Question

A particle starts from the origin of co-ordinates at time $t=0$ and moves in the $x-y$ 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. $\sqrt{\frac{2\alpha }{\beta }}$
• B. $\frac{\alpha }{2\beta }$
• C. $\sqrt{\frac{\alpha }{2\beta }}$
• D. Variable

Answer 1

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

Given,
Equation of motion is $y=\beta x^2$
$x_0=0,y_0=0$
Differentiate both sides,
$\frac{dy}{dt}=2\beta x\frac{dx}{dt}=2\beta x{v}_x$
$\frac{d^{2}y}{d{t}^2}=2\beta [x.a_x+v_x^2]$
$\alpha =2\beta [x{a}_x+v_x^2]$
$v_x=\text{constant},a_x=0$
$a=2\beta v_x^2$
$v=\sqrt{\frac{\alpha }{2\beta }}$