Given that the speed of the particle at the highest point of the rim is 8 m/s, and the wheel is rolling without slipping, the speed at any point on the rim is the sum of the velocity of the center of the wheel and the velocity due to the rotational motion.
Let:
- \( V_B = 8 \, \text{m/s} \) (speed at the highest point of the rim),
- \( V = 4 \, \text{m/s} \) (speed of the center of the wheel),
- \( V_P = \sqrt{2}V \) (velocity at point \( P \)).
Since the wheel is rolling without slipping, the speed at point \( P \) (which is the same level as the center of the wheel) will be: \[ V_P = \sqrt{2} \times 4 = 4\sqrt{2} \, \text{m/s} \]
Thus, the correct answer is (1).
A sportsman runs around a circular track of radius $ r $ such that he traverses the path ABAB. The distance travelled and displacement, respectively, are:
The motion of an airplane is represented by the velocity-time graph as shown below. The distance covered by the airplane in the first 30.5 seconds is km.