Question

A particle starts from the origin of coordinates at time $t=0$ and moves in the $xy$ plane with a constant acceleration $\alpha$ in the y-direction. Its equation of motion is $y=\beta x^2.$ Its velocity component in the x-direction is

• A. variable
• B. $\sqrt{\frac{2\alpha }{\beta }}$
• C. $\frac{\alpha }{\beta }$
• D. $\sqrt{\frac{\alpha }{2\beta }}$

Answer 1

The correct option is D $\sqrt{\frac{\alpha }{2\beta }}$

$\text{Step 1: Differentiating to get relation between Vx and Vy}$

$y=\beta x^2$

Differentiating w.r.t time $\left(t\right)$

$\frac{dy}{dt}=\beta 2x\frac{dx}{dt}$

$\Rightarrow V_y=2\beta x{V}_x$
$\text{Step 1: Again Differentiating to get relation of accelerations}$

$\frac{d{V}_y}{dt}=2\beta V_x\frac{dx}{dt}+2\beta x\frac{d{V}_x}{dt}$

$=2\beta V_x^2+0$ [$\frac{d{V}_x}{dt}=a_x=0;$ because the particle has constant acceleration along y-direction]

As per the question

$\frac{d{V}_y}{dt}=\alpha =2\beta V_x^2$

$\Rightarrow V_x=\sqrt{\frac{\alpha }{2\beta }}$