Answer
The position vector:
${\bf{r}}\left( t \right) = \left( {t,1,\frac{1}{6}{t^3}} \right)$
The velocity vector:
${\bf{v}}\left( t \right) = \left( {1,0,\frac{1}{2}{t^2}} \right)$
Work Step by Step
Find the velocity vector:
${\bf{v}}\left( t \right) = \smallint {\bf{a}}\left( t \right){\rm{d}}t = \smallint \left( {0,0,t} \right){\rm{d}}t = \left( {0,0,\frac{1}{2}{t^2}} \right) + {{\bf{c}}_0}$
The initial condition ${\bf{v}}\left( 0 \right) = \left( {1,0,0} \right)$ gives
$\left( {1,0,0} \right) = \left( {0,0,0} \right) + {{\bf{c}}_0}$
${{\bf{c}}_0} = \left( {1,0,0} \right)$
Thus,
${\bf{v}}\left( t \right) = \left( {0,0,\frac{1}{2}{t^2}} \right) + \left( {1,0,0} \right)$
${\bf{v}}\left( t \right) = \left( {1,0,\frac{1}{2}{t^2}} \right)$
Find the position vector:
${\bf{r}}\left( t \right) = \smallint {\bf{v}}\left( t \right){\rm{d}}t = \smallint \left( {1,0,\frac{1}{2}{t^2}} \right){\rm{d}}t = \left( {t,0,\frac{1}{6}{t^3}} \right) + {{\bf{c}}_1}$
The initial condition ${\bf{r}}\left( 0 \right) = \left( {0,1,0} \right)$ gives
$\left( {0,1,0} \right) = \left( {0,0,0} \right) + {{\bf{c}}_1}$
${{\bf{c}}_1} = \left( {0,1,0} \right)$
Thus,
${\bf{r}}\left( t \right) = \left( {t,0,\frac{1}{6}{t^3}} \right) + \left( {0,1,0} \right)$
${\bf{r}}\left( t \right) = \left( {t,1,\frac{1}{6}{t^3}} \right)$
