Answer
At $t=10$ s, the position of the mass is at ${\bf{r}}\left( {10} \right) = \left( {45, - 20} \right)$.
Work Step by Step
We have constant force ${\bf{F}} = \left( {5,2} \right)$ and $m=10$. By Newton's second law: ${\bf{F}} = m{\bf{a}}$. So,
$\left( {5,2} \right) = 10{\bf{a}}$, ${\ \ \ }$ ${\bf{a}} = \left( {\frac{1}{2},\frac{1}{5}} \right)$
1. Find the velocity vector
We have
${\bf{v}}\left( t \right) = \smallint {\bf{a}}\left( t \right){\rm{d}}t = \smallint \left( {\frac{1}{2},\frac{1}{5}} \right){\rm{d}}t = \left( {\frac{1}{2}t,\frac{1}{5}t} \right) + {{\bf{c}}_0}$
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,
${\bf{v}}\left( t \right) = \left( {\frac{1}{2}t,\frac{1}{5}t} \right) + \left( {2, - 3} \right)$
${\bf{v}}\left( t \right) = \left( {\frac{1}{2}t + 2,\frac{1}{5}t - 3} \right)$
2. Find the position vector
We have
${\bf{r}}\left( t \right) = \smallint {\bf{v}}\left( t \right){\rm{d}}t = \smallint \left( {\frac{1}{2}t + 2,\frac{1}{5}t - 3} \right){\rm{d}}t$
${\bf{r}}\left( t \right) = \left( {\frac{1}{4}{t^2} + 2t,\frac{1}{{10}}{t^2} - 3t} \right) + {{\bf{c}}_1}$
The initial condition ${\bf{r}}\left( 0 \right) = \left( {0,0} \right)$ gives
$\left( {0,0} \right) = \left( {0,0} \right) + {{\bf{c}}_1}$
${{\bf{c}}_1} = \left( {0,0} \right)$
Thus, the position of the mass at $t$ seconds is
${\bf{r}}\left( t \right) = \left( {\frac{1}{4}{t^2} + 2t,\frac{1}{{10}}{t^2} - 3t} \right)$
At $t=10$ s, the position of the mass is at ${\bf{r}}\left( {10} \right) = \left( {45, - 20} \right)$.