如何求“过河”最值问题,一直是中学的一个难点.求解这类问题,笔者总结了一种方法———“水陆等效法”,易懂,简便,实用.例1如图1所示,一个人发现S处有一溺水者,溺水者离岸的距离SB=10m,而发现者在河岸A处,距B点AB=20m.此人在岸上跑动的速度为v1=5m/s.而在水中的速度为v2=1m/s, How to solve the problem of “crossing the river” has always been a difficulty in middle schools. To solve this kind of problem, the author summarized a method—“water-land equivalent method”, which is easy to understand, simple, and practical. Example 1 is shown in Figure 1. A person found that there was a drowning person at S. The distance from the drowning man to the shore was SB=10m, and the discoverer was at the river bank A, and the distance from point B was AB=20m. The speed at which the man drove on the shore was v1=5m/s. The speed in the water is v2=1m/s,