Two buses P and Q start from a point at the same time and move in a straight line and their positions are represented by XP(t) = α t + β t² and XQ(t) = ft - t². At what time, both the buses have same velocity? (a) (α - f)/(1 + β) (b) (α + f)/(2(β - 1)) (c) (α + f)/(2(1 + β)) (d) (f - α)/(2(1 + β))

Answer: (d). vP = α + 2β t, vQ = f - 2t. α + 2β t = f - 2t ⇒ 2t(1 + β) = f - α So the answer is (d) t = (f - α)/(2(1 + β)).

Two buses P and Q start from a point at the same time and move in a… | JEE Main 2022 (25 Jun, Shift 2)

Q21JEE Main 2022 · 25 JUN · SHIFT 2Easy

Velocity, Acceleration & Variable Motion

Two buses P and Q start from a point at the same time and move in a straight line and their positions are represented by $X_P(t) = \alpha t + \beta t^2$ and $X_Q(t) = ft - t^2$ . At what time, both the buses have same velocity?

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Answer: (d)

$v_P = \alpha + 2\beta t$ , $v_Q = f - 2t$ .
$\alpha + 2\beta t = f - 2t \Rightarrow 2t(1 + \beta) = f - \alpha$
So the answer is (d) $t = \frac{f - \alpha}{2(1 + \beta)}$ .

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Velocity, Acceleration & Variable Motion question from JEE Main 2022 (25 Jun, Shift 2)

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