The train is in the same direction as A, and the relative speed: vehicle speed-human speed;

The train faces B, and the relative speed: speed+people speed.

The train passes by people, that is, from the front to the rear, and the journey is the distance of a car body.

So:

8× (vehicle speed-human speed) = 7× (vehicle speed+human speed),

That is, vehicle speed = 15× human speed.

When the human speed is regarded as 1, the vehicle speed is 15.

Train body length = 8× (15-1) =12, or, 7× (15+1) =1/kloc-.

As shown in the figure,

When the rear of the car leaves A (point A), the distance between the front of the car and A is 1 12, and it meets B after 5 minutes (300 seconds) (point B).

The locomotive has gone: 15× 300 = 4500, and the position five minutes ago from the rear (point A): 4500+112 = 4612.

During this time, A also left at the same time, leaving after 5 minutes: 1× 5× 60 = 300,

Then within 7 seconds of the train meeting B, they left again: 1× 7× 2 = 14.

Therefore, when the train leaves B, the distance between them is: 46 12-300- 14 = 4298.

The rest of the road, two people walking in the opposite direction, will meet again: 4298/(1+1) = 2149 (seconds).

[Note: ① I don't have a specific unit of speed here, all of which are "1", so there is no specific unit of distance and length. If for the sake of understanding, you can also assume that the speed is one meter per second, then the distance and length are meters. (2) How many roads B walked before the train met B, there is no need to discuss]