A clock a minute-hand `10 cm` long. Find the average velocity between `6.00 AM` to `6.30 AM` for the tip of minute-hand.

Options

A. `22/21 cm "min"^(-1)`
B. `2/21 cm "min"^(-1)`
C. `12/21 cm "min"^(-1)`
D. `2/3 cm "min"^(-1)`

Explanation

To find the average velocity of the tip of the minute hand of a clock from 6:00 AM to 6:30 AM, we can follow these steps:

### Step 1: Understand the movement of the minute hand
The minute hand of the clock moves in a circular path. From 6:00 AM to 6:30 AM, the minute hand moves from the 12 o'clock position to the 6 o'clock position.

### Step 2: Calculate the displacement
The displacement is the shortest distance between the initial and final positions of the minute hand.
- At 6:00 AM, the tip of the minute hand is at the 12 o'clock position (0 degrees).
- At 6:30 AM, the tip of the minute hand is at the 6 o'clock position (180 degrees).

The distance between these two points (12 o'clock and 6 o'clock) can be visualized as a straight line through the center of the clock.

Since the length of the minute hand is 10 cm, the displacement can be calculated as:
- Displacement = 2 * radius = 2 * 10 cm = 20 cm.

### Step 3: Calculate the time interval
The time interval from 6:00 AM to 6:30 AM is 30 minutes.

### Step 4: Calculate the average velocity
Average velocity is defined as the total displacement divided by the total time taken.

Using the formula:
\[ \text{Average Velocity} = \frac{\text{Displacement}}{\text{Time}} \]

Substituting the values:
\[ \text{Average Velocity} = \frac{20 \text{ cm}}{30 \text{ min}} = \frac{2}{3} \text{ cm/min} \]

### Final Answer
The average velocity of the tip of the minute hand from 6:00 AM to 6:30 AM is \( \frac{2}{3} \) cm/min.

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