A particle is moving on a circular path with constant speed v then the change in its velocity after it has described an angle of `60^(@)` will be

Options

A. `vsqrt(2)`
B. `(v)/(2)`
C. `vsqrt(3)`
D. v

Explanation

To solve the problem of finding the change in velocity of a particle moving in a circular path after it has described an angle of \(60^\circ\), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial and Final Velocities**:
- The particle is moving with a constant speed \(v\).
- Initially, let's denote the initial velocity vector as \(\vec{v_i} = v \hat{i}\) (pointing in the positive x-direction).

2. **Finding the Final Velocity**:
- After the particle has traveled an angle of \(60^\circ\), its final velocity vector can be represented in terms of its components.
- The final velocity \(\vec{v_f}\) can be expressed as:
\[ \vec{v_f} = v \cos(60^\circ) \hat{i} + v \sin(60^\circ) \hat{j} \]
- Using the values of \(\cos(60^\circ) = \frac{1}{2}\) and \(\sin(60^\circ) = \frac{\sqrt{3}}{2}\):
\[ \vec{v_f} = v \cdot \frac{1}{2} \hat{i} + v \cdot \frac{\sqrt{3}}{2} \hat{j} = \frac{v}{2} \hat{i} + \frac{\sqrt{3}v}{2} \hat{j} \]

3. **Calculating 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 we have:
\[ \Delta \vec{v} = \left(\frac{v}{2} \hat{i} + \frac{\sqrt{3}v}{2} \hat{j}\right) - (v \hat{i}) \]
- Simplifying this:
\[ \Delta \vec{v} = \left(\frac{v}{2} - v\right) \hat{i} + \frac{\sqrt{3}v}{2} \hat{j} = \left(-\frac{v}{2}\right) \hat{i} + \frac{\sqrt{3}v}{2} \hat{j} \]

4. **Finding the Magnitude of the Change in Velocity**:
- The magnitude of the change in velocity \(|\Delta \vec{v}|\) can be calculated using the Pythagorean theorem:
\[ |\Delta \vec{v}| = \sqrt{\left(-\frac{v}{2}\right)^2 + \left(\frac{\sqrt{3}v}{2}\right)^2} \]
- Calculating this:
\[ |\Delta \vec{v}| = \sqrt{\frac{v^2}{4} + \frac{3v^2}{4}} = \sqrt{\frac{4v^2}{4}} = \sqrt{v^2} = v \]

5. **Conclusion**:
- The change in velocity after the particle has described an angle of \(60^\circ\) is \(v\).

### Final Answer:
The change in velocity after the particle has described an angle of \(60^\circ\) is \(v\).