A particle P moves with speed V along AB and BC, sides of a square ABCD. Another particle Q also starts at A and moves with the same speed but along AD and DC of the same square ABCD. Then their respective changes in velocities are

Options

A. equal in magnitude but different in direction
B. different in magnitude but same in directions
C. different both in magnitude and direction
D. same both in magnitude and direction

Answer

To solve the problem, we need to analyze the motion of both particles P and Q as they move along the sides of the square ABCD.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a square ABCD.
- Particle P moves along sides AB and BC.
- Particle Q moves along sides AD and DC.
- Both particles have the same speed \( V \).

2. **Identifying the Initial Velocities**:
- For particle P moving from A to B (along AB), the velocity vector can be represented as:
\[ \vec{V_P} = V \hat{i} \]
(assuming rightward along the x-axis).
- When particle P reaches point B and starts moving towards C (along BC), its velocity vector becomes:
\[ \vec{V_P} = V \hat{j} \]
(upward along the y-axis).
- For particle Q moving from A to D (along AD), the velocity vector is:
\[ \vec{V_Q} = -V \hat{j} \]
(downward along the negative y-axis).
- When particle Q reaches point D and starts moving towards C (along DC), its velocity vector becomes:
\[ \vec{V_Q} = -V \hat{i} \]
(leftward along the negative x-axis).

3. **Calculating Changes in Velocity**:
- For particle P:
- Change in velocity when moving from AB to BC:
\[ \Delta \vec{V_P} = \vec{V_P (BC)} - \vec{V_P (AB)} = V \hat{j} - V \hat{i} \]
This results in:
\[ \Delta \vec{V_P} = V (\hat{j} - \hat{i}) \]
- For particle Q:
- Change in velocity when moving from AD to DC:
\[ \Delta \vec{V_Q} = \vec{V_Q (DC)} - \vec{V_Q (AD)} = -V \hat{i} - (-V \hat{j}) \]
This results in:
\[ \Delta \vec{V_Q} = -V \hat{i} + V \hat{j} = V (\hat{j} - \hat{i}) \]

4. **Comparing Changes in Velocity**:
- The changes in velocity for both particles are:
- For P: \( \Delta \vec{V_P} = V (\hat{j} - \hat{i}) \)
- For Q: \( \Delta \vec{V_Q} = V (\hat{j} - \hat{i}) \)
- Both changes in velocity have the same magnitude \( V \) and the same direction \( (\hat{j} - \hat{i}) \).

5. **Conclusion**:
- The respective changes in velocities for particles P and Q are equal in magnitude and direction.

### Final Answer:
The correct option is: **Same both in magnitude and direction**.