A uniform solid cylinder is rolling without slipping on a...

For a cylinder rolling without slipping on a horizontal surface, the velocity of the center of mass is $v_B = \omega R$, where $\omega$ is the angular velocity and $R$ is the radius.

The velocity of any point on the cylinder is the vector sum of the translational velocity of the center of mass and the tangential velocity due to rotation.

For point A (at the highest position):
The translational velocity is $v_B$ (forward).
The tangential velocity is $\omega R = v_B$ (forward).
Thus, the speed of point A is $v_A = v_B + v_B = 2v_B$.

For point C (on the periphery at the same horizontal level as B):
The translational velocity is $v_B$ (forward).
The tangential velocity is $\omega R = v_B$ (downward).
Since these two velocity vectors are perpendicular, the speed of point C is $v_C = \sqrt{v_B^2 + v_B^2} = \sqrt{2}v_B$.

Comparing $v_A$ and $v_C$:
$v_A = 2v_B = \sqrt{2}(\sqrt{2}v_B) = \sqrt{2}v_C$.

Therefore, the correct relation is $v_A = \sqrt{2} v_C$.

Answer: $v_A = \sqrt{2} v_C$