Answer
${\bf{r}}\left( t \right) = \left( {\frac{1}{{12}}{t^4} - \frac{1}{2}{t^2} - t + 1,\frac{1}{6}{t^3} + \frac{1}{2}{t^2} + t,\frac{1}{{20}}{t^5}} \right)$
Work Step by Step
Find the velocity vector:
${\bf{v}}\left( t \right) = \smallint {\bf{r}}{\rm{''}}\left( t \right){\rm{d}}t = \smallint \left( {{t^2} - 1,t + 1,{t^3}} \right){\rm{d}}t$
$ = \left( {\frac{1}{3}{t^3} - t,\frac{1}{2}{t^2} + t,\frac{1}{4}{t^4}} \right) + {{\bf{c}}_0}$
The initial condition ${\bf{v}}\left( 0 \right) = \left( { - 1,1,0} \right)$ gives
$\left( { - 1,1,0} \right) = {{\bf{c}}_0}$
Thus,
${\bf{v}}\left( t \right) = \left( {\frac{1}{3}{t^3} - t,\frac{1}{2}{t^2} + t,\frac{1}{4}{t^4}} \right) + \left( { - 1,1,0} \right)$
${\bf{v}}\left( t \right) = \left( {\frac{1}{3}{t^3} - t - 1,\frac{1}{2}{t^2} + t + 1,\frac{1}{4}{t^4}} \right)$
Find the position vector:
${\bf{r}}\left( t \right) = \smallint {\bf{v}}\left( t \right){\rm{d}}t = \smallint \left( {\frac{1}{3}{t^3} - t - 1,\frac{1}{2}{t^2} + t + 1,\frac{1}{4}{t^4}} \right){\rm{d}}t$
$ = \left( {\frac{1}{{12}}{t^4} - \frac{1}{2}{t^2} - t,\frac{1}{6}{t^3} + \frac{1}{2}{t^2} + t,\frac{1}{{20}}{t^5}} \right) + {{\bf{c}}_1}$
The initial condition ${\bf{r}}\left( 0 \right) = \left( {1,0,0} \right)$ gives
$\left( {1,0,0} \right) = {{\bf{c}}_1}$
Thus,
${\bf{r}}\left( t \right) = \left( {\frac{1}{{12}}{t^4} - \frac{1}{2}{t^2} - t,\frac{1}{6}{t^3} + \frac{1}{2}{t^2} + t,\frac{1}{{20}}{t^5}} \right) + \left( {1,0,0} \right)$
Hence, the solution is
${\bf{r}}\left( t \right) = \left( {\frac{1}{{12}}{t^4} - \frac{1}{2}{t^2} - t + 1,\frac{1}{6}{t^3} + \frac{1}{2}{t^2} + t,\frac{1}{{20}}{t^5}} \right)$
