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NEET
Exam
NEET
Subject
Physics
Chapter
Motion in One Dimension

NEET · Physics · Motion in One Dimension

The velocitydisplacement (v-s) graph shows the motion of particle...

The velocitydisplacement (v−s)(v-s) graph shows the motion of particle moving in a straight linè. The velocitydisplacement (v-s) graph shows the motion of particle moving in a straight linè. Velocity-displacement
Velocity-displacement graph is a circle of radius 2 m2 \mathrm{~m} and centre is at (2,0)m(2,0) \mathrm{m}. The value of acceleration for this particle at a point (2−2,2)m(2-\sqrt{2}, \sqrt{2}) \mathrm{m} will be ms−2\mathrm{ms}^{-2}.

Options

  1. A2\sqrt{2}
  2. B4
  3. C2
  4. D3

Correct answer

A. 2\sqrt{2}

Step-by-step solution

As, graph is a circle with centre (2,0)(2,0) and radius 2 , its equation is (s−2)2+(v−0)2=22(s−2)2+v2=4\begin{aligned} (s-2)^2+(v-0)^2 & =2^2 \\ (s-2)^2+v² & =4 \end{aligned}Differentiatingwithrespecttotime,wegetwith respect to time, we get2(s−2)dsdt+2vdvdt=02(s−2)v+2v\cdota=0\begin{aligned}2(s-2) \frac{d s}{d t}+2 v \frac{d v}{d t}& =0 \\ 2(s-2) v+2 v \cdota & =0 \end{aligned}At,At,s=2-\sqrt{2}andandv=\sqrt{2} 2(2−2−2)×2+22a=0⇒−4+22a=0⇒a=422=2 ms−2\begin{aligned}& 2(2-\sqrt{2}-2) \times \sqrt{2}+2 \sqrt{2}a=0 \\ \Rightarrow & -4+2 \sqrt{2}a=0 \Rightarrow a=\frac{4}{2 \sqrt{2}}=\sqrt{2} \mathrm{~ms}^{-2} \end{aligned}$
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