{a}}\left( t \right){\rm{d}}t = \smallint \left( {6t + 2,4 - 12t} \right){\rm{d}}t$
$ = \left( {3{t^2} + 2t,4t - 6{t^2}} \right) + {{\bf{c}}_0}$
The initial condition ${\bf{v}}\left( 0 \right) = \left( {2,3} \right)$ gives
$\left( {2,3} \right) = \left( {0,0} \right) + {{\bf{c}}_0}$
${{\bf{c}}_0} = \left( {2,3} \right)$
Thus,
${\bf{v}}\left( t \right) = \left( {3{t^2} + 2t,4t - 6{t^2}} \right) + \left( {2,3} \right)$
${\bf{v}}\left( t \right) = \left( {3{t^2} + 2t + 2, - 6{t^2} + 4t + 3} \right)$
Step 2. Find the position vector
We have
${\bf{r}}\left( t \right) = \smallint {\bf{v}}\left( t \right){\rm{d}}t = \smallint \left( {3{t^2} + 2t + 2, - 6{t^2} + 4t + 3} \right){\rm{d}}t$
${\bf{r}}\left( t \right) = \left( {{t^3} + {t^2} + 2t, - 2{t^3} + 2{t^2} + 3t} \right) + {{\bf{c}}_1}$
The initial condition ${\bf{r}}\left( 0 \right) = \left( {4,6} \right)$ gives
$\left( {4,6} \right) = \left( {0,0} \right) + {{\bf{c}}_1}$
${{\bf{c}}_1} = \left( {4,6} \right)$
Thus,
${\bf{r}}\left( t \right) = \left( {{t^3} + {t^2} + 2t, - 2{t^3} + 2{t^2} + 3t} \right) + \left( {4,6} \right)$
${\bf{r}}\left( t \right) = \left( {{t^3} + {t^2} + 2t + 4, - 2{t^3} + 2{t^2} + 3t + 6} \right)$
The position of the mass at $t=2$ is
${\bf{r}}\left( 2 \right) = \left( {20,4} \right)$
