Two skaters have weight in the ratio `4:5` and are 9m apart, on a smooth frictionless surface. They pull on a rope stretched between them. The ratio of the distance covered by them when they meet each other will be

Options

A. `5:4`
B. `4:5`
C. `25:16`
D. `16:25`

Explanation

To solve the problem, we need to find the ratio of the distances covered by two skaters when they meet while pulling on a rope. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Define the weights of the skaters**:
Let the weight of the first skater be \( W_1 = 4m \) and the weight of the second skater be \( W_2 = 5m \), where \( m \) is a common factor.

2. **Determine the masses**:
Since weight is proportional to mass (Weight = Mass × Gravity), we can express the masses as:
- Mass of the first skater, \( m_1 = \frac{W_1}{g} = \frac{4m}{g} \)
- Mass of the second skater, \( m_2 = \frac{W_2}{g} = \frac{5m}{g} \)

3. **Use Newton's Second Law**:
According to Newton's second law, the acceleration of each skater can be expressed as:
\[ a_1 = \frac{F}{m_1} \quad \text{and} \quad a_2 = \frac{F}{m_2} \]
where \( F \) is the force exerted by each skater on the rope.

4. **Calculate the ratio of accelerations**:
The ratio of the accelerations \( \frac{a_1}{a_2} \) can be calculated as:
\[ \frac{a_1}{a_2} = \frac{F/m_1}{F/m_2} = \frac{m_2}{m_1} = \frac{5m/g}{4m/g} = \frac{5}{4} \]

5. **Determine the distances covered**:
Since both skaters start from rest, we can use the kinematic equation for distance:
\[ s_1 = \frac{1}{2} a_1 t^2 \quad \text{and} \quad s_2 = \frac{1}{2} a_2 t^2 \]

6. **Find the ratio of distances**:
The ratio of the distances covered by the skaters when they meet can be expressed as:
\[ \frac{s_1}{s_2} = \frac{\frac{1}{2} a_1 t^2}{\frac{1}{2} a_2 t^2} = \frac{a_1}{a_2} \]
Substituting the ratio of accelerations:
\[ \frac{s_1}{s_2} = \frac{5}{4} \]

7. **Conclusion**:
Therefore, the ratio of the distances covered by the two skaters when they meet is \( 5:4 \).

### Final Answer:
The ratio of the distance covered by the two skaters when they meet is \( 5:4 \).