Question

A particle moves from the point $(2.0\hat{i}+4.0\hat{j})m$, at $t=0$, with an initial velocity $(5.0\hat{i}+4.0\hat{j})m/s$. It is acted upon by a constant force which produces a constant acceleration $(4.0\hat{i}+4.0\hat{j})m-s^{-2}$. The distance of the particle from the origin at time $2s$ is found to be $n\sqrt{2}$, then $n$ is

Answer 1

$\overrightarrow{r_o}=(2.0\hat{i}+4.0\hat{j})m$
$\overrightarrow{v_o}=(5.0\hat{i}+4.0\hat{j})m/s$, $\overrightarrow{a}=(4.0\hat{i}+4.0\hat{j})m/s^2$
Along $x$-axis, $S_{ox}=2m,V_{ox}=5m/s$
$a_x=4m/s^2$

$S_x=S_{ox}+v_{ox}t+\frac{1}{2}{a}_x{t}^2$
$=2+5\times 2+\frac{1}{2}\times 4\times 2^2=20m$

Along $y$-axis,
$S_{oy}=4m,V_{oy}=4m/s,a_y=4m/s^2$
$S_y=S_{oy}+V_{oy}t+\frac{1}{2}{a}_y{t}^2$
$=4+4\times 2+\frac{1}{2}\times 4\times 2^2=20m$

$S=\sqrt{S_x^2+S_y^2}=\sqrt{{20}^2+{20}^2}=20\sqrt{2}$
$⟹n=20$