Question

A particle projected from origin moves in $x-y$ plane with a velocity $\vec{v}=3\vec{i}+6x\hat{j}$, where $\vec{i}$ and $\hat{j}$ are the unit vector along $x$ and $y$ axis. Find 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

Answer: (A)

Method 1: $\vec{V}=3\hat{i}+6x\hat{j}$
also $\vec{v}=\frac{dx}{dt}\hat{i}+\frac{dy}{dt}\hat{j}$
$\Rightarrow \frac{dx}{dt}=3$
$\int dx=\int 3dt$
$x=3t$
$\frac{dy}{dt}=6x$
$dy=6x\times dt$
$\int dy=\int 6\times 3tdt$
$=18\int tdt\Rightarrow 18\times \frac{t^2}{2}$
$y=9t^2$
$=9\times \frac{x^2}{9}$
$y=x^2$
Method 2:
$V_x\hat{i}+V_y\hat{j}=\vec{V}$
$V_x=3$
$V_y=6x$
We know
$\frac{d_y}{d_x}=\tan \theta =\frac{V_y}{V_x}$
$\frac{d_y}{d_x}=\frac{6x}{3x}$
${\int }_{0}dy={\int }_{0}2xdx$
$y=x^2$