The distance between cities A and B is 480 km. Two trains left these cities at the same time towards each other.

The distance between cities A and B is 480 km. Two trains left these cities at the same time towards each other. After 3 hours of movement, they had to walk 60 km before the meeting. Find the speed of each train if the distance between cities A and B in one of them is 2 hours faster than the second.

Let us denote by t the time during which the faster train travels the distance between these two cities, equal to 480 km.

From the condition of the problem it is known that the distance between cities A and B one of the trains travels 2 hours faster than the second, therefore, the slower train travels this distance in t + 2 hours.

According to the condition of the problem, having left at the same time towards each other, two trains after 3 hours were at a distance of 60 km from each other, therefore, we can draw up the following equation:

480 / t + 480 / (t + 2) + 60 = 480,

solving which, we get:

3 * 480 / t + 3 * 480 / (t + 2) = 480 – 60;

3 * 480 / t + 3 * 480 / (t + 2) = 420;

24 / t + 24 / (t + 2) = 7;

24 * (t + 2) + 24t = 7t * (t + 2);

24t + 48 + 24t = 7t ^ 2 + 14t;

48 + 48t = 7t ^ 2 + 14t;

7t ^ 2 + 14t – 48t – 48 = 0;

7t ^ 2 – 34t – 48 = 0;

t = (17 ± √ (289 + 7 * 48)) / 7 = (17 ± √625) / 7 = (17 ± 25) / 7;

t = (17 + 25) / 7 = 42/7 = 6;

Therefore, the speed of the first train is 480/6 = 80 km / h, and the speed of the second train is 480 / (6 + 2) = 280/8 = 60 km / h.

Answer: the speed of the first train is 80 km / h, and the speed of the second train is 60 km / h.