A particle is moving on a circular path with constant speed v. The magnitude of the change in its velocity after it has described an angle of `90^(@)` is :

Options

A. v
B. `sqrt(2) v`
C. 0
D. `sqrt(3)` v

Explanation

To solve the problem of finding the magnitude of the change in velocity of a particle moving in a circular path after it has described an angle of 90 degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial and Final Positions**:
- The particle starts at a position on the circular path and moves 90 degrees to reach a new position.

2. **Define Initial and Final Velocity Vectors**:
- Let’s assume the particle starts at the top of the circle (0 degrees) and moves to the right (90 degrees).
- At the initial position (0 degrees), the velocity vector \( \vec{v_i} \) can be represented as:
\[ \vec{v_i} = v \hat{j} \]
- At the final position (90 degrees), the velocity vector \( \vec{v_f} \) can be represented as:
\[ \vec{v_f} = -v \hat{i} \]

3. **Calculate the Change in Velocity**:
- The change in velocity \( \Delta \vec{v} \) is given by:
\[ \Delta \vec{v} = \vec{v_f} - \vec{v_i} \]
- Substituting the values of \( \vec{v_f} \) and \( \vec{v_i} \):
\[ \Delta \vec{v} = (-v \hat{i}) - (v \hat{j}) = -v \hat{i} - v \hat{j} \]

4. **Magnitude of the Change in Velocity**:
- To find the magnitude of \( \Delta \vec{v} \), we use the Pythagorean theorem:
\[ |\Delta \vec{v}| = \sqrt{(-v)^2 + (-v)^2} = \sqrt{v^2 + v^2} = \sqrt{2v^2} = v\sqrt{2} \]

5. **Final Result**:
- Therefore, the magnitude of the change in velocity after the particle has described an angle of 90 degrees is:
\[ |\Delta \vec{v}| = v\sqrt{2} \]