If the velocity of a moving car is reduced to half its original velocity, its kinetic energy would

The correct option is

D

Become one fourth

$v^{\prime }=\frac{v}{2}$

$K{E}^{\prime }=\frac{1}{2}m{\left(v^{\prime }\right)}^2$

$K{E}^{\prime }=\frac{1}{2}m{\left(\frac{v}{2}\right)}^2$

$K{E}^{\prime }=\frac{1}{8}m{v}^2=\frac{KE}{4}$

Kinetic energy is directly proportional to velocity squared (K.E. = $\frac{1}{2}m{v}^2$). So, if velocity is halved, then kinetic energy becomes one fourth.