Answer
$$\mathbf{r}=90 t^{2} \mathbf{i}+\left(90 t^{2}-\frac{16}{3} t^{3}+100\right) \mathbf{j}$$
Work Step by Step
Given \begin{aligned} \frac{d \mathbf{r}}{d t} &=\int (180 t) \mathbf{i}+\left(180 t-16 t^{2}\right) \mathbf{j} \\ \mathbf{r}(0) &=100 \mathbf{j} \end{aligned}
Then
\begin{align*}
\mathbf{r}&=\int\left[(180 t) \mathbf{i}+\left(180 t-16 t^{2}\right) \mathbf{j}\right] d t\\
&=90 t^{2} \mathbf{i}+\left(90 t^{2}-\frac{16}{3} t^{3}\right) \mathbf{j}+\mathbf{C}
\end{align*}
Since $\mathbf{r}(0) =100 \mathbf{j} $, then
\begin{align*}
\mathbf{r}(0)&=90(0)^{2} \mathbf{i}+\left[90(0)^{2}-\frac{16}{3}(0)^{3}\right] \mathbf{j}+\mathbf{C}\\
100 \mathbf{j}&=\mathbf{C}
\end{align*}
Hence $$\mathbf{r}=90 t^{2} \mathbf{i}+\left(90 t^{2}-\frac{16}{3} t^{3}+100\right) \mathbf{j}$$
