Q104 / 200 · NEET 2012 (Screening) · Physics

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

  • A
    $24 \mathrm{~ms}^{-2}$
  • B
    zero
  • C
    $6 \mathrm{~ms}^{-2}$
  • Correct answer
    D
    $12 \mathrm{~ms}^{-2}$

Solution

Given, $x=8+12 t-t^3$ We know $v=\frac{d x}{d t}$ and $a=\frac{d v}{d t}$ So, $v=12-3 t^2$ and $a=-6 t$ At $t=2 \mathrm{~s}$ $y=0$ and $a=-6 t$ So, $a=-12 \mathrm{~m} / \mathrm{s}^2$ So, retardation of the particle $=12 \mathrm{~m} / \mathrm{s}^2$.

Hence, the correct answer is option: 'D'.

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