# A particle is moving in a circle of radius R with constant speed . The time period of particle is T=1 Second in a time t = T/6 , If the difference between average speed and magnitude of average velocity of the partical is 2m/sec, find the radius of the circle(in meteres)

Dear Student,

Please find below the solution to the asked query:

Given that the time period of the particle is T. Then, the angular velocity is,

So, in time t = T/6, the particle will be having the angular displacement,

$\theta =\omega t=\frac{2\pi }{T}×\frac{T}{6}=\frac{\pi }{3}$
Therefore, the distance travelled by the body is,

Displacement of the body is,

$dispacement=R$

So, the magnitude of velocity is,

$\left|V\right|=\frac{Displacement}{time}=\frac{R}{T}{6}}=\frac{6R}{T}$

The magnitude of speed is,

$\left|S\right|=\frac{Dis\mathrm{tan}ce}{time}=\frac{\pi R}{3}}{T}{6}}=\frac{2\pi R}{T}$

Therefore, the difference of the magnitude of the speed and the magnitude of the velocity is,

Hope this information will clear your doubts about the topic.

If you have any more doubts just ask here on the forum and our experts will try to help you out as soon as possible.

Regards

• 67
What are you looking for?