A motorcycle started from point A to point B. After 2 hours, a car left A for B and arrived at B at the same time

A motorcycle started from point A to point B. After 2 hours, a car left A for B and arrived at B at the same time as the motorcycle. If a car and a motorcycle had left A and B towards each other at the same time, they would have met 1 hour and 20 minutes after the departure. How long was the journey from A to B on the motorcycle if the speed of the motorcycle and car are constant?

Let the speed of the car be x km / h, and the speed of the motorcycle y km / h. If a car and a motorcycle simultaneously left A and B towards each other, then the distance between the car and the motorcycle in an hour after the start of movement or their speed of convergence would be (x + y) km / h, and the distance from point A to point B would be 4 ∙ (x + y) / 3 km, since they would have met 1 hour 20 minutes after departure (1 hour 20 minutes = 4/3 hours). From the condition of the problem it is known that a motorcycle left point A to point B and after 2 hours a car left A to B, while their approach speed was (x – y) km / h, and the distance from point A to point B was 2 ∙ (x – y) km, since the car arrived at B simultaneously with the motorcycle, that is, it caught up with it in 2 hours. Knowing this, we compose the equation:
4 ∙ (x + y) / 3 = 2 ∙ (x – y);
x = 2.2 ∙ y.
Then the entire path from A to B will be 2 ∙ (2.2 ∙ y – y) = 2.4 ∙ y (km). To find how long the bike was traveling from A to B, you need to divide the distance between points by the speed of the bike:
(2.4 ∙ y): y = 2.4 (hours).
Answer: The motorcycle was on the way for 2.4 hours, moving from point A to point B.