Two boys are standing at points A and B on ground, where distance AB = x. The boy at B stars running perpendicular to AB with velocity v₁. The boy at A starts running simultaneously with velocity v and meets the other boy in time t. The value of t is (a) [(x)/(v - v₁)]1/2 (b) [(x)/(v₁ - v)]1/2 (c) [(x²)/(v² - v₁²)]1/2 (d) [(x²)/(v₁² - v²)]1/2

Answer: (c). Meeting point C: BC = v₁ t, AC = vt, right angle at B. (vt)² = x² + (v₁ t)² t = [(x²)/(v² - v₁²)]1/2 So the answer is (c).

Two boys are standing at points A and B on ground, where distance AB… | MHT CET 2025 (25 Apr, Shift 1)

Q8MHT CET 2025 · 25 APR · SHIFT 1Medium

Relative Velocity in Two Dimensions

Two boys are standing at points A and B on ground, where distance $AB = x$ . The boy at B stars running perpendicular to $AB$ with velocity $v_1$ . The boy at A starts running simultaneously with velocity v and meets the other boy in time $t$ . The value of $t$ is

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Answer: (c)

Meeting point C: $BC = v_1 t$ , $AC = vt$ , right angle at B.
$(vt)^2 = x^2 + (v_1 t)^2$
$t = \left[\frac{x^2}{v^2 - v_1^2}\right]^{1/2}$
So the answer is (c).

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Relative Velocity in Two Dimensions question from MHT CET 2025 (25 Apr, Shift 1)

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