Multiple choice

A train starts from rest with uniform acceleration ${a_{0.}}$ After some time driver crosses a telegraph pole with velocity ${v_1}\;and\;$ guard in the last compartment crosses the same pole with velocity ${v_2}\;and\;$ The velocity of train when mid-point of train crosses the same pole is.

  1. $\dfrac{{{v_1} + {v_2}}}{2}$
  2. $\sqrt {\dfrac{{{v_1}^2 - {v_2}^2}}{2}} $
  3. $\sqrt {\dfrac{{{v_1}^2 + {v_2}^2}}{2}} $
  4. $\dfrac{{{v_2}}}{2}$
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C Correct answer
Explanation

Using v^2 = u^2 + 2as, let L be the length of the train. v1^2 = 0 + 2a(x) and v2^2 = 0 + 2a(x + L). The midpoint is at x + L/2. v_mid^2 = 2a(x + L/2) = 2ax + aL = v1^2 + (v2^2 - v1^2)/2 = (v1^2 + v2^2)/2.

AI explanation

Let the length of the train be L and its uniform acceleration be a. For the driver crossing the pole, use the kinematic equation v^2 = u^2 + 2as to get v1^2 = 2aL. For the guard crossing the pole at the end of the train, the total distance traveled by the train is 2L, so v2^2 = 2a(2L) = 4aL. When the midpoint of the train crosses the pole, the train has moved a distance of 1.5L, making its velocity v^2 = 2a(1.5L) = 3aL. Substituting aL from the first two equations gives v^2 = (v1^2 + v2^2) / 2, so the velocity is sqrt((v1^2 + v2^2) / 2).

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