Two boys are standing at the ends A and B of a ground where…

Physics Motion in a Straight Line Uniform Motion Single Correct MCQ

Published on: September 12, 2026

Two boys are standing at the ends A and B of a ground where AB = a . The boy at B starts running in a direction perpendicular to AB with velocity \(V_1 \cdot V_1.\) The boy at A starts running simultaneously with velocity V and catches the other boy in a time t , where t is

\(\frac{a}{\sqrt{v^{2}+v_{1}^{2}}}\)

\(\sqrt{\frac{a^{2}}{v^{2} - v_{1}^{2}}}\)

\(\frac{a}{v - v_1}\)

\(\frac{a}{(v+v_1)}\)

Text Solution

Verified by Experts

The correct answer is:

Let two boys meet at point C after time ' t ' from the starting. Then AC = vt , BC = v_1 t

(AC)^2 = (AB)^2 + (BC)^2 ⇒ ⇒ v^{2} = a^{2} + v_{1}^{2}

By solving we get \(t = \sqrt{\frac{a^{2}}{v_{1}^{2} - v^{2}}}\)

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