小学奥数:列车过桥问题!

2025-01-04 14:13:38
推荐回答(2个)
回答1:

由题意可知列车与甲同向而行,与乙相向而行。
设步行速度为每秒x米,列车速度为每秒y米,列车长度为a米。
列车从甲身边开过时有关系式:a+10x=10y (1)
列车从乙身边开过时有关系式:a-9x=9y (2)
(1)式-(2)式得到:y=19x (3)。
火车尾离开 甲3分钟后,列车头离甲的距离为180y+a-180x。也就是此时甲乙相距的距离。再到列车尾离开乙又过了9秒,这9秒内甲乙两人各走了9x米,则两者距离又近了18x米。也就是说这个时候两人的距离是s=180y+a-180x-18x (4)。由(1)、(3)式知a=10y-10x=180x。将a代入(4)得s=180y-18x=180*19x-18x=3402x米。两人相向而行,到相遇所用时间为3402x/2x=1701秒=28.35分钟。
算出1611秒答案的忽略了列车的长度。

回答2:

3分=180秒
由题可得火车是和甲同向而行的,和乙是相向而行的
全列车从甲身边开过用了10秒,全列车从乙身边开过用了9秒,
那么(火车的速度-甲的速度)×10=(火车的速度+乙的速度)×9
因为两人速度相同
所以火车速度=行人速度的19倍
当火车和乙相遇时,经过了180秒,这时候甲乙之间的路程需要行人行走:(19-1)×180=3240秒
再经过9秒,那么这时候甲乙之间的距离是需要行人走:3240-9×2=3222秒
3222÷2=1611秒
答全列车从乙身边开过用了9秒,火车离开乙1611秒后两人相遇

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