A particle is moving on a circular path of radius r with uniform speed v. What is the displacement of the particle after it has described an angle of `60^(@)`?

Options

A. `r sqrt(2)`
B. `rsqrt(3)`
C. `r`
D. `2r`

Explanation

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

### Step-by-Step Solution:

1. **Understand the Circular Path**:
The particle is moving in a circular path with radius \( r \). The initial position of the particle can be considered at point A on the circle.

2. **Determine the Final Position**:
After moving \( 60^\circ \) along the circular path, the particle reaches point B. The angle \( 60^\circ \) is measured from the center of the circle.

3. **Visualize the Triangle**:
The displacement is the straight-line distance from point A to point B. The points A, B, and the center of the circle (let's call it O) form an isosceles triangle \( OAB \), where \( OA = OB = r \) (the radius of the circle).

4. **Calculate the Angle at the Center**:
The angle at the center \( \angle AOB \) is \( 60^\circ \). The angles at points A and B (let's call them \( \angle OAB \) and \( \angle OBA \)) are equal because the triangle is isosceles. Therefore, each of these angles is:
\[ \angle OAB = \angle OBA = \frac{180^\circ - 60^\circ}{2} = 60^\circ \]

5. **Use the Cosine Rule**:
To find the displacement \( AB \), we can use the cosine rule in triangle \( OAB \):
\[ AB^2 = OA^2 + OB^2 - 2 \cdot OA \cdot OB \cdot \cos(\angle AOB) \]
Substituting \( OA = OB = r \) and \( \angle AOB = 60^\circ \):
\[ AB^2 = r^2 + r^2 - 2 \cdot r \cdot r \cdot \cos(60^\circ) \]
Since \( \cos(60^\circ) = \frac{1}{2} \):
\[ AB^2 = r^2 + r^2 - 2 \cdot r^2 \cdot \frac{1}{2} \]
\[ AB^2 = 2r^2 - r^2 = r^2 \]
Therefore, the displacement \( AB \) is:
\[ AB = \sqrt{r^2} = r \]

6. **Final Answer**:
The displacement of the particle after it has described an angle of \( 60^\circ \) is \( r \).