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Question

(D) Let $S$ be the total length of the train and $a$ be the uniform acceleration.Using the equation of motion $v^2 - u^2 = 2aS$,we have:$v^2 - u^2 = 2

Vedclass · Accepted Answer

$\sqrt{\frac{u^2 + v^2}{2}}$

Vedclass · Answer

(D) Let $S$ be the total length of the train and $a$ be the uniform acceleration.Using the equation of motion $v^2 - u^2 = 2aS$,we have:$v^2 - u^2 = 2aS \implies aS = \frac{v^2 - u^2}{2}$.Let $V_c$ be the velocity of the middle wagon as it passes the pole. The distance covered by the train from the engine to the middle wagon is $S/2$.Applying the equation of motion for this distance:$V_c^2 - u^2 = 2a(S/2) = aS$.Substituting the value of $aS$:$V_c^2 - u^2 = \frac{v^2 - u^2}{2}$.$V_c^2 = u^2 + \frac{v^2 - u^2}{2} = \frac{2u^2 + v^2 - u^2}{2} = \frac{u^2 + v^2}{2}$.Therefore,$V_c = \sqrt{\frac{u^2 + v^2}{2}}$.

$A$ goods train accelerating uniformly on a straight railway track approaches an electric pole standing on the side of the track. Its engine passes the pole with velocity $u$ and the guard's room passes with velocity $v$. The middle wagon of the train passes the pole with a velocity of:

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