Two boys are standing at points A and B on ground where distance AB = a. The boy at point B starts running perpendicular to line AB with velocity ' V₁ '. The boy at point A starts running simultaneously with velocity 'V' and catches the other boy in time 't'. The value of 't' is (a) [(a²)/((V² - V₁²))]1/2 (b) [(a²)/((V₁² - V²))]1/2 (c) [(a²)/((V² - V₁²))] (d) [(a²)/((V₁² - V²))]

Answer: (a). They meet at a point C with BC = V₁ t, AC = Vt and AB ⊥ BC. (Vt)² = a² + (V₁ t)² t = [(a²)/(V² - V₁²)]1/2 So the answer is (a).

Two boys are standing at points A and B on ground where distance AB =… | MHT CET 2026 (16 Apr, Shift 1)

Q3MHT CET 2026 · 16 APR · SHIFT 1Medium

Relative Velocity in Two Dimensions

Two boys are standing at points A and B on ground where distance AB = a. The boy at point B starts running perpendicular to line AB with velocity ' $V_1$ '. The boy at point A starts running simultaneously with velocity 'V' and catches the other boy in time 't'. The value of 't' is

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Answer: (a)

They meet at a point C with $BC = V_1 t$ , $AC = Vt$ and $AB \perp BC$ .
$(Vt)^2 = a^2 + (V_1 t)^2$
$t = \left[\frac{a^2}{V^2 - V_1^2}\right]^{1/2}$
So the answer is (a).

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Relative Velocity in Two Dimensions question from MHT CET 2026 (16 Apr, Shift 1)

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