The accleration of a particle as seen from two frames `S_(1)` and `S_(2)` have equal magnitudes 4 `m//s^(2)`

Options

A. the frames must be at rest with respect to each other
B. the frames may be moving with respect to each other but neither should be acclerated with respect to the other
C. the accleration of `S_(2)` with respect of `S_(1)` may be either zero or 8 `m//s^(2)`.
D. the acceleration of `S_(2)` with respect to `S_(1)` may have any value between zero and `8 m//s^(2)`

Answer

To solve the problem, we need to analyze the relationship between the accelerations of the two frames of reference, \( S_1 \) and \( S_2 \), as well as the particle's acceleration in these frames.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The acceleration of a particle as seen from frame \( S_1 \) is \( a_{p/S_1} = 4 \, \text{m/s}^2 \).
- The acceleration of the same particle as seen from frame \( S_2 \) is \( a_{p/S_2} = 4 \, \text{m/s}^2 \).

2. **Using the Relative Acceleration Formula**:
- The relative acceleration of frame \( S_2 \) with respect to frame \( S_1 \) can be expressed as:
\[ a_{S_2/S_1} = a_{p/S_1} + a_{S_1/p} \]
- Here, \( a_{S_1/p} \) is the acceleration of frame \( S_1 \) with respect to the particle.

3. **Calculating the Relative Acceleration**:
- Since both frames observe the particle with the same acceleration of \( 4 \, \text{m/s}^2 \), we can denote:
\[ a_{S_1/p} = 4 \, \text{m/s}^2 \]
- Therefore, we can write:
\[ a_{S_2/S_1} = a_{p/S_1} + a_{p/S_2} = 4 + 4 = 8 \, \text{m/s}^2 \]

4. **Considering the Angle Between Accelerations**:
- The actual relative acceleration can vary depending on the angle \( \theta \) between the two acceleration vectors. The formula for the resultant acceleration is:
\[ a_{S_2/S_1} = \sqrt{a_{p/S_1}^2 + a_{p/S_2}^2 + 2 \cdot a_{p/S_1} \cdot a_{p/S_2} \cdot \cos(\theta)} \]
- Plugging in the values:
\[ a_{S_2/S_1} = \sqrt{4^2 + 4^2 + 2 \cdot 4 \cdot 4 \cdot \cos(\theta)} \]
\[ = \sqrt{16 + 16 + 32 \cdot \cos(\theta)} \]
\[ = \sqrt{32 + 32 \cdot \cos(\theta)} \]

5. **Finding the Range of the Resultant Acceleration**:
- The value of \( \cos(\theta) \) can vary from -1 to 1:
- If \( \cos(\theta) = 1 \) (when the accelerations are in the same direction):
\[ a_{S_2/S_1} = \sqrt{32 + 32} = \sqrt{64} = 8 \, \text{m/s}^2 \]
- If \( \cos(\theta) = -1 \) (when the accelerations are in opposite directions):
\[ a_{S_2/S_1} = \sqrt{32 - 32} = \sqrt{0} = 0 \, \text{m/s}^2 \]

6. **Conclusion**:
- Therefore, the acceleration of \( S_2 \) with respect to \( S_1 \) can take any value between \( 0 \, \text{m/s}^2 \) and \( 8 \, \text{m/s}^2 \).

### Final Answer:
The correct option is:
**D: The acceleration of \( S_2 \) with respect to \( S_1 \) may have any value between \( 0 \) and \( 8 \, \text{m/s}^2 \).**