Two pedestrians left points A and B at the same time towards each other. Their meeting took place at 10 o’clock.

Two pedestrians left points A and B at the same time towards each other. Their meeting took place at 10 o’clock. The pedestrian leaving A walked 2 km more before the meeting. Continuing on his way, he arrived at B at 10:40 am. The second pedestrian arrived at A at 11:30. Find the distance from A to B. Using the equation system.

1. The speed of the first pedestrian is: V1 km / h;

2. The speed of the second pedestrian is: V2 km / h;

3. Distance between points A and B: S km;

4. The meeting of pedestrians took place at: Tb = 10 hours;

5. The first pedestrian covered the distance: S1 km;

6. After the meeting, he came to point B in: T1 = 10 hours 40 minutes – 10 hours = 40 minutes = 2/3 hours;

7. Its speed is equal to: V1 = S2 / T1 = (3 * (S – S1) / 2) km / h;

8. The second pedestrian walked to the meeting: S2 = (S1 – 2) km;

9. He came to point A after the meeting in: T2 = 11 hours 30 minutes – 10 hours = 1.5 hours;

10. Its speed is: V2 = S1 / T2 = S1 / 1.5 km / h;

11. Let’s compose the equation of pedestrian traffic before the meeting:

Tb = S1 / V1 = S2 / V2;

S1 / (3 * (S – S1) / 2) = (S1 – 2) / S1 / 1.5; (S – S1 = S2 = S1 – 2)

2 * S1 / (3 * (S1 – 2) = 3 * (S1 – 2) / (2 * S1);

4 * S1² = 9 * (S1 – 2) ²;

5 * S1² – 36 * S1 + 36 = 0;

S11.2 = (36 + – √ (36² – 4 * 5 * 36) / (2 * 5) = (36 + – 24) / 10;

S11 = (36 – 24) / 10 = 1.2 (does not satisfy the condition of the problem S2 = S1 – 2);

S1 = (36 + 24) / 10 = 6 km;

S2 = S1 – 2 = 6 – 2 = 4 km;

S = S1 + S2 = 6 + 4 = 10 km.

Answer: the distance between points A and B is 10 km.