A particle describes an angle `theta` in a circular path with a constant speed `v`. Find the `a` charge in the velocity of the particle and `b` average acceleration of the particle during the motion in the curve (circle).
.

Explanation

a. As the particle moves from `P` to `Q`. The velocty turns through an angle `theta`. Then,
`|Deltavecv|=sqrt(v_(1)^(2)+v_(2)^(2)-2v_(1)v_(2) cos theta)`
`= sqrt(v^(2)+v^(2)-2 vv cos theta)=2v sin ((theta)/(2))`
b. The time of motion is `Deltat=R theta //v`. Then, average acceleration is `|veca_(av)|=vec v Delta`.
Substituting `|Delta vecv|` and `Delta`, we have `|veca_(aV)|=(2v sin ((theta)/(2)))/((R theta)/(v))=sin ((theta)/(2))`.