Answer
$\vec{r}=(72.0m)\hat{i}+(90.7\mathrm{m})\hat{\mathrm{j}}.$
Work Step by Step
The components of acceleration, $\left\{\begin{array}{l}
a_{x}=3t\\
a_{y}=4
\end{array}\right.$
indicate that they vary with time (not constant), so we use
$\left[\begin{array}{lll}
\vec{v}=\frac{d\vec{r}}{dt} & (4- 10) & \text{instantaneous velocity} \\
\vec{v}=v_{x}\hat{\mathrm{i}}+v_{y}\hat{\mathrm{j}}+v_{z}\hat{\mathrm{k}}, & (4- 11) & \text{instantaneous velocity}\\
& & v_{x}=dx/dt, v_{y}=dy/dt,v_{z}=dz/dt.\\
& & \\
\vec{a}=\frac{d\vec{v}}{dt} & (4- 16) & \text{instantaneous acceleration} \\
\vec{a}=a_{x}\hat{i}+a_{y}\hat{j}+a_{z}\hat{k} & (4- 17) & \text{instantaneous acceleration} \\
& & a_{x}=dv_{x}/dt, a_{y}=dv_{y}/dt, \mathrm{a}\mathrm{n}\mathrm{d} a_{z}=dv_{z}/dt.
\end{array}\right]$
Integrating the components of $\vec{a}$, we find components of $\vec{v}$
$\left[\begin{array}{ll}
v_{x}=\frac{3t^{2}}{2}+v_{0x} & =\frac{3}{2}t^{2}+5.00 \\
v_{y}=\frac{4t^{2}}{2}=2.0t^{2}+v_{0y} & =2.00t^{2.}+2.00
\end{array}\right]$
Integrating the components of $\vec{v}$, we find components of $\vec{r}$
$\left[\begin{array}{ll}
r_{x}=\frac{\frac{3}{2}t^{3}}{3.00}+(5.00)t+r_{0x} & =\frac{t^{3}}{2}+(5.00)t+20.0 \\
r_{y}=\frac{2t^{3}}{3}+(2.00)t+r_{0y} & =\frac{2t^{3}}{3}+(2.00)t+40.0
\end{array}\right]$
(a)
At $t=4.00\mathrm{s}$,
$\left\{\begin{array}{ll}
r_{x}=\frac{(4.00)^{3}}{2}+(5.00)(4.00)+20.0 & =(72.0m) \\
r_{y}=\frac{2(4.00)^{3}}{3}+(2.00)(4.00)+40.0 & =(90.7\mathrm{m})
\end{array}\right\}$
$\vec{r}=(72.0m)\hat{i}+(90.7\mathrm{m})\hat{\mathrm{j}}.$
