1, 42 minutes = 0.7 hours

∵ The fault point of this car is 0/5km away from the airport/kloc-,and only five people can be sent at a time, with an average speed of 60km/h.

∴( 1) After the first group of people were sent away by car, the second group of people waited for the car to return to pick them up;

Then the time to drive the first group of people to the airport = 15 ÷ 60 = 0.25 hours;

∴ Time for the car to return to pick up the second batch of people and then go to the airport = 15× 2 ÷ 60 = 0.5 hours;

* * * Time = 0.25+0.5 = 0.75 hours > 0.7 hours

So it is impossible for all eight fans to arrive at the airport within the specified time.

(2) While the first group of people were sent away by car, the second group of people walked to the airport at an average speed of 5km/h, met a returning car on the way and got on the bus.

Then the time to drive the first group of people to the airport = 15 ÷ 60 = 0.25 hours;

At this time, the distance between the second group and the airport is =15-5× 0.25 =13.75km;

∴ Time taken for the car to return to meet the second group of people =13.75 ÷ (60+5) =1/52 hours;

At this time, the distance between the second group and the airport =13.75-5×11/52 =165/13km;

∴ Time to arrive at the airport after the car picks up the second batch of people =165/13 ÷ 60 =1/52 hours;

* * * Time = 0.25+11/52+1/52 = 35/52 hours < 0.7 hours.

So all eight fans can arrive at the airport within the specified time.

Therefore, only the second method can make all eight fans arrive at the airport within the specified time.

2.( 1)∵ Two trains travel in opposite directions, and it takes 9 seconds from meeting to all staggering (two ends meet and two ends leave).

The length of trains A and B are 144m and 180m respectively.

∴ Sum of the distance traveled by two trains =144+180 = 324m.

∴ sum of two train speeds = 324 ÷ 9 = 36m/s

Car A travels 4 meters more per second than car B.

∴ A car speed = (36+4) ÷ 2 = 20m/s

Train speed B = (36-4) ÷ 2 = 16m/s

(2) If driving in the same direction, the front of car A will overtake car B from the rear of car B..

At this time, it is quite still with car B. For the length of cars A and B, car A walks at a speed of 4 meters per second.

∴ Time = (144+180) ÷ 4 = 81sec.

3. Set the speed of armor as X km/h. ..

45 minutes = 0.75 hours

The distance between A and B is 25.5 kilometers. A arrives at B and stays for 45 minutes (B hasn't arrived at B yet). The speed of A train is three times that of B train, which is 1 km more.

Then I returned to A from B and met B on the way. At this time, the departure time from them is 3 hours.

∴ A Walking time = 3-0.75 = 2.25 hours;

The speed of b is (x-1)/3 km/h.

The distance a and b traveled when they met, the distance between a and b = 2 times.

∴2.25x+(x- 1)/3×3=2×25.5=5 1

∴ x = 16 km/h

∴ (x-1)/3 = (16-1)/3 = 5km/h.

Therefore, the speed of A is 16 km/h, and the speed of B is 5 km/h. ..