A wheel rolls along a plane surface such that its centre moves at a...

For a wheel in pure rolling, the velocity of its centre is $v_0$ and its angular velocity is $\omega = \dfrac{v_0}{R}$, where $R$ is the radius of the wheel.

The velocity of any point on the wheel is the vector sum of the velocity of the centre and the velocity of the point with respect to the centre.

Let the wheel move in the horizontal direction with velocity $\vec{v}_{c} = v_0 \hat{i}$.

For a particle on the rim at the same level as the centre, its position vector relative to the centre is horizontal. Hence, its velocity relative to the centre is vertical, given by $\vec{v}_{pc} = \pm \omega R \hat{j} = \pm v_0 \hat{j}$.

The velocity of the particle with respect to the ground is $\vec{v}_{p} = \vec{v}_{c} + \vec{v}_{pc} = v_0 \hat{i} \pm v_0 \hat{j}$.

The speed of the particle is the magnitude of its velocity:

$|\vec{v}_{p}| = \sqrt{v_0^2 + (\pm v_0)^2} = \sqrt{2}v_0$