Answer
The particle's location at $t=2$:
${\bf{r}}\left( 2 \right) = \left( {3,3,\frac{{16}}{3}} \right)$
Work Step by Step
Find the position vector:
${\bf{r}}\left( t \right) = \smallint {\bf{v}}\left( t \right){\rm{d}}t = \smallint \left( {1,t,2{t^2}} \right){\rm{d}}t = \left( {t,\frac{1}{2}{t^2},\frac{2}{3}{t^3}} \right) + {\bf{c}}$,
where ${\bf{c}}$ is a constant vector.
At $t=0$, we have ${\bf{r}}\left( 0 \right) = \left( {1,1,0} \right)$. So,
$\left( {1,1,0} \right) = {\bf{c}}$
Thus,
${\bf{r}}\left( t \right) = \left( {t,\frac{1}{2}{t^2},\frac{2}{3}{t^3}} \right) + \left( {1,1,0} \right)$
${\bf{r}}\left( t \right) = \left( {t + 1,\frac{1}{2}{t^2} + 1,\frac{2}{3}{t^3}} \right)$
The particle's location at $t=2$:
${\bf{r}}\left( 2 \right) = \left( {3,3,\frac{{16}}{3}} \right)$
