Two cars move uniformly towards each othar, the distance between them decreases at the rate 50 m/s. If they move in same direction with different speeds, the distance between them increases at the rate 10 m/s. The speed of two cars will be

Options

A. 30 m/s and 20 m/s
B. 40 m/s and 10 m/s
C. 15 m/s and 35 m/s
D. 25 m/s and 25 m/s

Answer

To solve the problem, we need to analyze the motion of the two cars under the given conditions.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- Let the speed of car A be \( v_1 \) and the speed of car B be \( v_2 \).
- When the cars move towards each other, the distance between them decreases at a rate of 50 m/s. This can be expressed as:
\[ \frac{ds}{dt} = -50 \, \text{m/s} \]
- When the cars move in the same direction, the distance between them increases at a rate of 10 m/s:
\[ \frac{ds}{dt} = 10 \, \text{m/s} \]

2. **Setting Up the Equations**:
- For the first scenario (cars moving towards each other):
\[ v_1 + v_2 = 50 \quad \text{(Equation 1)} \]
- For the second scenario (cars moving in the same direction):
\[ v_1 - v_2 = 10 \quad \text{(Equation 2)} \]

3. **Solving the Equations**:
- We can add Equation 1 and Equation 2:
\[ (v_1 + v_2) + (v_1 - v_2) = 50 + 10 \]
This simplifies to:
\[ 2v_1 = 60 \]
Therefore:
\[ v_1 = 30 \, \text{m/s} \]

- Now, substituting \( v_1 \) back into Equation 1 to find \( v_2 \):
\[ 30 + v_2 = 50 \]
Thus:
\[ v_2 = 50 - 30 = 20 \, \text{m/s} \]

4. **Final Result**:
- The speeds of the two cars are:
- Speed of car A (\( v_1 \)) = 30 m/s
- Speed of car B (\( v_2 \)) = 20 m/s