Question

A particle crossing the origin of co-ordinates at time $t = 0$, moves in the $xy$-plane with a constant acceleration a in the $y$-direction. If its equation of motion is $ y=b{{x}^{2}} $ ($b$ is a constant), its velocity component in the $x$-direction is

• A. $ \sqrt{\frac{2b}{a}} $
• B. $ \sqrt{\frac{A}{2b}} $
• C. $ \sqrt{\frac{a}{b}} $
• D. $ \sqrt{\frac{b}{a}} $
• E. $ \sqrt{ba} $

Answer 1

Answer: (B)

$ y=b{{x}^{2}} $ $ \frac{dy}{dt}=2bx\frac{dx}{dt} $ ...(i)
$ \frac{dy}{dt}=at $
$ (\because {{v}_{y}}={{u}_{y}}+{{a}_{y}}t) $
$ at=2bx\frac{dx}{dt} $
$ atdt=2bx\,dx $ Take integration of both sides
$ \int{atdt}=\int{2bx}\,dx $
$ \frac{a{{t}^{2}}}{2}=b{{x}^{2}}+c $ ...(ii)
At $ t=0,\text{ }x=0 $
$ c\Rightarrow 0 $
Then, $ \frac{a{{t}^{2}}}{2}=b{{x}^{2}} $
$ x=\sqrt{\frac{a{{t}^{2}}}{2b}}=t\sqrt{\frac{a}{2b}} $
$ \therefore $ $ {{v}_{x}}=\frac{dx}{dt}=\sqrt{\frac{a}{2b}} $