Question

A particle moves along the curve $y=\frac{x^2}{2}.$ Here $x$ varies with time as $x=\frac{t^2}{2}.$ Where $x$ and $y$ are measured in meter and $t$ in second. At $t=2s$, the velocity of the particle (in $m{s}^{-1}$) is

• A. $2\hat{i}-4\hat{j}$
• B. $2\hat{i}+4\hat{j}$
• C. $4\hat{i}+2\hat{j}$
• D. $4\hat{i}-2\hat{j}$

Answer 1

The correct option is B $2\hat{i}+4\hat{j}$

$\text{Step 1: Calculate}$ $V_x$

Velocity in $x$ direction is given by:

$V_x=\frac{dx}{dt}=\frac{d}{dt}\left(\frac{t^2}{2}\right)$ $=tm/s$

at $t=2s$, $V_x=2m/s$

$\text{Step 2: Calculate}$ $V_y$

$y=\frac{x^2}{2}=\frac{t^4}{8}$

Velocity in $y$ direction is given by:

$V_y=\frac{dy}{dt}=\frac{d}{dt}\left(\frac{t^4}{8}\right)$ $=\frac{t^3}{2}m/s$

at $t=2s$, $V_y=4m/s$

$\therefore V=2\hat{i}+4\hat{j}m/s$

Hence the velocity of the particle at $t=2s$ is $V=2\hat{i}+4\hat{j}m/s$. Option $B$ is correct.