A particle is moving in a circular path with a uniform speed.v. Show that, when the particle traverses through an angle of `120^@`, the change in its velocity is `sqrt(3)v`.

Explanation

The initial velocity `vecv_1` and the final velocity `vecv_2` are shown in Fig.2.26. in Fig.2.27, they are drawn from the same initial point.


Here, `|vecv_1|=|vecv_2|=v`.
Change in velocity `vecv_2-vecv_1`
`|vec(v)_(2)-vec(v)_(1)|=sqrt(v_(1)^(2)+v_(2)^(2)-2v_(1)v_(2)cos 120^(@))`
`=sqrt(v^(2)+v^(2)-2v*v*(-1/2))=sqrt(3v^(2))=sqrt(3v)`