Math Problem Statement

'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.

Solution

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 2P11\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 vA=4m/sv_A = 4 \, \text{m/s}.
  • The speed of 'B' is vB=7m/sv_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: vrelative=vA+vB=4+7=11m/s.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: t1=Distance between A and Bvrelative=6011seconds.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: dB=vBt1=76011=42011meters.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 XX and YY.

  • 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: dB=242011=84011meters.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: dB=342011=126011meters.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 2P11\frac{2P}{11} meters. Therefore, the total distance covered by 'B' by the third meeting is 126011meters\frac{1260}{11} \, \text{meters}, which corresponds to this distance.

Hence, the total distance covered by 'B' by the third meeting is 126011\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.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Relative Motion
Speed
Distance

Formulas

-

Theorems

-

Suitable Grade Level

Grades 9-12