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JEE Main 202429 JanuaryMorning Shift
Exam
JEE Main
Year
2024
Date
29 January
Shift
Morning Shift
Subject
Physics
Chapter
Motion in One Dimension

JEE Main 2024 · 29 January · Morning Shift · Physics · Motion in One Dimension

A body starts moving from rest with constant acceleration covers...

A body starts moving from rest with constant acceleration covers displacement S1{S}_{1} in first (p−1)( p - 1 ) seconds and S2{S}_{2} in first pp seconds. The displacement S1+S2{S}_{1} + {S}_{2} will be made in time :

Options

  1. A(2p+1)s( 2p + 1 ) s
  2. B(2p2−2p+1)s\sqrt{\left(2{p}^{2} - 2p + 1\right)}s
  3. C(2p−1)s( 2p - 1 ) s
  4. D(2p2−2p+1)s\left(2{p}^{2} - 2p + 1\right)s

Correct answer

B. (2p2−2p+1)s\sqrt{\left(2{p}^{2} - 2p + 1\right)}s

Step-by-step solution

For the displacement S1{S}_{1} of the body, it can be written that

S1=12a(p−1)2...(1){S}_{1} = \dfrac{1}{2}a ( p - 1{)}^{2} . . . \left(1\right)

And, for the displacement S2{S}_{2} of the body, it follows that

S2=12a(p)2...(2){S}_{2} = \dfrac{1}{2}a ( p{)}^{2} . . . \left(2\right)

If tt be the required time, it follows that

S1+S2=12at2...(3){S}_{1} + {S}_{2} = \dfrac{1}{2}a{t}^{2} . . . \left(3\right)

Equations (1), (2) and (3) imply that

(p−1)2+p2=t2⇒t=2p2−2p+1s( p - 1{)}^{2} + {p}^{2} = {t}^{2} ⇒ t = \sqrt{2{p}^{2} - 2p + 1}s

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