Seconds speed mathematics
The tortoise is wrong. He said it was impossible for the rabbit to catch up with him again. Rabbit just lags behind him for a while. Think with normal thinking, regardless of various realistic or external factors such as slowing down after running. Even if a fast runner catches up with a slow runner from behind, if the distance is infinite, will the fast runner not catch up? That's impossible. In scientific terms, this story is based on Zeno's paradox in ancient Greece. If the rabbit wants to catch up with the tortoise, it must first reach the original position of the tortoise. When it reaches this position, the tortoise crawls forward to a new position, and the rabbit must run to this new position. This process can continue indefinitely. Such an "infinite process" certainly needs an infinite number of time periods, but the sum of these infinite time periods can be a "finite value". How long will it take? Let's assume that the rabbit runs at a speed of V = 10 m/s and the tortoise runs at a speed of V = 1 m/s. The rabbit runs 100 m for the first time, 10 sec for the second time and 0. 1 sec for the third time. In this "infinite process", the total time is s =10+1+0.1+... This is a problem of summation of infinite proportional series. The first term of this series is a = 10, and the common ratio is q = 0. 1, so s = a/(1-q) =10/(1-0/). During this period, the total distance that rabbits run is10×100/9 =1000/9 (meters); The total crawling distance of tortoise is/kloc-0 /×100/9 =100/9 (m). The difference between the two numbers (1000/9)-(100/9) =100 (meters), that is,11. )