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Highly selective Backlog Qs for JEE MainPhysicsMotion in One Dimension

A particle moves with constant acceleration along a straight line. If v 1 , v 2 and v 3 are the average velocities in the three successive intervals t 1 , t 2 and t 3 of time, then the correct relation is

Options

  1. Av 1 - v 2 v 2 - v 3 = t 1 - t 2 t 2 + t 3
  2. Bv 1 - v 2 v 2 - v 3 = t 1 - t 2 t 1 - t 3
  3. Cv 1 - v 2 v 2 - v 3 = t 1 - t 2 t 2 - t 3
  4. Dv 1 - v 2 v 2 - v 3 = t 1 + t 2 t 2 + t 3

Correct answer

D. v 1 - v 2 v 2 - v 3 = t 1 + t 2 t 2 + t 3

Step-by-step solution

Let (u ) be the initial velocity. ( v₁=u+a t₁, v₂=u+d (t₃+t₂ ) and ) (v₃=u+a (t₁+t₃+t₃ ) ) Now, (v₁= u w₁ 2 , u (u+a t₁ ) 2 =u+ 1 2 ) at, (v₁= v₁+c₂ 2 +u+a t₁+ 1 2 a t₂ ) (v₃= v₂+v₂ 2 =u+a t₁+a t₂+ 1 2 a t₃ ) So, ( (y₁-v₂ )=- 1 2 a (t₃+t₂ ) ) and ( (v₂-v₃ )=- 1 2 ) a ( (t₂+t₃ ) ) ( (v₁-v₂ ) (v₂-v₃ ) = (t₁+t₂ ) (t₂+t₃ ) )

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