In the above question, the magnitude of the average acceleration over the arc of `60^(@)` is

Options

A. `10 m//s^(2)`
B. `11 m//s^(2)`
C. `11.5 m//s^(2)`
D. `12 m//s^(2)`

Explanation

To solve the problem of finding the magnitude of the average acceleration over the arc of 60 degrees for an automobile moving around a curve, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Radius of the curve, \( R = 300 \) meters
- Constant speed of the automobile, \( V = 60 \) meters/second
- Angle of the arc, \( \theta = 60^\circ \)

2. **Calculate Change in Velocity (\( \Delta V \)):**
- The formula for change in velocity when moving along an arc is given by:
\[ \Delta V = 2V \sin\left(\frac{\theta}{2}\right) \]
- Substitute the values:
\[ \Delta V = 2 \times 60 \times \sin\left(\frac{60^\circ}{2}\right) = 2 \times 60 \times \sin(30^\circ) \]
- Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ \Delta V = 2 \times 60 \times \frac{1}{2} = 60 \text{ m/s} \]

3. **Calculate Angular Velocity (\( \omega \)):**
- The angular velocity can be calculated using the formula:
\[ \omega = \frac{V}{R} \]
- Substitute the values:
\[ \omega = \frac{60}{300} = \frac{1}{5} \text{ rad/s} \]

4. **Calculate the Time Taken (\( T \)) to Cover the Arc:**
- The time taken to cover the arc can be calculated using:
\[ T = \frac{\theta}{\omega} \]
- Convert \( \theta \) from degrees to radians:
\[ \theta = 60^\circ = \frac{\pi}{3} \text{ radians} \]
- Substitute the values:
\[ T = \frac{\frac{\pi}{3}}{\frac{1}{5}} = \frac{5\pi}{3} \text{ seconds} \]

5. **Calculate Average Acceleration (\( A_{avg} \)):**
- The average acceleration can be calculated using the formula:
\[ A_{avg} = \frac{\Delta V}{T} \]
- Substitute the values:
\[ A_{avg} = \frac{60}{\frac{5\pi}{3}} = 60 \times \frac{3}{5\pi} = \frac{180}{5\pi} = \frac{36}{\pi} \text{ m/s}^2 \]
- Using \( \pi \approx 3.14 \):
\[ A_{avg} \approx \frac{36}{3.14} \approx 11.46 \text{ m/s}^2 \]

6. **Final Answer:**
- The magnitude of the average acceleration over the arc of \( 60^\circ \) is approximately \( 11.46 \text{ m/s}^2 \).