This classmate arrived at the head of the team, and the relative speed was12-8 = 4 km/h; When returning from the beginning to the end, the relative speed is12+8 = 20km/h.

Equation: S/4+S/20 = 14.4/60.

Solve the equation and get s = 0.8km.

A: The team is 0.8 kilometers long.

2、( 1- 1/50)( 1- 1/49)( 1- 1/48)……( 1- 1/4)( 1- 1/3)

=49/50×48/49×47/48……3/4×2/3

=2/50

= 1/25

3. A: The shortest time is 7/9 hours.

Scheme: student B goes, and the teacher sends student A away12.5km; A student goes straight ahead, and the teacher picks up B student and arrives at the station at the same time as A student.

Commentary: Finally, the motorcycle and another classmate went to the railway station, because they didn't waste any time. Two students always want a 5km/h and a 45km/h, so it is the shortest time to arrive at the same time.

Suppose it took the teacher x hours to send a student, then B walked 5x kilometers, and the teacher took A 45x kilometers, and the distance between them was 40x kilometers.

At this time, when the teacher comes back by bike, it becomes a problem for the teacher to meet the teacher. The required time is 40x/(45+5) = 4x/5h. During this period, both A and B walked 5(4x/5)=4x km.

Finally, the teacher needs to walk (15-5x-4x) km with B and (15-45x-4x) km with A, and the time spent by them is the same = (15-5x-4x)/45 = (/kloc-). The sum of three times is x+(4x/5)+(15-5x-4x)/45 = 7/9 hours = 140/3 minutes < 50 minutes, and you can get to the train station.

So the teacher took a 45x =12.5km. ..