A body starts moving from rest with constant acceleration covers displacement S₁ in first (p-1) seconds and S₂ in first p seconds. The displacement S₁ + S₂ will be made in time: (a) (2p+1)s (b) (2p-1)s (c) (2p² - 2p + 1)s (d) √((2p² - 2p + 1)s)

Answer: (d). From rest, S = 1/2at²: S₁ = 1/2a(p - 1)², S₂ = 1/2ap². S₁ + S₂ = 1/2a[(p - 1)² + p²] = 1/2a(2p² - 2p + 1) Set equal to 1/2aT²: T = √(2p² - 2p + 1) s So the answer is (d) √(2p² - 2p + 1) s.

A body starts moving from rest with constant acceleration covers… | JEE Main 2024 (29 Jan, Shift 1)

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Equations of Kinematics

A body starts moving from rest with constant acceleration covers displacement $S_1$ in first $(p-1)$ seconds and $S_2$ in first $p$ seconds. The displacement $S_1 + S_2$ will be made in time:

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Answer: (d)

From rest, $S = \frac{1}{2}at^2$ : $S_1 = \frac{1}{2}a(p - 1)^2$ , $S_2 = \frac{1}{2}ap^2$ .
$S_1 + S_2 = \frac{1}{2}a\left[(p - 1)^2 + p^2\right] = \frac{1}{2}a(2p^2 - 2p + 1)$
Set equal to $\frac{1}{2}aT^2$ : $T = \sqrt{2p^2 - 2p + 1}\text{ s}$
So the answer is (d) $\sqrt{2p^2 - 2p + 1}$ s.

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