The centre of a wheel rolling on a plane surface moves with a speed  ${\text{v}}_0,\text{A}$  particle on the rim of the wheel at the same level as the centre will be moving at speed.

Given that

The centre of rolling wheel move along two plane with speed $={\text{v}}_{\text{0}}$

${\text{V}}_0=$Speed of rim at units

${\text{V}}_0=\text{R}\omega$

Velocity at points A and B at same level $=\text{v\hspace{0.17em}};-\text{v}$

Since with constant than velocity of point A = velocity due to translation + velocity due to rotation

$=\frac{\text{v}}{\text{R}}=\frac{{\text{v}}_0}{\text{R}}$

$\Rightarrow \text{v}={\text{v}}_{\text{0}}$

$={\overrightarrow{v}}_0+{\overrightarrow{v}}_0\left(\text{or}\right)90°$

$\therefore {\text{v}}_{\text{A}}=\sqrt{2}{\text{v}}_{\text{0}}$