According to the meaning of the question, it is a catch-up problem. The bus should stop when the express train is 8km away from it, and the catch-up time = distance ÷ speed difference. It is known that the bus has been driving for two hours when the express train leaves, so the distance between the two cars is 40×2=80km. Because the bus must stop when the express train is 8km away from it, the catching-up distance is 80-8=72km. Then the catch-up time =72÷(58-40)=4 hours, and the departure time of the express train is known as 1 1 point, so the bus stop is 1 1+4= 15 point. This problem can be listed as a general formula: (40×2-8)÷(58-40)=4.

2. When Party A set out, Party B had walked 4×3= 12km, while Party A walked 10km, and Party A's speed was 5km/h, 10÷5=2 hours, that is, both of them walked for 2 hours after starting from Party A.. At this time, Otsuichi left 65433. That is to say, it will take another 5 hours for A to catch up with B. The general formula is [4× 3 ~ 4× (10 ÷ 5)-10] ÷ 6-4 ~ = 5.

3. Arithmetic method is not as intuitive and simple as equation. If you solve this problem by equation, let the total distance be your x, and according to the meaning of the problem, you can get the equation: X/4.5+ 1=X/6+3, and the solution is X=36. If we use arithmetic to solve problems, we might as well understand the total distance as "1". In this case, the time for A to complete the course is 1 ÷ 4.5 = 2/9, and the time for B to complete the course is 1/6. According to the meaning of the question, A stopped for 3 hours and arrived at 65438+ later than B.

4. The car has been driving for 2 hours before the failure and 4 hours after the failure. Given that the vehicle speed is 40km/h, the distance between Party A and Party B is 40× (2+4) = 240. According to the meaning of the question, the tractor did not stop while driving, and the speed was always 30km/h, so the driving time of the tractor was 240. The general formula is: 40× (2+4) (30) (2+4) = 2.

5. Before the rabbit set off again, it was * * * gone 10+200=2 10 minutes, and it was known that the speed of the tortoise was 30m/min, so the distance that the tortoise walked was 30× 210 = 6300m, and the rabbit set off for the first time and walked/kloc. This is the catch-up distance, and the catch-up time = catch-up distance ÷ speed difference, and the catch-up time =3000÷(330-30)= 10min, so when the rabbit catches up with the tortoise, the tortoise walks again 10min, so the distance the tortoise walks is 6300+. The general formula is: 8000-30× 210-(30× 210-10× 330) ÷ (330-30 )× 30 =1400.

6. Bell arrived first, two minutes early. According to the meaning of the question, the speed of the lion is 500m/min, and it runs half the distance in one minute, so the total distance is 500× 1×2= 1000 meters. By the time the lion jumps into the water to play, the bear has run 100 meters, so its remaining distance is 900 meters. In view of its constant speed, it arrives.