Answer
The position of the mass at $t=3$ is ${\bf{r}}\left( 3 \right) = \left( {37,21} \right)$.
Work Step by Step
We have the force ${\bf{F}} = \left( {24t,16 - 8t} \right)$ and $m=4$. By Newton's second law: ${\bf{F}} = m{\bf{a}}$. So,
$\left( {24t,16 - 8t} \right) = 4{\bf{a}}$, ${\ \ \ }$ ${\bf{a}} = \left( {6t,4 - 2t} \right)$
1. Find the velocity vector
We have
${\bf{v}}\left( t \right) = \smallint {\bf{a}}\left( t \right){\rm{d}}t = \smallint \left( {6t,4 - 2t} \right){\rm{d}}t = \left( {3{t^2},4t - {t^2}} \right) + {{\bf{c}}_0}$
The initial condition ${\bf{v}}\left( 0 \right) = \left( {0,0} \right)$ gives
$\left( {0,0} \right) = \left( {0,0} \right) + {{\bf{c}}_0}$
${{\bf{c}}_0} = \left( {0,0} \right)$
Thus,
${\bf{v}}\left( t \right) = \left( {3{t^2},4t - {t^2}} \right)$
2. Find the position vector
We have
${\bf{r}}\left( t \right) = \smallint {\bf{v}}\left( t \right){\rm{d}}t = \smallint \left( {3{t^2},4t - {t^2}} \right){\rm{d}}t$
${\bf{r}}\left( t \right) = \left( {{t^3},2{t^2} - \frac{1}{3}{t^3}} \right) + {{\bf{c}}_1}$
The initial condition ${\bf{r}}\left( 0 \right) = \left( {10,12} \right)$ gives
$\left( {10,12} \right) = \left( {0,0} \right) + {{\bf{c}}_1}$
${{\bf{c}}_1} = \left( {10,12} \right)$
Thus, the position of the mass at $t$ seconds is
${\bf{r}}\left( t \right) = \left( {{t^3},2{t^2} - \frac{1}{3}{t^3}} \right) + \left( {10,12} \right)$
${\bf{r}}\left( t \right) = \left( {{t^3} + 10,2{t^2} - \frac{1}{3}{t^3} + 12} \right)$
The position of the mass at $t=3$ is ${\bf{r}}\left( 3 \right) = \left( {37,21} \right)$.
