Question:

A particle is moving with a uniform speed $v$ in a circular path of radius $r$ with the centre at $O$. When the particle moves from a point $P$ to $Q$ on the circle such that $? POQ=\theta$, then the magnitude of the change in velocity is

Updated On: Feb 15, 2024
  • $2v\,sin\left(2\theta\right)$
  • zero
  • $2v\,sin\left(\frac{\theta}{2}\right)$
  • $2v\,cos\left(\frac{\theta}{2}\right)$
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The Correct Option is C

Solution and Explanation


$\therefore$ Change in magnitude of velocity
$|\Delta v | =\sqrt{v^{2}+v^{2}-2 v^{2} \cos \theta}$
$=2 v \sin \frac{\theta}{2}$
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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