Question

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

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

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, $\vec{r}_{0}=0$,
$\vec{v}_{0}=5\hat{i}\,ms^{-1}$,
$\vec{a}=\left(3\hat{i}+2\hat{j}\right)ms^{-2}$
$\therefore \vec{r}=5t\,\hat{i}+\frac{1}{2}\left(3\hat{i}+2\hat{j}\right)t^{2}$
$=\left(5t+1.5t^{2}\right)\hat{i}+1t^{2}\,\hat{j}\,...\left(i\right)$
Compare it with
$\vec{r}=x\hat{\hat{i}}+y\hat{j}$, we get
$x = 5t+ 1.5 t^{2}$,
$y= 1t^{2}$
$\because x=84\,m$
$\therefore 84=5t+1.5t^{2}$
On solving, we get $t = 6\,s$
At $t = 6\, s$,
$ y = (1) (6)^2$
$ = 36\, m$