To solve the problem, we will break it down into two parts as requested.
### Part (a): Finding the Time Period of the Particle
1. **Identify the formula for time period (T)**:
The time period \( T \) of a particle moving in a circle is given by the formula:
\[
T = \frac{2\pi r}{v}
\]
where \( r \) is the radius of the circle and \( v \) is the constant speed of the particle.
2. **Substitute the known values**:
Given:
- Radius \( r = 4 \, \text{cm} \)
- Speed \( v = 1 \, \text{cm/s} \)
Substitute these values into the formula:
\[
T = \frac{2\pi \times 4 \, \text{cm}}{1 \, \text{cm/s}} = 8\pi \, \text{s}
\]
3. **Calculate the numerical value**:
Using \( \pi \approx 3.14 \):
\[
T \approx 8 \times 3.14 = 25.12 \, \text{s}
\]
Thus, the time period \( T \) is approximately **25.12 seconds**.
### Part (b): Finding Average Speed, Average Velocity, and Average Acceleration
1. **Average Speed**:
Since the particle is moving with constant speed, the average speed over any interval will be equal to the constant speed:
\[
\text{Average Speed} = v = 1 \, \text{cm/s}
\]
2. **Average Velocity**:
Average velocity is defined as the displacement divided by the time taken.
- **Displacement**: In the time interval from \( t = 0 \) to \( t = \frac{T}{4} \), the particle moves a quarter of the circle. The displacement is the straight line distance from the starting point to the point after \( T/4 \). This forms a right triangle with both legs equal to the radius \( r = 4 \, \text{cm} \):
\[
\text{Displacement} = \sqrt{(4 \, \text{cm})^2 + (4 \, \text{cm})^2} = \sqrt{32} = 4\sqrt{2} \, \text{cm}
\]
- **Time Taken**: The time taken for this interval is \( \frac{T}{4} = \frac{25.12 \, \text{s}}{4} = 6.28 \, \text{s} \).
Now, calculate the average velocity:
\[
\text{Average Velocity} = \frac{\text{Displacement}}{\text{Time}} = \frac{4\sqrt{2} \, \text{cm}}{6.28 \, \text{s}} \approx 0.9 \, \text{cm/s}
\]
3. **Average Acceleration**:
Average acceleration is defined as the change in velocity divided by the time taken.
- **Initial Velocity**: At \( t = 0 \), the velocity is directed along the tangent to the circle.
- **Final Velocity**: At \( t = \frac{T}{4} \), the velocity is directed tangentially at the new position, which is 90 degrees from the initial position.
The change in velocity can be calculated using vector subtraction:
\[
\Delta v = v_f - v_i
\]
Since the velocities are perpendicular, we can use the Pythagorean theorem:
\[
|\Delta v| = \sqrt{(1 \, \text{cm/s})^2 + (1 \, \text{cm/s})^2} = \sqrt{2} \, \text{cm/s}
\]
Now, average acceleration:
\[
\text{Average Acceleration} = \frac{|\Delta v|}{\Delta t} = \frac{\sqrt{2} \, \text{cm/s}}{6.28 \, \text{s}} \approx 0.23 \, \text{cm/s}^2
\]
### Final Answers:
- (a) Time Period \( T \approx 25.12 \, \text{s} \)
- (b) Average Speed \( = 1 \, \text{cm/s} \), Average Velocity \( \approx 0.9 \, \text{cm/s} \), Average Acceleration \( \approx 0.23 \, \text{cm/s}^2 \)
