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 that, $s = \sqrt{t}$
$\therefore\ \ \frac{ds}{dt} = \frac{1}{2\sqrt{t}}$
$\Rightarrow \ \ v = \frac{1}{2\sqrt{t}}\ \ \Rightarrow \ \ \frac{dv}{dt} = - \frac{1}{2.2t^{3/2}}$
$\Rightarrow \ \ \ a = - \frac{2}{\left( 2\sqrt{t} \right)^{3}}$
$\Rightarrow \ \ a = - 2v^{3}$
$\Rightarrow \ \ a \propto v^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.