A particle moves in a circle of radius `R = 21/22 m` with constant speed `1 m//s.` Find,
(a) magnitude of average velocity and
(b) magnitude of average acceleration in 2 s.

Explanation

To solve the problem, we will break it down into two parts as requested:

### Part (a): Magnitude of Average Velocity

1. **Determine the Time for One Complete Revolution**:
The circumference \( C \) of the circle is given by the formula:
\[ C = 2\pi R \]
Substituting \( R = \frac{21}{22} \, \text{m} \):
\[ C = 2\pi \left(\frac{21}{22}\right) \approx 3 \, \text{m} \]
The speed \( v \) of the particle is \( 1 \, \text{m/s} \). Therefore, the time \( T \) for one complete revolution is:
\[ T = \frac{C}{v} = \frac{3}{1} = 3 \, \text{s} \]

2. **Calculate the Angle Covered in 2 Seconds**:
Since the particle completes \( 360^\circ \) in \( 3 \, \text{s} \), the angle \( \theta \) covered in \( 2 \, \text{s} \) is:
\[ \theta = \frac{360^\circ}{3 \, \text{s}} \times 2 \, \text{s} = 240^\circ \]

3. **Determine the Displacement**:
The displacement \( s \) can be calculated using the formula for the chord length in a circle:
\[ s = 2R \sin\left(\frac{\theta}{2}\right) \]
Substituting \( R = \frac{21}{22} \, \text{m} \) and \( \theta = 240^\circ \):
\[ s = 2 \left(\frac{21}{22}\right) \sin\left(120^\circ\right) = 2 \left(\frac{21}{22}\right) \left(\frac{\sqrt{3}}{2}\right) = \frac{21\sqrt{3}}{22} \, \text{m} \]

4. **Calculate Average Velocity**:
The average velocity \( V_{\text{avg}} \) is given by:
\[ V_{\text{avg}} = \frac{s}{t} = \frac{\frac{21\sqrt{3}}{22}}{2} = \frac{21\sqrt{3}}{44} \, \text{m/s} \]

### Part (b): Magnitude of Average Acceleration

1. **Determine the Change in Velocity**:
The initial and final velocities are both \( v = 1 \, \text{m/s} \) but at angles \( 0^\circ \) and \( 240^\circ \) respectively. The change in velocity \( \Delta \vec{v} \) can be calculated using the vector form:
\[ \Delta \vec{v} = \vec{v_f} - \vec{v_i} \]
The components of the velocities can be expressed as:
\[ \vec{v_i} = (1, 0) \quad \text{and} \quad \vec{v_f} = (1 \cos 240^\circ, 1 \sin 240^\circ) = \left(-\frac{1}{2}, -\frac{\sqrt{3}}{2}\right) \]
Thus,
\[ \Delta \vec{v} = \left(-\frac{1}{2} - 1, -\frac{\sqrt{3}}{2} - 0\right) = \left(-\frac{3}{2}, -\frac{\sqrt{3}}{2}\right) \]

2. **Calculate the Magnitude of Change in Velocity**:
The magnitude of \( \Delta \vec{v} \) is given by:
\[ |\Delta \vec{v}| = \sqrt{\left(-\frac{3}{2}\right)^2 + \left(-\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{9}{4} + \frac{3}{4}} = \sqrt{3} \]

3. **Calculate Average Acceleration**:
Average acceleration \( a_{\text{avg}} \) is given by:
\[ a_{\text{avg}} = \frac{|\Delta \vec{v}|}{t} = \frac{\sqrt{3}}{2} \, \text{m/s}^2 \]

### Final Answers:
(a) The magnitude of average velocity is \( \frac{21\sqrt{3}}{44} \, \text{m/s} \).

(b) The magnitude of average acceleration is \( \frac{\sqrt{3}}{2} \, \text{m/s}^2 \).