Q 11-02-106JEE MainJEE Main 2021 (25 Feb, Shift 1)Easy
An engine of a train, moving with uniform acceleration, passes the signal-post with velocity $u$ and the last compartment with velocity $v$. The velocity with which middle point of the train passes the signal post is:
Answer: (B) $\sqrt{\frac{v^2+u^2}{2}}$
Relative to the train, the signal post moves with uniform acceleration $a$ over the train's length $L$: $v^2 = u^2 + 2aL$.
At the middle point: $v_m^2 = u^2 + 2a\dfrac{L}{2} = u^2 + \dfrac{v^2 - u^2}{2} = \dfrac{u^2 + v^2}{2}$, so $v_m = \sqrt{\dfrac{v^2+u^2}{2}}$.
Solution by Sreeraj P, M.Sc Physics