Two particles A and B get 4m closer each second while traveling in opposite direction They get 0.4 m closer every second while traveling in same direction. The speeds of A and B are respectively :

Options

A. ` 2.2 ms ^(-1) and 0.4 ms^(-1)`
B. ` 2.2 ms ^(-1) and 1.8ms ^(-1)`
C. ` 4 ms ^(-1 ) and 0.4 ms ^(-1) `
D. None of these

Explanation

To solve the problem, we need to find the speeds of two particles A and B based on their relative motion in two different scenarios: when they are moving towards each other and when they are moving in the same direction.

### Step-by-Step Solution:

1. **Understanding the Problem:**
- When particles A and B move in opposite directions, they get 4 meters closer each second.
- When they move in the same direction, they get 0.4 meters closer each second.

2. **Setting Up the Equations:**
- Let \( V_A \) be the speed of particle A and \( V_B \) be the speed of particle B.
- When moving in opposite directions, the relative speed is the sum of their speeds:
\[ V_A + V_B = 4 \quad \text{(Equation 1)} \]
- When moving in the same direction, the relative speed is the difference of their speeds:
\[ V_A - V_B = 0.4 \quad \text{(Equation 2)} \]

3. **Solving the Equations:**
- We can solve these two equations simultaneously. Start by adding Equation 1 and Equation 2:
\[ (V_A + V_B) + (V_A - V_B) = 4 + 0.4 \]
This simplifies to:
\[ 2V_A = 4.4 \]
Therefore, we find:
\[ V_A = \frac{4.4}{2} = 2.2 \, \text{m/s} \]

4. **Finding the Speed of Particle B:**
- Now substitute \( V_A \) back into Equation 1 to find \( V_B \):
\[ 2.2 + V_B = 4 \]
Rearranging gives:
\[ V_B = 4 - 2.2 = 1.8 \, \text{m/s} \]

5. **Final Answer:**
- The speeds of particles A and B are:
\[ V_A = 2.2 \, \text{m/s} \quad \text{and} \quad V_B = 1.8 \, \text{m/s} \]