Question

A particle at rest is projected from the origin moves in $x-y$ plane with a velocity of $\overrightarrow{v}=(3\hat{i}+6x\hat{j})\text{m/s}$, where $\hat{i}$ and $\hat{j}$ are unit vectors along $+x$ and $+y$ axis. Which of the following equations represent the equation of path followed by the particle?

• A. $y=x^2$
• B. $y=\frac{1}{x^2}$
• C. $y=2x^2$
• D. $y=\frac{1}{x}$

Answer 1

The correct option is A $y=x^2$

Given $\overrightarrow{v}=(3\hat{i}+6x\hat{j})\text{m/s}$
Hence, velocity in $+x$ direction is $v_x=3\text{m/s}$ and in $+y$ direction is $v_y=6x\text{m/s}$
$\because \overrightarrow{v}=v_x\hat{i}+v_y\hat{j}$
Now,
$\frac{dx}{dt}=3$
$\Rightarrow {\int }_0^{x}dx={\int }_0^{t}3dt$
$\Rightarrow x=3t$ ...........$\left(i\right)$
Similarly,
$\frac{dy}{dt}=6x$
$\Rightarrow \frac{dy}{dt}=6\times 3t$ $[$ from $\left(i\right)]$
$\Rightarrow \frac{dy}{dt}=18t$
$\Rightarrow {\int }_0^{y}dy={\int }_0^{t}18tdt$
$\Rightarrow y=\frac{18t^2}{2}$
$\Rightarrow y=9t^2$
$\Rightarrow y=9{\left(\frac{x}{3}\right)}^2$ $[$ from $\left(i\right)]$
$\Rightarrow y=x^2$