Q148 / 180 · COMEDK 2013 · Physics

The displacement $x$ (in $\mathrm{m}$) of a particle of mass $\mathrm{m}$ (in $\mathrm{kg}$ ) moving in one dimension under the action of a force, is related to time $t$ (in $\mathrm{s}$) by $t=\sqrt{x}+3$. The displacement of the particle when its velocity is zero, will be

  • A
    $4 \mathrm{~m}$
  • Correct answer
    B
    $0$
  • C
    $6 \mathrm{~m}$
  • D
    $2 \mathrm{~m}$

Solution

Given, $t=\sqrt{x}+3$ $\Rightarrow \quad \sqrt{x}=t-3$ or $\quad x=(t-3)^{2}$ Velocity, $v=\frac{d x}{d t}=\frac{d}{d t}(t-3)^{2}=2(t-3)$ When velocity of particle is zero, then $\begin{aligned} 2(t-3) &=0 \\ \Rightarrow &=3 \mathrm{~s} \\ & \text { At } t=3 \mathrm{~s}, x=(3-3)^{2}=0 \end{aligned}$

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

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