This can be said to be the most classic travel problem. You don't need to analyze the specific distance that the puppy ran, just pay attention to that it takes 20 seconds for A and B to meet. During these 20 seconds, the puppy has been running, so the distance it runs is 120 meters.
Speaking of this classic question, there are many stories. Here's a classic mathematician gossip post: john von neumann was once asked a question familiar to elementary school students in China: Two people walked towards each other, and a dog ran in the middle, asking how many roads the dog had walked after they met. The trick is to find the time of meeting first and then multiply it by the speed of the dog. Neumann, of course, immediately gave the answer. The questioner said disappointedly, you must have heard of this trick. Neumann was surprised: What trick? I just calculate every time the dog runs, and then calculate the infinite series?
Suppose you stand between A and B and want to take a taxi to B. You see an empty car coming from far away, and no one else is competing with you for the empty car on the whole road. We assume that the speed of the car and the speed of the person are fixed, and the speed is greater than the speed of the person. In order to get to the destination faster, should we walk towards the car or follow it a little further?
When this question is raised in various crowded situations, everyone's views will often be immediately divided into two completely different factions, each with its own reasons. Some people say that because the speed is faster than that of people, I want to get on the bus as soon as possible and make full use of the speed advantage of the car, so I want to walk up empty and meet the car in advance. The other faction said that in order to reach my destination as soon as possible, I should make full use of my time and run to my destination non-stop. Therefore, I should walk some way to my destination by myself, and then let the taxi take me the rest of the way.
In fact, the answer is surprisingly simple, and the time spent by the two schemes is obviously the same. As long as you think about it from the taxi's point of view, the problem becomes obvious: no matter where people get on the bus, the taxi will drive from A to B anyway, so the time you arrive at B is always equal to the time the taxi travels all the way, plus the time it may pick up people on the way. From the point of view of saving trouble, it is best to stand still.
However, many people have found a bug in this problem: in some extreme cases, it may be better to go in the direction of the car, because you may go straight to the finish line, and at this time the taxi has not caught up with you!
Someone started from the foot of the mountain at eight o'clock in the morning, walked up the mountain along the mountain road and reached the top of the mountain at eight o'clock in the evening. However, he did not move at a constant speed, sometimes slowly, sometimes quickly, and sometimes even stopped. The next day, he set off from the top of the mountain at eight o'clock in the morning and went down the mountain along the original road, sometimes fast and sometimes slow, and finally reached the foot of the mountain at eight o'clock in the evening. Try to explain that this person must have passed the same point on the mountain road at the same time these two days.
This topic is also a classic among classics. Overlap this person's two-day trip to one day. In other words, imagine a person walking from the foot of the mountain to the top of the mountain, and another person walking from the top of the mountain to the foot of the mountain on the same day. The two men are bound to meet somewhere on the road. This means that this person passed through here at the same time for two days.
4. Which is faster for a ship to go back and forth between A and B in still water or between A and B in flowing water?
This is a classic question. The answer is that ships travel faster in still water. Note that the average of the actual speed of the ship downstream is its still water speed, but from the conclusion of the last question, the actual total average speed will be less than this average. Therefore, it takes a longer total time for a ship to travel back and forth in running water.
Considering an extreme situation can make the answer to the question extremely obvious and have an absurd comedy effect. Suppose the ship is just upstream. If the speed of water is equal to the speed of the ship, it will reach the turning point at twice the original speed. But it will never come back.
5. A, B and C three-way running 100 meter. Every time Party A wins10m, Party B wins10m. How many meters is a beat of C?
The answer is 19 meter. ? B wins C 10 meter? When B reached the finish line, C only reached 90 meters. ? A wins B 10 meter? When A reaches the finish line, B only reaches 90 meters, while C should still be at 8 1 meter. So a beat of c times 19 meters.
6. My brother runs 100 meters, and my brother runs 1 meter. The second time, my brother stepped back from the starting line 1 m to compete with my brother, so who will win?
The answer is that my brother still won. The time for my brother to run 100 meter is equal to the time for my brother to run 99 meters. The second time, my brother started from-1 m, and my brother started from 0 m. The two will be tied at the 99th meter. The rest 1 meter, my brother overtook my brother and won.
7. If you go up the mountain at a speed of 2 meters per second and down the mountain at a speed of 6 meters per second (suppose you go up and down the mountain by the same mountain road). So, what's your average speed during the whole journey?
This is one of the most easily mistaken problems in primary school travel, and it is also a problem that children can't understand anyway. The answer is not 4 meters per second, but 3 meters per second. Assuming that the whole journey is s meters, the uphill time is S/2, the downhill time is S/6, the total round-trip distance is 2S, and the total round-trip time is S/2+S/6, so the average speed of the whole journey is 2S/(S/2+S/6) = 3.
In fact, we can easily see that if the first half speed is A and the second half speed is B, then the total average speed should be less than (a+b)/2. This is because you will spend more time on the slow half beat, thus reducing the average speed. In fact, the total average speed should be the harmonic average of a and b, that is, 2/( 1/a+ 1/b). It is easy to prove that the harmonic average is always less than or equal to the arithmetic average.
8. You need to walk from Terminal 1 to Terminal 2. The road is divided into two sections, one is flat and the other is automatic conveyor belt. Assuming that your walking speed is constant, the actual walking speed on the conveyor belt is your speed on the flat ground plus the speed of the conveyor belt. If it takes two seconds to stop and do something during the whole process (such as squatting down to tie your shoelaces), where should you spend these two seconds to get to your destination faster? Many people may think the two schemes are the same, right? However, the real answer is that it will be faster to spend these two seconds on the conveyor belt. This is because the conveyor belt can provide you with some extra speed, so you will want to stay on the conveyor belt for a longer time and make full use of its benefits. So if you have to stop for a while, stop on the conveyor belt for a while.
More related articles on mathematics in Xiaoshengchu: