Question:

A particle moves in a circular path with uniform speed v. When it turns by 90°, find ratio of \(\frac{v}{<\overrightarrow{v}>}\)?

Updated On: Sep 27, 2024
  • \(\frac{\pi}{\sqrt{2}}\)

  • \(\frac{2\pi}{\sqrt{2}}\)
  • \(\frac{\pi}{2}\)
  • \(2\pi\)
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The Correct Option is A

Solution and Explanation

\(|<\overrightarrow{v}>|=\frac{|displacement|}{time}=\frac{\sqrt{2}R}{\frac{\pi R}{v}}=\frac{\sqrt{2}v}{\pi}\)
\(\frac{v}{<\overrightarrow{v}>}=\frac{v}{(\frac{\sqrt{2}v}{\pi})}=\frac{\pi}{\sqrt{2}}\)

So, the correct answer is (A): \(\frac{\pi}{\sqrt{2}}\)

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Concepts Used:

Motion in a Plane

It is a vector quantity. A vector quantity is a quantity having both magnitude and direction. Speed is a scalar quantity and it is a quantity having a magnitude only. Motion in a plane is also known as motion in two dimensions. 

Equations of Plane Motion

The equations of motion in a straight line are:

v=u+at

s=ut+½ at2

v2-u2=2as

Where,

  • v = final velocity of the particle
  • u = initial velocity of the particle
  • s = displacement of the particle
  • a = acceleration of the particle
  • t = the time interval in which the particle is in consideration