2 pedestrians went from city A to B to meet each other, the distance between cities is 19 km.

2 pedestrians went from city A to B to meet each other, the distance between cities is 19 km. After a certain time, they met 9 km from the city A. How fast a pedestrian walked from city A, if it is known that his speed is 1 km / h faster than the other and he made a stop for 30 minutes along the way.

Let the speed of the second pedestrian moving from the city B be equal to x km / h, then the speed of the first pedestrian moving from the city A is equal to (x + 1) km / h. The first pedestrian walked 9 km before the meeting, and the second walked 19 – 9 = 10 kilometers. The first pedestrian was on the way (including stopping time) (9 / (x + 1) + 1/2) hours (to find the time, the distance traveled must be divided by the speed; 30 min = 1/2 h), and the second pedestrian was on the way 10 / x hours. The travel time for the first and second pedestrians is the same. Let’s make an equation and solve it.

9 / (x + 1) + 1/2 = 10 / x;

(9 * 2x) / (2x (x + 1) + (1 * x (x + 1)) / (2x (x + 1)) = (10 * 2 (x + 1)) / (2x (x + one));

9 * 2x + x (x + 1) = 10 * 2 (x + 1);

18x + x ^ 2 + x = 20x + 20;

x ^ 2 + 19x – 20x – 20 = 0;

x ^ 2 – x – 20 = 0;

D = b ^ 2 – 4ac;

D = (- 1) ^ 2 – 4 * 1 * (- 20) = 1 + 80 = 81; √D = 9;

x = (- b ± √D) / (2a);

x1 = (1 + 9) / 2 = 10/2 = 5 (km / h) – speed of the second pedestrian;

x2 (1 – 9) / 2 = – 8/2 = – 4 – the speed cannot be negative.

x + 1 = 5 + 1 = 6 (km / h) – the speed of the first pedestrian.

Answer. The speed of the first pedestrian from city A is 6 km / h.