Q. If the distance s covered by a particle in time $t$ is proportional to the cube root of its velocity, then the acceleration is
A) A constant
B) $\propto s^{3}$
C) $\propto \frac{1}{s^{3}}$
D) $\propto s^{5}$
Explanation
(d)
Since,$s^{3} \propto v\ \Rightarrow \frac{ds}{dt} = ks^{3}$…….(i)
$\Rightarrow \ \ \ \ \ \ \ \frac{d^{2}s}{dt^{2}} = 3ks^{2}\frac{ds}{dt\ }$
$\Rightarrow \ \ \ \ \ \ \ \ \frac{d^{2}s}{dt} = 3k^{2}s^{5}\ \ \ \ \ \ \ \ \left\lbrack from\ Eq.(i) \right\rbrack$
Hence, acceleration of particle is proportional to $s^{5}$.
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.