A bicycle wheel is rolling along a level road with a linear speed...

For a wheel in pure rolling on a stationary horizontal surface, the velocity of any point on the wheel is the vector sum of the translational velocity of the center of mass ($v_0$) and the tangential velocity due to rotation about the center of mass ($\omega R$).

Since the wheel is rolling without slipping, the velocity of the lowest point $A$ (point of contact) is zero. Thus, $v_A = v_0 - \omega R = 0$, which gives $\omega R = v_0$.

The velocity of the topmost point $C$ is the sum of the translational and rotational velocities, both pointing in the same direction: $v_C = v_0 + \omega R = v_0 + v_0 = 2v_0$.

For point $B$, the translational velocity is $v_0$ horizontally, and the rotational velocity is $\omega R = v_0$ vertically downwards. The net speed of $B$ is $v_B = \sqrt{v_0^2 + v_0^2} = \sqrt{2}v_0$.

The speed of the center $O$ is simply $v_0$. Since $\sqrt{2}v_0 > v_0$, the speed of $B$ is greater than the speed of $O$.

Therefore, the speed of $A$ is zero, the speed of $C$ is $2v_0$, and the speed of $B$ is greater than the speed of $O$.