Question

Two objects A and B are separated horizontally by distance ' a '. Object B moves in a direction perpendicular to distance ' a ' with velocity ' \( V_1 \) '. Simultaneously object A covers shortest distance with velocity ' V ' and meets object B in time ' t '. The time ' t ' is given by

• A. \( \frac{a}{V+V_1} \)
• B. \( \sqrt{\frac{a^2}{V_1^2+V^2}} \)
• C. \( \frac{a}{\sqrt{V^2-V_1^2}} \)
• D. \( \frac{a}{V-V_1} \)

Hint:

- Match perpendicular velocities for interception - Resolve velocity into components

Answer 1

The Correct Option is C

Concept: For shortest path interception, object A must move such that its velocity component perpendicular to initial line equals velocity of B.

Step 1: Resolve velocity of A.
Let A move at angle $\theta$ with horizontal. \[ \text{Vertical component of A} = V \sin\theta = V_1 \]

Step 2: Find horizontal component.
\[ V \cos\theta = \sqrt{V^2 - V_1^2} \]

Step 3: Time to meet.
Horizontal distance covered = $a$: \[ t = \frac{a}{V \cos\theta} \]

Step 4: Substitute.
\[ t = \frac{a}{\sqrt{V^2 - V_1^2}} \]