'A' and 'B' start walking towards each other from point 'X' and 'Y', respectively. 'A' is walking at 4 m/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.

Given the problem, let’s solve it step by step:

### Step 1: Assign variables to key pieces of information
- Let the total distance between point 'X' and point 'Y' be \( d = 60 \) meters.
- 'A' is walking at a speed of 4 m/s.
- 'B' is walking at a speed of 7 m/s.
- They meet for the third time after both have traveled back and forth between the two points.
- The distance covered by 'A' by the time they meet for the third time is \( \frac{2P}{11} \), where \( P \) is some unknown quantity we need to find.
  
### Step 2: Determine relative speed
When 'A' and 'B' are walking towards each other, their relative speed is the sum of their individual speeds:
\[
v_{\text{relative}} = 4 + 7 = 11 \, \text{m/s}
\]
Thus, they cover 11 meters every second together.

### Step 3: Time taken to meet for the first time
The time taken for them to meet for the first time, denoted \( t_1 \), is the time required to cover the entire distance \( d = 60 \) meters:
\[
t_1 = \frac{d}{v_{\text{relative}}} = \frac{60}{11} \, \text{seconds}.
\]

### Step 4: Distances covered by 'A' and 'B' at the first meeting
- In \( t_1 \) seconds, the distance covered by 'A' is:
\[
d_A = 4 \times t_1 = 4 \times \frac{60}{11} = \frac{240}{11} \, \text{meters}.
\]
- The distance covered by 'B' is:
\[
d_B = 7 \times t_1 = 7 \times \frac{60}{11} = \frac{420}{11} \, \text{meters}.
\]

### Step 5: After the first meeting
After meeting for the first time, 'A' turns around and walks back towards 'X', while 'B' turns around and walks back towards 'Y'. They will meet again during this back-and-forth motion.

For each meeting, the process of calculating the distances covered by both 'A' and 'B' follows similar logic, but with alternating directions of travel.

### Step 6: Given information and goal
We know that by the third meeting, the distance covered by 'A' is given as \( \frac{2P}{11} \). Our goal is to find the total distance covered by 'B' in that time.

Let's denote the time taken for the third meeting as \( t_{\text{total}} \). The total distance covered by 'A' by that time is:
\[
d_A = \frac{2P}{11}.
\]
Using the speed of 'A', we can express the total time taken as:
\[
t_{\text{total}} = \frac{d_A}{v_A} = \frac{\frac{2P}{11}}{4} = \frac{2P}{44} \, \text{seconds}.
\]

### Step 7: Distance covered by 'B'
Since 'B' is traveling at 7 m/s, the total distance covered by 'B' by the time they meet for the third time is:
\[
d_B = 7 \times t_{\text{total}} = 7 \times \frac{2P}{44} = \frac{14P}{44} = \frac{7P}{22} \, \text{meters}.
\]

Thus, the total distance covered by 'B' by the time they meet for the third time is \( \frac{7P}{22} \).

Would you like to know more details, or have any questions?

Here are 5 relative questions for practice:
1. How do you calculate the relative speed between two people moving towards each other?
2. What happens to the time taken for each successive meeting when two people walk back and forth between two points?
3. If 'A' was walking at 5 m/s instead of 4 m/s, how would the time of the first meeting change?
4. How would you calculate the total time until 'B' meets 'A' for the fourth time?
5. How can you find the number of times two people meet if you know their speeds and the distance between them?

**Tip:** When solving problems involving relative motion, always calculate relative speed first to simplify the time calculations for each meeting.

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

This problem explores relative motion concepts involving two individuals, 'A' and 'B', walking towards each other. With a total distance of 60 meters, speeds of 4 m/s and 7 m/s respectively, and multiple meetings, learn how to calculate the total distance covered by 'B' by their third meeting. Suitable for Grades 9-11.

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