As shown in the figure, a particle is moving with constant speed s...

The given data is

$v = \pim{s}^{- 1}$

By using trigonometric ratios in the triangle shown above the displacement can be found out by

$\sin 60 ° = \dfrac{AB}{2R} ⇒ AB = R\sqrt{3}$

The formula for distance is

$d = vt ⇒ t = \dfrac{d}{v} ⇒ t = \dfrac{120 °}{360 °} \times \dfrac{2\piR}{\pi} = \dfrac{2R}{3}$

So, the average velocity is

${v}_{avg} = \dfrac{\sqrt{3}R}{\dfrac{2R}{3}} = \dfrac{3\sqrt{3}}{2}$

$= 1 . 5\sqrt{3}m{s}^{- 1}$