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JEE Main 202125 JulyEvening Shift
Exam
JEE Main
Year
2021
Date
25 July
Shift
Evening Shift
Subject
Physics
Chapter
Motion in One Dimension

JEE Main 2021 · 25 July · Evening Shift · Physics · Motion in One Dimension

The relation between time t and distance x for a moving body is...

The relation between time tt and distance xx for a moving body is given as t=mx2+nxt = m{x}^{2} + nx, where mm and nn are constants. The retardation of the motion is: (When vv stands for velocity)

Options

  1. A2mv32m{v}^{3}
  2. B2mnv32mn{v}^{3}
  3. C2nv32n{v}^{3}
  4. D2n2v32{n}^{2}{v}^{3}

Correct answer

A. 2mv32m{v}^{3}

Step-by-step solution

t=mx2+nxt = m{x}^{2} + nx

1v=dtdx=2mx+n\dfrac{1}{v} = \dfrac{\text{d}t}{\text{d}x} = 2mx + n

V=12mx+nV = \dfrac{1}{2mx + n}

dvdt=−2m(2mx+n)2(dxdt)\dfrac{\text{d}v}{\text{d}t} = - \dfrac{2m}{( 2mx + n{)}^{2}}\left(\dfrac{\text{d}x}{\text{d}t}\right)

a=−(2m)v3a = - ( 2m ) {v}^{3}

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