# 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.

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