Two particle A and B start moving due to their mutual interaction only. If at any time `t`, `vec(a)_(A)` and `vec(a)_(B)` are their respective accelerations, `vec(v)_(A)` and `vec(v)_(B)` are their respective velocities, and upto that time `W_(A)` and `W_(B)` are the work done on A and B respectively by the mutual force, `m_(A)` and `m_(B)` are their masses respectively, then which of the following is always correct.

Options

A. `vecv_A + vecv_B = 0`
B. ` m_Avecv_A + m_Bvecv_B = 0`
C. `W_A + W_B = 0`
D. `veca_A + veca_B = 0`

Explanation

To solve the problem, we need to analyze the situation of two particles A and B interacting with each other. We will derive the correct relationship based on the principles of physics, particularly focusing on the conservation of momentum.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have two particles A and B that are interacting with each other. The only forces acting on them are mutual forces (action-reaction pairs).
- Since there are no external forces acting on the system, we can consider it as an isolated system.

2. **Initial Conditions**:
- At time `t = 0`, both particles start from rest. Therefore, their initial velocities are:
\[ \vec{v}_A(0) = 0 \quad \text{and} \quad \vec{v}_B(0) = 0 \]

3. **Applying the Law of Conservation of Momentum**:
- According to the law of conservation of momentum, the total momentum of an isolated system remains constant if no external forces are acting on it.
- The momentum of particle A at time `t` is given by:
\[ \vec{P}_A = m_A \vec{v}_A \]
- The momentum of particle B at time `t` is given by:
\[ \vec{P}_B = m_B \vec{v}_B \]

4. **Total Momentum at Time t**:
- The total momentum of the system at time `t` is:
\[ \vec{P}_{\text{total}} = \vec{P}_A + \vec{P}_B = m_A \vec{v}_A + m_B \vec{v}_B \]

5. **Setting the Initial Momentum**:
- Since both particles started from rest, the initial momentum of the system is:
\[ \vec{P}_{\text{initial}} = 0 \]

6. **Equating Initial and Final Momentum**:
- By the conservation of momentum, we have:
\[ \vec{P}_{\text{total}} = \vec{P}_{\text{initial}} \implies m_A \vec{v}_A + m_B \vec{v}_B = 0 \]

7. **Conclusion**:
- Rearranging the equation gives us:
\[ m_A \vec{v}_A + m_B \vec{v}_B = 0 \]
- This indicates that the momentum of particle A is equal in magnitude and opposite in direction to the momentum of particle B, which is consistent with the conservation of momentum.

### Final Answer:
The correct relationship that is always true is:
\[ m_A \vec{v}_A + m_B \vec{v}_B = 0 \]