Given, $\vec{u}=10.0 \mathrm{jms}^{-1}$ at $t=0$;
$$
\vec{a}=\frac{d}{d t}(\vec{v})=(8.0 \hat{i}+2.0 \hat{j}) \mathrm{ms}^{-2}
$$
Integrating the above relation with the change of time, from 0 to $t$, velocity changes from $u$ to $v$, we get,
$$
\begin{aligned}
\vec{v}-\vec{u} &=(8.0 \hat{i}+2.0 \hat{j}) t \text { or, } \\
\vec{v} &=\vec{u}+(8.0 \hat{i}+2.0 \hat{j}) t \\
\vec{v} &=\frac{d}{d t}(\vec{r}) \text { or, } d r=\vec{v} d t
\end{aligned}
$$
so, $\overrightarrow{d r}=(u+8.0 t \hat{i}+2.0 t \hat{j}) d t$
Integrating the above relation with the change of time, from 0 to $t$, displacement changes from 0 to $r$, we get,
$$
\begin{aligned}
\vec{r}=\left(\vec{u} t+1 / 28.0 t^2 \hat{i}+1 / 2.0 t^2 \hat{j}\right)
\end{aligned}
$$
Here, it is given, $x=4 t^2$ and $y=10 t+t^2$
$$
\Rightarrow t=\left(\frac{x}{4}\right)^{\frac{1}{2}}=2 \mathrm{~s} .
$$
(a) At $x=16 \mathrm{~m}, t=\left(\frac{16}{4}\right)^{\frac{1}{2}}=2 \mathrm{~s}$. and,
$y=10 \times 2+2^2=24 \mathrm{~m}$
(b) Velocity of the particle at time $t$ is $v=10 \hat{j}+8 t \hat{i}+2 t \hat{j}$, when, $t=2 \mathrm{~s}$, then,
$$
\begin{aligned}
&v=10 \hat{j}+8 \times 2 t \hat{i}+2 \times 2 \hat{j}=16 \hat{i}+14 \hat{j} \\
&\text { Speed }=|\vec{v}|=\sqrt{16^2+14^2}=21.26 \mathrm{~ms}^{-1}
\end{aligned}
$$