A table clock has its minute hand 5 cm long. Find the average velocity of the minute hand between 6:00 a.m. to 6:30 a.m.

Options

A. `11 xx 10^(-5) m//s`
B. `5.5 xx 10^(-5) m//s`
C. `9 xx 10^(-5) m//s`
D. `18 xx 10^5 m//s`

Explanation

To find the average velocity of the minute hand of a clock between 6:00 a.m. and 6:30 a.m., we can follow these steps:

### Step 1: Understand the Problem
We need to calculate the average velocity of the minute hand, which involves determining the displacement and the time taken.

### Step 2: Determine the Length of the Minute Hand
The length of the minute hand is given as 5 cm.

### Step 3: Calculate the Displacement
At 6:00 a.m., the minute hand points at the 12 on the clock. At 6:30 a.m., the minute hand points at the 6. The displacement is the straight line distance between these two points.

- The distance from the 12 to the 6 is the diameter of the circle traced by the minute hand.
- The diameter can be calculated as:
\[ \text{Diameter} = 2 \times \text{Length of minute hand} = 2 \times 5 \text{ cm} = 10 \text{ cm} \]

### Step 4: Convert Displacement to Meters
To convert the displacement from centimeters to meters:
\[ 10 \text{ cm} = \frac{10}{100} \text{ m} = 0.1 \text{ m} \]

### Step 5: Determine the Time Interval
The time from 6:00 a.m. to 6:30 a.m. is 30 minutes. We need to convert this time into seconds:
\[ 30 \text{ minutes} = 30 \times 60 \text{ seconds} = 1800 \text{ seconds} \]

### Step 6: Calculate Average Velocity
The formula for average velocity (\( V_{avg} \)) is given by:
\[ V_{avg} = \frac{\text{Total Displacement}}{\text{Total Time}} \]
Substituting the values we found:
\[ V_{avg} = \frac{0.1 \text{ m}}{1800 \text{ s}} = \frac{1}{18000} \text{ m/s} \]

### Step 7: Simplify the Result
Calculating this gives:
\[ V_{avg} \approx 5.56 \times 10^{-5} \text{ m/s} \]

### Final Answer
The average velocity of the minute hand between 6:00 a.m. and 6:30 a.m. is approximately \( 5.56 \times 10^{-5} \text{ m/s} \).

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