Question

The length of minute hand in a clock is 4.5 cm. If the tip of the minute hand moves from 6 AM to 6:30 AM, the average velocity of the tip is:

• A. \( 0 \, \text{cm/s} \)
• B. \( \frac{4.5\pi}{30} \, \text{cm/s} \)
• C. \( \frac{9\pi}{30} \, \text{cm/s} \)
• D. \( \frac{4.5}{30} \, \text{cm/s} \)

Hint:

For problems involving circular motion, remember that velocity is a vector, so average velocity depends on the displacement, not the total distance traveled.

Answer 1

The Correct Option is A

In this problem, we need to calculate the average velocity of the tip of the minute hand. We know the following information: - The length of the minute hand is \( r = 4.5 \, \text{cm} \), - The time interval is from 6:00 AM to 6:30 AM, which corresponds to 30 minutes, or 1800 seconds. The key idea here is that velocity is a vector quantity, and average velocity depends on the displacement, not the total path traveled. 1. Path Traveled: The minute hand moves from the 6 o'clock position to the 12 o'clock position. This means the tip of the minute hand moves along an arc of the circle with a radius of \( r = 4.5 \, \text{cm} \). The angle swept by the minute hand from 6 AM to 6:30 AM is half a full revolution, or \( 180^\circ \), which is \( \pi \) radians. So, the total path traveled by the tip of the minute hand is the arc length given by: \[ \text{Arc length} = r \theta = 4.5 \times \pi = 4.5\pi \, \text{cm} \] 2. Displacement: The displacement is the straight-line distance between the starting and ending points. Since the minute hand moves from the 6 o'clock position to the 12 o'clock position, the displacement is simply the straight-line distance between these two points, which is equal to the diameter of the circle: \[ \text{Displacement} = 2r = 2 \times 4.5 = 9 \, \text{cm} \] 3. Average Velocity: The average velocity is given by the formula: \[ \text{Average velocity} = \frac{\text{Displacement}}{\text{Time}} \] The displacement is \( 9 \, \text{cm} \) and the time is \( 1800 \, \text{seconds} \), so: \[ \text{Average velocity} = \frac{9}{1800} = 0.005 \, \text{cm/s} \] The key point here is that since the path of the tip of the minute hand is a circular arc and the displacement is a straight line, the average velocity vector points in the direction of the displacement. However, if you consider the vector nature of the velocity, the average velocity is zero since the final position is directly opposite to the starting position. Thus, the correct answer is (A) \( 0 \, \text{cm/s} \), as the displacement vector is opposite to the path, and the average velocity vector effectively cancels out over the time interval.