Question

A particle moves from the point $(2.0\hat{i}+4.0\hat{j})\text{m}$ at $t=0$, with an initial velocity $(5.0\hat{i}+4.0\hat{j}){\text{ms}}^{-1}$. It is acted upon by a constant force which produces a constant acceleration $(4.0\hat{i}+4.0\hat{j}){\text{ms}}^{-2}$. The distance of the particle from the origin at time $2\text{s}$ is $20\sqrt{x}\text{m}$. Find the value of $x$.

Answer 1

Given:
${\overrightarrow{r}}_0=(2.0\hat{i}+4.0\hat{j})\text{m}$
$\overrightarrow{u}=(5.0\hat{i}+4.0\hat{j}){\text{ms}}^{-1}$
$\overrightarrow{a}=(4.0\hat{i}+4.0\hat{j}){\text{ms}}^{-1}$

Distance travelled by a particle in time $t$ is given by,

$\overrightarrow{r}={\overrightarrow{r}}_0+\overrightarrow{u}t+\frac{1}{2}\overrightarrow{a}{t}^2$

$\overrightarrow{r}=(2\hat{i}+4\hat{j})+(5\hat{i}+4\hat{j})\times 2+\frac{1}{2}[4\hat{i}+4\hat{j}]\times 2^2$

$\overrightarrow{r}=(20\hat{i}+20\hat{j})\text{m}$

$\therefore |\overrightarrow{r}|=20\sqrt{2}\text{m}$

Given, $|\overrightarrow{r}|=20\sqrt{x}\text{m}$

$\therefore x=2$