Q. A particle moves in a straight line so that $s = \sqrt{t}$, then its acceleration is proportional to
A) (velocity)$\ ^{3}$
B) Velocity
C) $(velocity)^{2}$
D) $(velocity)^{3/2}$
Explanation
(a)
Given ,$s = \sqrt{t}$
$\therefore\ \ \frac{ds}{dt} = \frac{1}{2\sqrt{t\ }}\ \ \ \ \ \ \Rightarrow v = \frac{1}{2\sqrt{t\ }}$
$\Rightarrow \ \frac{dv}{dt} = - \frac{1}{2.2t^{3/2}}$ $\Rightarrow a = \frac{- 2}{\left( 2\sqrt{t} \right)^{3}}$
$\Rightarrow \ \ \ a = - 2v^{3}\ \ \Rightarrow a \propto v^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.