Q. A particle located at $x = 0$ at time $t = 0$, starts moving along the positive $x$-direction with a velocity ‘$v$’ that varies as $v = a\sqrt{x}$. The displacement of the particle varies with time as
A) $t$
B) $t^{1/2}$
C) $t^{3}$
D) $t^{2}$
Explanation
(d) $v = \alpha\sqrt{x} \Rightarrow \frac{dx}{dt} = \propto \sqrt{x} \Rightarrow \frac{dx}{\sqrt{x}} = \alpha dt$
By integrating both sides $\int_{}^{}x^{- 1/2}dx = \int_{}^{}{\alpha\ dt}$
$\Rightarrow \frac{\sqrt{x}}{1/2} = \alpha t \Rightarrow \sqrt{x} = \frac{1}{2}\alpha t \Rightarrow x = \frac{1}{4}\alpha^{2}t^{2}\ \therefore x \propto t^{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.