Q: A body starts from rest with uniform acceleration and acquires a velocity V in time T. The instantaneous kinetic energy of the body the body in time t is proportional to:

(A) $\frac{V}{T} t $

(B) $\frac{V^2}{T} t^2 $

(C) $\frac{V^2}{T^2} t $

(D) $\frac{V^2}{T^2} t^2 $

Solution : $ \displaystyle K.E = \frac{1}{2}mv^2$

Since , v = 0 + at = at

$ \displaystyle K.E= \frac{1}{2}m(at)^2$

$ \displaystyle a= \frac{\Delta v}{\Delta t}= \frac{V}{T} $

Hence , $ \displaystyle K.E= \frac{1}{2}m (\frac{Vt}{T})^2$