This is an open question. If eight people can catch the train, we should try to let the car go when others leave.

Option 1: the car can be designed to send the first four people, while the others keep moving forward at the same time. After the car is delivered to the railway station, it will come back to pick up the rest of them.

Scheme 2: firstly, the first group of people will be sent to the place near the railway station by car, so that the first group of people will go and the second group will also go; Then the car goes back to pick up the second group of people, so that the second group of people and the first group of people can arrive at the railway station at the same time.

Solution:

There are two feasible schemes to catch the train:

When the first four people were sent away by the car, the others kept moving forward at the same time. After the car is delivered to the railway station, it will come back to pick up the rest of them.

Assuming that it takes x hours for the car to return and x hours for pedestrians to meet, there are:

60x+5x= 15×2，

X=6/ 13, so * * * time:

6/13+(15-5 * 6/13)/60 = 26/39 hours =40 (minutes);

(2) First, the first group of people are sent to the place near the railway station by car, so that the first group of people can leave and the second group of people are also leaving; Then the car went back to pick up the second group of people and let the second group of people and the first group of people go to the railway station at the same time. In this scheme, everyone either rode or walked, and no one wasted time standing still, so the two groups walked the same distance successively.

Let this distance be x kilometers, then the distance of each group by car is 15-x kilometers, and the time of * * * is x/5+( 15-x)/60 hours.

When the car sent the first group to the airport x kilometers away, when the second group came back, the second group had already walked x kilometers.

At this time, the distance traveled by the car is15-x+15-2x = 30-3x (km);

Since the time for the car to travel 30-3x kilometers is equal to the time for the second group to travel X kilometers, there are:

(30-3x)/60=x/5，

Solution: x=2 (km).

The time used is:

2/5+( 15-2)/60=37/60 hours =37 minutes.

37 minutes < 40 minutes,

The second scheme is more time-saving.