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 \(A\beta = \alpha\) . The boy at B starts running in a direction perpendicular to AB with velocity \vec{v}_1. The boy at A starts running simultaneously with velocity \vec{v} and catches the other boy in a time t , where t is

\(0 : \sqrt{v^2 - 1 \cdot v^2}\)

\(\sqrt{\frac{v_0^2}{v^2 - v_1^2}}\)

\(\alpha'(v - v_1)\)

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 \(AC \quad vt\) , \(\beta C \quad v_1 t\)

(AC)^2 = (AB)^2 + (BC)^2 ⇒ ⇒ \(\nu^2 \tau^2 = \sigma^2 + v_1^2 t^2\)

By solving we get \(t = \frac{l}{\sqrt{v^2 - v_1^2}}\)

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