A particle is moving in a circle of radius `r` centred at `O` with constant speed `v`. What is the change in velocity in moving from `A "to" B (/_ AOB = 40^@)` ?

Options

A. `2 v sin 20^@`
B. `4 v sin 40^@`
C. `2 v sin 40^@`
D. `v sin 20^@`

Explanation

To find the change in velocity of a particle moving in a circle from point A to point B, where the angle ∠AOB is 40 degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial and Final Velocities**:
- Let the initial velocity at point A be \( \vec{v_1} \) and the final velocity at point B be \( \vec{v_2} \).
- Since the particle is moving with constant speed \( v \), we have:
\[ |\vec{v_1}| = |\vec{v_2}| = v \]

2. **Determine the Angle Between the Velocities**:
- The angle between the two velocity vectors \( \vec{v_1} \) and \( \vec{v_2} \) is given as \( \theta = 40^\circ \).

3. **Use the Formula for Change in Velocity**:
- The change in velocity \( \Delta \vec{v} \) can be calculated using the formula:
\[ |\Delta \vec{v}| = |\vec{v_2} - \vec{v_1}| \]
- This can be expressed using the cosine rule:
\[ |\Delta \vec{v}| = \sqrt{|\vec{v_1}|^2 + |\vec{v_2}|^2 - 2 |\vec{v_1}| |\vec{v_2}| \cos \theta} \]

4. **Substituting the Values**:
- Since \( |\vec{v_1}| = |\vec{v_2}| = v \):
\[ |\Delta \vec{v}| = \sqrt{v^2 + v^2 - 2 v v \cos(40^\circ)} \]
- Simplifying this:
\[ |\Delta \vec{v}| = \sqrt{2v^2 - 2v^2 \cos(40^\circ)} \]
- Factor out \( 2v^2 \):
\[ |\Delta \vec{v}| = \sqrt{2v^2(1 - \cos(40^\circ))} \]

5. **Using the Identity for Sine**:
- We can use the identity \( 1 - \cos(\theta) = 2 \sin^2\left(\frac{\theta}{2}\right) \):
\[ 1 - \cos(40^\circ) = 2 \sin^2(20^\circ) \]
- Therefore:
\[ |\Delta \vec{v}| = \sqrt{2v^2 \cdot 2 \sin^2(20^\circ)} = \sqrt{4v^2 \sin^2(20^\circ)} \]

6. **Final Expression for Change in Velocity**:
- Taking the square root:
\[ |\Delta \vec{v}| = 2v \sin(20^\circ) \]

### Conclusion:
The change in velocity as the particle moves from point A to point B is:
\[ |\Delta \vec{v}| = 2v \sin(20^\circ) \]