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NEET 2012
Exam
NEET
Year
2012
Subject
Physics
Chapter
Motion in One Dimension

NEET 2012 · Physics · Motion in One Dimension

The motion of a particle along a straight line is described by...

The motion of a particle along a straight line is described by equation x=8+12t−t3x=8+12 t-t^3 where, xx is in metre and tt in second. The retardation of the particle when its velocity becomes zero, is

Options

  1. A24 ms−224 \mathrm{~ms}^{-2}
  2. Bzero
  3. C6 ms−26 \mathrm{~ms}^{-2}
  4. D12 ms−212 \mathrm{~ms}^{-2}

Correct answer

D. 12 ms−212 \mathrm{~ms}^{-2}

Step-by-step solution

Given, x=8+12t−t3x=8+12 t-t^3 We know v=dxdtv=\frac{d x}{d t} and a=dvdta=\frac{d v}{d t} So, v=12−3t2v=12-3 t^2 and a=−6ta=-6 t At t=2 st=2 \mathrm{~s} y=0y=0 and a=−6ta=-6 t So, a=−12 m/s2a=-12 \mathrm{~m} / \mathrm{s}^2 So, retardation of the particle =12 m/s2=12 \mathrm{~m} / \mathrm{s}^2.
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