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. It's equation of motion is $y=\beta x^2$. It's velocity component in the $x$- direction is

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

Answer 1

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

$a_y=\alpha$
$\Rightarrow \frac{d{v}_y}{dt}=\alpha$
$\Rightarrow v_y={\int }^{}\alpha dt=\alpha t+constant$
At $t=0$, $v_y=0$
$\Rightarrow constant=0$
$\Rightarrow v_y=\alpha t$
$\Rightarrow \frac{dy}{dt}=\alpha t$ ............(1)
$\Rightarrow 2\beta x\frac{dx}{dt}=\alpha t$
$\Rightarrow 2\beta x{v}_x=\alpha t$
$v_x=\frac{\alpha t}{2\beta x}$ ............(2)
From (1)
$y={\int }^{}\alpha tdt=\alpha \frac{t^2}{2}+c^{\prime }$
At $t=0$, $y=0$
$\Rightarrow c^{\prime }=0$
$\Rightarrow y=\alpha \frac{t^2}{2}$
$\beta x^2=\alpha \frac{t^2}{2}$
$\Rightarrow t=\sqrt{\frac{2\beta x^2}{\alpha }}$ ....(3)
Substituting t from (3) into (2)
$v_x=\frac{\alpha \sqrt{\frac{2\beta x^2}{\alpha }}}{2\beta x}=\sqrt{\frac{\alpha }{2\beta }}$