Question

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 v₁. The boy at A starts running simultaneously with velocity v and catches the other boy in a time t, where t is :

• A. \(\frac{a}{\sqrt{v^2+v^2_1}}\)
• B. \(\sqrt{\frac{a^2}{v^2-v_1^2}}\)
• C. \(\frac{a}{v-v_1}\)
• D. \(\frac{a}{v+v_1}\)

Answer 1

The Correct Option is B

To solve this problem, we'll consider the scenario where two boys start running simultaneously, and one catches the other. Here, one boy is at point A, and the other at point B, with distance \(AB = a\). The boy at B runs in a direction perpendicular to AB with a velocity \(v_1\), and the boy at A needs to catch him with a velocity \(v\).

First, let's analyze the movement of both boys:

1. The boy at point B runs perpendicular to line AB with velocity \(v_1\). Thus, in time \(t\), he will have covered a distance \(v_1t\) perpendicular to AB.
2. For the boy at point A to catch the boy at point B, the distance he covers in time \(t\) should account for both the initial separation \(a\) and the distance covered by the first boy in the perpendicular direction.

This forms a right triangle with:

• One side as \(a\) (horizontal distance)
• Other side as \(v_1t\) (vertical distance)
• The hypotenuse as the path taken by the boy starting from A, which equals \(vt\).

According to the Pythagorean theorem:

c^2 = a^2 + b^2

Substituting the respective distances:

(vt)^2 = a^2 + (v_1t)^2

Solving this equation for \(t\):

v^2t^2 = a^2 + v_1^2t^2

Simplifying:

t^2(v^2 - v_1^2) = a^2

Thus:

t = \sqrt{\frac{a^2}{v^2 - v_1^2}}

This matches the correct option. Therefore, the correct answer is:

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