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 Question:
This is a classic relative kinematics "intercept" problem. Girl B runs along a straight line perpendicular to the initial separation vector AB. Girl A must angle her run perfectly so that she intercepts Girl B. We need to find the time $t$ it takes for them to meet.

Step 2: Key Formula or Approach:
For Girl A to intercept Girl B, the component of Girl A's velocity that is parallel to Girl B's velocity must exactly match Girl B's velocity. This ensures they stay "aligned" as they run.
Let the vector $\vec{AB}$ be along the x-axis. Girl B runs along the y-axis.
Velocity of B: $\vec{V_B} = V_1 \hat{j}$
Velocity of A: $\vec{V_A} = V_2 \cos\theta \hat{i} + V_2 \sin\theta \hat{j}$ (where $\theta$ is the angle A makes with line AB).
To intercept, the y-components must be equal: $V_2 \sin\theta = V_1$.
The relative velocity driving them together along the x-axis is closing the distance $b$.

Step 3: Detailed Explanation:
From the matching velocity components constraint:
$$\sin\theta = \frac{V_1}{V_2}$$
The speed at which Girl A is physically closing the distance $b$ along the line AB is her x-component of velocity:
$$V_{closing} = V_2 \cos\theta$$
Using the trigonometric identity $\cos\theta = \sqrt{1 - \sin^2\theta}$:
$$V_{closing} = V_2 \sqrt{1 - \left(\frac{V_1}{V_2}\right)^2}$$
$$V_{closing} = V_2 \sqrt{\frac{V_2^2 - V_1^2}{V_2^2}} = V_2 \frac{\sqrt{V_2^2 - V_1^2}}{V_2}$$
$$V_{closing} = \sqrt{V_2^2 - V_1^2}$$
The time $t$ required to cover the initial separation distance $b$ at this closing speed is:
$$t = \frac{\text{Distance}}{\text{Closing Speed}} = \frac{b}{\sqrt{V_2^2 - V_1^2}}$$

Step 4: Final Answer:
The required time is $\frac{b}{\sqrt{V_2^2 - V_1^2}}$, matching option (d).