题目内容
(14分)如图7所示,一修路工在长为x=100 m的隧道中,突然发现一列火车出现在离右隧道口(A)x0=200 m处,修路工所处的位置在无论向左还是向右跑恰好能安全脱离危险的位置.问这个位置离隧道右出口距离是多少?他奔跑的最小速度至少应是火车速度的多少倍?![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241418507542934.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241418507542934.jpg)
40 m
倍
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141851097308.png)
设人奔跑的最小速度是v,火车速度是v0,这个位置离隧道右出口A的距离为x1,离隧道左出口B的距离为x2,则由题意可得:
=
,
=
,x1+x2=x
已知x0=200 m,x=100 m.
解以上三个方程得:
x1=40 m,v=
v0.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141851144394.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141851191425.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141851238426.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141851284525.png)
已知x0=200 m,x=100 m.
解以上三个方程得:
x1=40 m,v=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824141851097308.png)
![](http://thumb2018.1010pic.com/images/loading.gif)
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