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(C) For a wheel rolling on the ground with a translational speed $v = 2 \, m/s$,every point on the wheel has a translational velocity $v_t = v$ in the

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$2\sqrt{2} \, m/s$

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(C) For a wheel rolling on the ground with a translational speed $v = 2 \, m/s$,every point on the wheel has a translational velocity $v_t = v$ in the forward direction.Additionally,due to rotation about the center of mass,every point on the rim has a tangential velocity $v_R = v$ relative to the center.At the extremities of the horizontal diameter,the translational velocity $v_t$ is horizontal,and the rotational velocity $v_R$ is vertical (downward at the front point,upward at the back point).Since these two velocity vectors are perpendicular to each other,the resultant velocity $v_N$ is given by:$v_N = \sqrt{v_t^2 + v_R^2} = \sqrt{v^2 + v^2} = \sqrt{2}v$Substituting $v = 2 \, m/s$:$v_N = 2\sqrt{2} \, m/s$.

$A$ wheel is rolling along the ground with a speed of $2\ m/s$. The magnitude of the velocity of the points at the extremities of the horizontal diameter of the wheel is equal to:

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