A particle has an initial velocity vector $\overrightarrow{u}=4\hat{i}-3\hat{j}\text{m/s}$ and constant acceleration $\overrightarrow{a}=\hat{i}-2\hat{j}{\text{m/s}}^2$. Find its displacement at $t=10\text{s}$.

The correct option is D $90\hat{i}-130\hat{j}$
Given,
$u_x=4\text{m/s}$ $a_x=1{\text{m/s}}^2$
$u_y=-3\text{m/s}$ $a_y=-2{\text{m/s}}^2$

Displacement in $x$ direction =$S_x=u_{x}t+\frac{1}{2}{a}_x{t}^2$
$S_x=4\times 10+\frac{1}{2}\times 1\times {10}^2$
$S_x=90\text{m}$

Displacement in $y$ direction =$S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2$
$S_y=(-3)\times 10+\frac{1}{2}\times (-2)\times {10}^2$
$S_y=-130\text{m}$

Displacement $\overrightarrow{S}=90\hat{i}-130\hat{j}$

Hence, option $\text{(D)}$ is correct answer.