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MHT CET · Physics · Motion In Two Dimensions

Two boys are standing at points \(A\) and \(B\) on ground, where distance \(A B=x\). The boy at \(B\) stars running perpendicular to \(A B\) with velocity \(V_1\). The boy at A starts running simultaneously with velocity v and meets the other boy in time \(t\). The value of \(t\) is

  1. A \(\left[\frac{x}{v-v_1}\right]^{1 / 2}\)
  2. B \(\left[\frac{x}{v_1-v}\right]^{1 / 2}\)
  3. C \(\left[\frac{x^2}{v^2-v_1^2}\right]^{1 / 2}\)
  4. D \(\left[\frac{x^2}{v_1^2-v^2}\right]^{1 / 2}\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(\left[\frac{x^2}{v^2-v_1^2}\right]^{1 / 2}\)

Step-by-step Solution

Detailed explanation

Let the coordinates of A be \((0,0)\) and B be \((x,0)\). Position of boy B at time \(t\): \((x, V_1 t)\).