A body starts moving from rest with constant acceleration covers displacement $S_{1}$ in first ($p-1$) seconds and $\mathrm{S}_{2}$ in first p seconds. The displacement $\mathrm{S}_{1}+\mathrm{S}_{2}$ will be made in time:

(B) $\sqrt{\left(2 p^{2}-2 p+1\right)} s$

(D) $\left(2 p^{2}-2 p+1\right) s$

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · 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:

(A) $(2 p + 1) s$
(B) $(2 p^{2} - 2 p + 1) s$ ✓ Correct
(C) $(2 p - 1) s$
(D) $(2 p^{2} - 2 p + 1) s$

Answer: (B)

Solution

Starting from rest with constant acceleration

a

S_{1} = \frac{1}{2} a (p - 1)^{2}, S_{2} = \frac{1}{2} a p^{2}

S_{1} + S_{2} = \frac{1}{2} a [(p - 1)^{2} + p^{2}]

If this same displacement is covered from rest in time

t

S_{1} + S_{2} = \frac{1}{2} a t^{2}

, so

t^{2} = (p - 1)^{2} + p^{2} = 2 p^{2} - 2 p + 1 ⟹ t = 2 p^{2} - 2 p + 1 s

Key confirmed: B.

JEE Main · 2024 · 29 Jan 2024 · Shift 1 · Single correct option

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