Question

Two girls are standing at the ends 'A' and 'B' of a ground where $AB = b$. The girl at 'B' starts running perpendicular to 'AB' with velocity $V_1$. The girl at 'A' starts running simultaneously with velocity $V_2$ and in shortest distance meets the other girl in time 't'. The value of 't' is ______.

• A. $\frac{b}{\sqrt{V_1^2 + V_2^2}}$
• B. $\frac{b}{V_1 + V_2}$
• C. $\frac{b}{V_2 - V_1}$
• D. $\frac{b}{\sqrt{V_2^2 - V_1^2}}$

Hint:

This is geometrically equivalent to the classic "swimmer crossing a river" problem where the swimmer wants to reach the exact opposite bank (zero drift). The effective velocity across the river is always $\sqrt{V_{swimmer}^2 - V_{river}^2}$.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
This is a relative motion problem. For the shortest distance meeting, we consider the triangle formed by their paths.

Step 2: Formula Application:
Let they meet at point $C$. $BC$ is the distance covered by girl B: $d_B = V_1 t$. $AC$ is the distance covered by girl A: $d_A = V_2 t$. Since $BC \perp AB$, triangle $ABC$ is a right-angled triangle.

Step 3: Explanation:
Using Pythagoras theorem: $AC^2 = AB^2 + BC^2$ $(V_2 t)^2 = b^2 + (V_1 t)^2$ $t^2(V_2^2 - V_1^2) = b^2 \implies t = \frac{b}{\sqrt{V_2^2 - V_1^2}}$.

Step 4: Final Answer:
The time taken is $\frac{b}{\sqrt{V_2^2 - V_1^2}}$.