Question

A particle starts from origin at $t=0$ and moves in the $x-y$ plane with a constant acceleration $\alpha$ in the $y$ direction. The equation of motion is $y=k{x}^2$. Its velocity component along $x$ direction is

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

Answer 1

Answer: (D)

$\sqrt{\frac{\alpha }{2k}}$

For constant acceleration along $y-$axis, $v_y^2=u_y^2+2\alpha yor{v}_y^2=2\alpha y$
or $\sqrt{y}=\frac{v_y}{\sqrt{2\alpha }}$
As $y=k{x}^2$
or $x=\sqrt{\frac{y}{k}}$
Taking derivative w.r.t time, we get,
$v_x=\left(\frac{1}{\sqrt{k}}\right)\left(\frac{v_y}{2\sqrt{y}}\right)$
$=\left(\frac{1}{\sqrt{k}}\right)\left(\frac{{\sqrt{2\alpha }v}_y}{2v_y}\right)$
$=\sqrt{\frac{\alpha }{2k}}$