Here, $\vec{u}=10.0 \hat{j}\,\mathrm{m}\,\mathrm{s}^{-1}$ at $t=0$ $\vec{a}= \frac{d\vec{v}}{dt}= (8.0 \hat{i}+2.0 \hat{j})\,\mathrm{m}\,\mathrm{s}^{-2}$ So $,\, d\vec{v}= (8.0 \hat{i}+2.0 \hat{j})\, dt$ Integrating it within the limits of motion i.e. as time changes from 0 to $t$, velocity changes from $u$ to $v$, we have $\vec{v} - \vec{u} = (8.0 \hat{i}+2.0 \hat{j}) t$ or $\vec{v} = \vec{u} + 8.0 t \hat{i} + 2.0 t \hat{j}$ As $\vec{v} = \frac{d\vec{r}}{dt}$ or $d\vec{r} = \vec{v}\, t$ So, $d\vec{r} = ( \vec{u} + 8.0 t \hat{i} + 2.0 t \hat{j} )\, dt$Integrating it within the conditions of motion i.e. as time changes from 0 to t, displacement is from 0 to r, we have $\vec{r} = \vec{u} + \frac{1}{2} \times 8.0 t^{2} \hat{i} + \frac{1}{2} \times 2.0 t^{2} \hat{j}$or $x \hat{i} + y \hat{j}$ $=10 t \hat{j} + 4.0 t^{2} \hat{i} + t^{2} \hat{j}$ $=4.0 t^{2} \hat{i} + (10 t + t^{2}) \hat{j}$Here, we have, $x=4.0 t^{2}$ and $y=10 t + t^{2} \therefore t = \left(\frac{x}{4}\right)^{1/2}$(a) $ $ At $x=16 \,\mathrm{m}$ ; $t = \left(\frac{16}{4}\right)^{1/2} = 2\,\mathrm{s}$, $y = 10 \times 2 + 2^{2} = 24\,\mathrm{m}$(b) Velocity of the particle at time t is $\vec{v} = 10 \hat{j} + 8.0 t \hat{i} + 2.0 t \hat{j}$When $t = 2\,\mathrm{s}$, then$\vec{v} = 10 \hat{j} + 8.0 \times 2 \hat{i} + 2.0 \times 2 \hat{j}$ $=16 \hat{i} + 14 \hat{j}$Speed $= |\vec{v}| = \sqrt{16^{2}+14^{2}} = 21.26\,\mathrm{m}\,\mathrm{s}^{-1}$