Answer
${\bf{r}}\left( t \right) = \left( {2{t^2} - \frac{8}{3}{t^3} + t,{t^4} - \frac{1}{6}{t^3} + 1} \right)$
Work Step by Step
Find the velocity vector:
${\bf{v}}\left( t \right) = \smallint {\bf{r}}{\rm{''}}\left( t \right){\rm{d}}t = \smallint \left( {4 - 16t,12{t^2} - t} \right){\rm{d}}t$
$ = \left( {4t - 8{t^2},4{t^3} - \frac{1}{2}{t^2}} \right) + {{\bf{c}}_0}$
The initial condition ${\bf{v}}\left( 0 \right) = \left( {1,0} \right)$ gives
$\left( {1,0} \right) = {{\bf{c}}_0}$
Thus,
${\bf{v}}\left( t \right) = \left( {4t - 8{t^2},4{t^3} - \frac{1}{2}{t^2}} \right) + \left( {1,0} \right)$
${\bf{v}}\left( t \right) = \left( {4t - 8{t^2} + 1,4{t^3} - \frac{1}{2}{t^2}} \right)$
Find the position vector:
${\bf{r}}\left( t \right) = \smallint {\bf{v}}\left( t \right){\rm{d}}t = \smallint \left( {4t - 8{t^2} + 1,4{t^3} - \frac{1}{2}{t^2}} \right){\rm{d}}t$
$ = \left( {2{t^2} - \frac{8}{3}{t^3} + t,{t^4} - \frac{1}{6}{t^3}} \right) + {{\bf{c}}_1}$
The initial condition ${\bf{r}}\left( 0 \right) = \left( {0,1} \right)$ gives
$\left( {0,1} \right) = {{\bf{c}}_1}$
Thus,
${\bf{r}}\left( t \right) = \left( {2{t^2} - \frac{8}{3}{t^3} + t,{t^4} - \frac{1}{6}{t^3}} \right) + \left( {0,1} \right)$
Hence, ${\bf{r}}\left( t \right) = \left( {2{t^2} - \frac{8}{3}{t^3} + t,{t^4} - \frac{1}{6}{t^3} + 1} \right)$
