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MHT CET Medical202623 April 2026Morning ShiftPhysicsMotion in One DimensionActual

A train accelerates from rest at a constant rate 'p' for some time (attains velocity 'v') and then retards to rest at the constant rate 'q'. If the total distance covered by the train is 's', then the velocity 'v' of the train is

Options

  1. A( ( p+q 2pq ) s )^ 1 2
  2. B( ( p-q 2pq ) s )^ 1 2
  3. C( ( 2pq p+q ) s )^ 1 2
  4. D( ( 2pq p-q ) s )^ 1 2

Correct answer

C. ( ( 2pq p+q ) s )^ 1 2

Step-by-step solution

Let s₁ be the distance covered during acceleration and s₂ be the distance covered during retardation. Using the third equation of motion for the acceleration phase: v^2 = u^2 + 2as₁ v^2 = 0 + 2ps₁ s₁ = v^2 2p For the retardation phase: 0 = v^2 - 2qs₂ s₂ = v^2 2q Total distance s = s₁ + s₂ s = v^2 2p + v^2 2q s = v^2 2 ( 1 p + 1 q ) s = v^2 2 ( p+q pq ) Solving for v : v^2 = 2pq p+q s v = ( ( 2pq p+q ) s )^ 1 2 Answer: ( ( 2pq p+q ) s )^ 1 2

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