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NEET 2013
Exam
NEET
Year
2013
Subject
Physics
Chapter
Motion in One Dimension

NEET 2013 · Physics · Motion in One Dimension

The displacement ' x ' (in meter) of a particle of mass ' m ' (in...

The displacement ' xx ' (in meter) of a particle of mass ' mm ' (in kg) moving in one dimension under the action of a force, is related to time ' tt ' (in sec) by t=x+3t=\sqrt{x}+3. The displacement of the particle when its velocity is zero will be:

Options

  1. A4 m4 \mathrm{~m}
  2. B0 m0 \mathrm{~m} (zero)
  3. C6 m6 \mathrm{~m}
  4. D2 m2 \mathrm{~m}

Correct answer

B. 0 m0 \mathrm{~m} (zero)

Step-by-step solution

According to the question t=x+3⇒x=t−3t=\sqrt{x}+3 \Rightarrow \sqrt{x}=t-3 Squaring both sides we get x=(t−3)2x=(t-3)^2 Now velocity V=dxdt=ddt(t−3)2=2(t−3)V=\frac{d x}{d t}=\frac{d}{d t}(t-3)^2 =2(t-3) velocity becomes zero means 2(t−3)=0⇒t=3sec.At t=3sx=(3−3)2=0m\begin{aligned} & 2(t-3)=0 \Rightarrow t=3 \mathrm{sec} . \\ & \text{At } \begin{array}{l} t=3 \mathrm{ s} \\ \\ x=(3-3)^2=0 m \end{array} \end{aligned}
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