A train is moving at a constant speed V. When | vedclass

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(D) Let the speed of the first train be $V$ and the second train be $u$. Since they are moving in the same direction,the relative velocity of the firs

Vedclass · Accepted Answer

$(V - u)^2 / 2x$

Vedclass · Answer

(D) Let the speed of the first train be $V$ and the second train be $u$. Since they are moving in the same direction,the relative velocity of the first train with respect to the second train is $V_{rel} = V - u$.To avoid a collision,the first train must come to a relative rest with respect to the second train just as it reaches the distance $x$.Using the third equation of motion for relative motion: $v^2 - u_{rel}^2 = 2ax$.Here,the final relative velocity $v = 0$,the initial relative velocity $u_{rel} = (V - u)$,and the displacement is $x$.Substituting these values: $0^2 - (V - u)^2 = 2ax$.$- (V - u)^2 = 2ax$.$a = -\frac{(V - u)^2}{2x}$.The negative sign indicates retardation. Therefore,the minimum retardation required is $\frac{(V - u)^2}{2x}$.

$A$ train is moving at a constant speed $V$. When its driver observes another train in front of him on the same track and moving in the same direction with constant speed $u$. If the distance between the trains is $x$,what should be the minimum retardation of the train to avoid a collision?

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