Question

A particle moves along the positive branch of the curve $y=\frac{x^2}{2}$ where $x=\frac{t^2}{2}$,$x$ and $y$ are measured in metre and t in second. At $t=2s$, the velocity of the particle is

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

Answer 1

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

Given:
$y=\frac{x^2}{2}\dots \left(i\right)x=\frac{t^2}{2}\dots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$.
$y=\frac{1}{2}\cdot {\left(\frac{t^2}{2}\right)}^{2}y=\frac{t^4}{8}\dots \left(iii\right)$
Hence,
$v_x=\frac{dx}{dt}=\frac{d\left(\frac{t^2}{2}\right)}{dt}=\frac{2t}{2}=t{v}_y=\frac{dy}{dt}=\frac{d\left(\frac{t^4}{8}\right)}{dt}=\frac{t^3}{2}$

Therefore, at $t=2s$ velocity of the particle,
$v_x=2m/s,v_y=4m/s$
In vector form
$\overrightarrow{v}=v_x\hat{i}+v_y\hat{j}v=(2\hat{i}+4\hat{j})m/s$