Two girls are standing at the ends ' A ' and ' B ' of a ground where AB = b. The girl at ' B ' starts running in a direction perpendicular to ' AB ' with velocity ' V₁ '. The girl at ' A ' starts running simultaneously with velocity ' V₂ ' and in shortest distance meets the other girl in time ' t '. The value of ' t ' is (a) (b)/(√(V₁² + V₂²)) (b) (b)/(V₁ + V₂) (c) (b)/(V₂ - V₁) (d) (b)/(√(V₂² - V₁²))

Answer: (d). Meeting point C: BC = V₁ t, AC = V₂ t, right angle at B. (V₂ t)² = b² + (V₁ t)² t = (b)/(√(V₂² - V₁²)) So the answer is (d).

Two girls are standing at the ends ' A ' and ' B ' of a ground where… | MHT CET 2025 (25 Apr, Shift 2)

Q7MHT CET 2025 · 25 APR · SHIFT 2Medium

Relative Velocity in Two Dimensions

Two girls are standing at the ends ' $A$ ' and ' $B$ ' of a ground where $AB = b$ . The girl at ' B ' starts running in a direction 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

Show answer and solution

Answer: (d)

Meeting point C: $BC = V_1 t$ , $AC = V_2 t$ , right angle at B.
$(V_2 t)^2 = b^2 + (V_1 t)^2$
$t = \frac{b}{\sqrt{V_2^2 - V_1^2}}$
So the answer is (d).

Report a mistake

Relative Velocity in Two Dimensions question from MHT CET 2025 (25 Apr, Shift 2)

More questions on Relative Velocity in Two Dimensions