Question

A particle starts from origin at $t=0$ with a velocity $5\hat{i}m{s}^{-1}$ and moves in $x$-$y$ plane under the action of a force which produces a constant acceleration of $3\hat{i}+2\hat{j}m{s}^{-2}$ . The $y$-coordinate of the particle at the instant when its $x$-coordinate is $84m$ is

• A. $12m$
• B. $24m$
• C. $36m$
• D. $48m$

Answer 1

Answer: (C)

The position of the particle is given by
$\vec{r}={\vec{r}}_0+{\vec{v}}_{0}t+\frac{1}{2}\vec{a}{t}^2$
where, ${\vec{r}}_0$ is the position vector at $t=0$ and ${\vec{v}}_0$ is the velocity at $t=0$
Here, ${\vec{r}}_0=0$,
${\vec{v}}_0=5\hat{i}m{s}^{-1}$,
$\vec{a}=\left(3\hat{i}+2\hat{j}\right)m{s}^{-2}$
$\therefore \vec{r}=5t\hat{i}+\frac{1}{2}\left(3\hat{i}+2\hat{j}\right)t^2$
$=\left(5t+1.5t^2\right)\hat{i}+1t^2\hat{j}...\left(i\right)$
Compare it with
$\vec{r}=x\hat{\hat{i}}+y\hat{j}$, we get
$x=5t+1.5t^2$,
$y=1t^2$
$\because x=84m$
$\therefore 84=5t+1.5t^2$
On solving, we get $t=6s$
At $t=6s$,
$y=(1)(6{)}^2$
$=36m$