Question

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

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

Answer 1

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

$y=\beta x^2$
Differentiating w.r.t $t$,
$\frac{dy}{dt}=2\beta x\frac{dx}{dt}=2\beta x{v}_x$
Differenciating w.r.t $t$ again,
$\Rightarrow \frac{d^{2}y}{d{t}^2}=2\beta v_x\frac{dx}{dt}+2x\beta \frac{d{v}_x}{dt}$
$\Rightarrow \frac{d^{2}y}{d{t}^2}=2\beta v_x^2+2x\beta \frac{d{v}_x}{dt}$
$\frac{d{v}_x}{dt}=0$ ($\because$ There is no acceleration in the horizontal direction
$\frac{d^{2}y}{d{t}^2}=\alpha \text{(given)}\Rightarrow \alpha =2\beta v_x^2\therefore v_x=\sqrt{\frac{\alpha }{2\beta }}$
Alternate solution:
$y=\beta x^2$
Comparing with the equation of the trajectory,
$y=x\tan \theta -\frac{g{x}^2}{2u^2{\cos }^2\theta }$
$\tan \theta =0\Rightarrow \theta =0...\left(1\right)\frac{g}{2u^2{\cos }^2\theta }=\beta ...\left(2\right)$
acceleration in $y$ direction, $g=\alpha ..\left(3\right)$
Putting $\left(1\right)$ and $\left(3\right)$ in $\left(2\right)$, we get
$\frac{\alpha }{2u^2\left({\cos }^{2}0\right)}=\beta \Rightarrow u^2=\frac{\alpha }{2\beta }\Rightarrow u=\sqrt{\frac{\alpha }{2\beta }}$