'A' and 'B' start walking towards each other from point 'X' and 'Y', respectively. 'A' is walking at 4m / s and 'B' is walking at 7 m/s. If distance between 'A' and 'B' is 60 metres and they're travelling back and forth between 'X' and 'Y', then find the total distance covered by 'B' by the time he meets 'A' for the third time given that distance covered by 'A' in that time is (2P / 11) metres.

To solve this problem, we need to calculate the total distance covered by 'B' by the time he meets 'A' for the third time, given that the distance covered by 'A' is \( \frac{2P}{11} \) meters.

### Step 1: Define the situation and variables.

- The initial distance between 'A' and 'B' is 60 meters.
- The speed of 'A' is \( v_A = 4 \, \text{m/s} \).
- The speed of 'B' is \( v_B = 7 \, \text{m/s} \).

Let's analyze their movement over time.

### Step 2: First meeting.

- The relative speed at which 'A' and 'B' are approaching each other is:
  \[
  v_{\text{relative}} = v_A + v_B = 4 + 7 = 11 \, \text{m/s}.
  \]
- The time taken for them to meet for the first time is:
  \[
  t_1 = \frac{\text{Distance between A and B}}{v_{\text{relative}}} = \frac{60}{11} \, \text{seconds}.
  \]
- During this time, the distance covered by 'B' is:
  \[
  d_B = v_B \cdot t_1 = 7 \cdot \frac{60}{11} = \frac{420}{11} \, \text{meters}.
  \]

### Step 3: Second meeting.

Once 'A' and 'B' meet, they continue to travel back and forth between their initial positions \( X \) and \( Y \).

- The total distance covered by 'A' by the second meeting will be double the distance covered in the first meeting. Similarly, the total distance covered by 'B' will be twice the distance from the first meeting.
- Thus, the distance covered by 'B' up to the second meeting will be:
  \[
  d_B = 2 \cdot \frac{420}{11} = \frac{840}{11} \, \text{meters}.
  \]

### Step 4: Third meeting.

- Similarly, the distance covered by 'B' by the third meeting will be three times the distance from the first meeting, so:
  \[
  d_B = 3 \cdot \frac{420}{11} = \frac{1260}{11} \, \text{meters}.
  \]

### Step 5: Final distance.

We are given that the distance covered by 'A' by the third meeting is \( \frac{2P}{11} \) meters. Therefore, the total distance covered by 'B' by the third meeting is \( \frac{1260}{11} \, \text{meters} \), which corresponds to this distance.

Hence, the total distance covered by 'B' by the third meeting is \( \frac{1260}{11} \) meters.

---

### Do you want more details or have any questions?
### Related Questions:
1. How do we find the number of times 'A' and 'B' meet?
2. What would the situation look like if 'A' and 'B' had different initial positions?
3. Can you explain the relative speed concept in more detail?
4. How do the distances covered by 'A' and 'B' compare if they walk at different speeds?
5. What would change if 'A' and 'B' had the same speed?

### Tip:
When solving problems involving relative motion, always break the motion down into individual segments and use the concept of relative speed to simplify calculations.

Explore how to calculate the total distance covered by B by the time he meets A for the third time in a scenario involving relative motion. Understand the concepts of relative speed, distance, and time in this detailed solution. Suitable for advanced high school students.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation