Q. A box is moved along a straight line by a machine delivering constant power. The distance moved by the body in time t is proportional to

A) $1^{1/2}$
B) $t^{3/4}$
C) $t^{3/2}$
D) $t^{2}$
Explanation

(c) $P = Fv = m.\frac{dv}{dt} \bullet v$
$\int_{}^{}{v\ dv} = \int_{}^{}{\frac{p}{mdt};}\ \frac{{\ \ v}^{2}}{2} = \frac{pt}{m}\ $
$v = \sqrt{\frac{2p}{m}}t^{1/2\ ;}\frac{dx}{dt} = \sqrt{\frac{2p}{m}}t^{1/2}$
$\int_{}^{}{dx = \sqrt{\frac{2p}{m}}\ \int_{}^{}{t^{1/2}dt;}\ }$
$x = \sqrt{\frac{2p}{3}}\ \frac{t^{3/2}}{3/2} = \frac{2}{3}\sqrt{\frac{2p}{3}}t^{3/2}$
$x \propto t^{3/2}$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.