Question

A body moves in a plane, so that the displacement along $X$ and $Y-axes$ is $x=3t^3$ and $y=4t^3$ , then velocity is

• A. $9t$
• B. $15t$
• C. $15t^2$
• D. $25t^2$

Answer 1

The correct option is C

$15t^2$

Step 1. Given data

Displacement along $X-axis$, $x=3t^3$

Displacement along $Y-axis,y=4t^3$

Step 2. Finding the velocity, $v$

By using the formula of velocity,

$v=\frac{ds}{dt}$ [$s$ is the displacement and $t$ time]

The velocity along $X-axis$, $v_x=\frac{dx}{dt}$

$v_x=\frac{d}{dt}\left(3t^3\right)$

$v_x=9t^2$

The velocity along $Y-axis,v_y=\frac{dy}{dt}$

$v_y=\frac{d}{dt}\left(4t^3\right)$

$v_y=12t^2$

The velocity, $\overrightarrow{v}=9t^2\hat{i}+12t^2\hat{j}$

The magnitude of the velocity, $\left|\overrightarrow{v}\right|=\sqrt{{\left(9t^2\right)}^2+{\left(12t^2\right)}^2}$

$v=\sqrt{81t^4+144t^4}$

$v=\sqrt{225t^4}$

$v=15t^2$

Hence, the correct option is $\left(C\right)$.