An engine of a train,moving with uniform acce | vedclass

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(A) Let the total length of the train be $L = 2d$,where $d$ is the distance from the engine to the middle point,and also from the middle point to the

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let the total length of the train be $L = 2d$,where $d$ is the distance from the engine to the middle point,and also from the middle point to the last compartment.Let $a$ be the uniform acceleration of the train.When the engine passes the signal post,its velocity is $u$. When the middle point passes the signal post,let its velocity be $v^{\prime}$.Using the equation of motion $v_f^2 = v_i^2 + 2as$:For the first half of the train (from engine to middle point),we have:$(v^{\prime})^2 = u^2 + 2ad$  --- $(1)$For the second half of the train (from middle point to last compartment),we have:$v^2 = (v^{\prime})^2 + 2ad$  --- $(2)$From equation $(1)$,$2ad = (v^{\prime})^2 - u^2$.Substituting this into equation $(2)$:$v^2 = (v^{\prime})^2 + ((v^{\prime})^2 - u^2)$$v^2 = 2(v^{\prime})^2 - u^2$$2(v^{\prime})^2 = v^2 + u^2$$(v^{\prime})^2 = \frac{v^2 + u^2}{2}$$v^{\prime} = \sqrt{\frac{v^2 + u^2}{2}}$

An engine of a train,moving with uniform acceleration,passes a signal post with velocity $u$ and the last compartment passes the same signal post with velocity $v$. The velocity with which the middle point of the train passes the signal post is

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