Question

A particle moves along positive branch of a curve $y=\frac{x^2}{2}$, where $x=\frac{t^2}{2}$. $x$ and $y$ are measured in metres and $t$ in seconds. Velocity of particle at $t=2s$ is-

• A. $(2\hat{i}-4\hat{j})$ $m/sec$
• B. $(2\hat{i}+4\hat{j})$ $m/sec$
• C. $(2\hat{i}+2\hat{j})$ $m/sec$
• D. $(4\hat{i}-2\hat{j})$ $m/sec$

Answer 1

The correct option is B $(2\hat{i}+4\hat{j})$ $m/sec$

velocity at $t=2$

$v_x=\frac{dx}{dt}=t{|}_{t=2}=2$

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

$v_y=\frac{dy}{dt}=\frac{t^3}{2}{|}_{t=2}=4$

therefore velocity is $v=(2i+4j)m/sec$