题目内容
一条宽度为L的河,水流速度为
,已知船在静水中速度为
,那么:
(1)怎样渡河时间最短?
(2)若
,怎样渡河位移最小?
(3)若
,怎样渡河船漂下的距离最短?
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624567231.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624583236.gif)
(1)怎样渡河时间最短?
(2)若
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624598286.gif)
(3)若
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624614287.gif)
(1)船头与河岸垂直,![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624629334.gif)
(2)船头应指向河的上游,并与河岸成一定的角度θ,![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624645404.gif)
(3)船头与河岸的夹角应为![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624661402.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624629334.gif)
(2)船头应指向河的上游,并与河岸成一定的角度θ,
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624645404.gif)
(3)船头与河岸的夹角应为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624661402.gif)
:(1)小船过河问题,可以把小船的渡河运动分解为它同时参与的两个运动,一是小船运动,一是水流的运动,船的实际运动为合运动。如图4所示。设船头斜向上游与河岸成任意角θ。这时船速在垂直于河岸方向的速度分量为
,渡河所需要的时间为
,可以看出:L、v船一定时,t随sinθ增大而减小;当
时,
(最大)。所以,船头与河岸垂直
。
图4
(2)如图5所示,渡河的最小位移即河的宽度。为了使渡河位移等于L,必须使船的合速度v的方向与河岸垂直,即使沿河岸方向的速度分量等于0。这时船头应指向河的上游,并与河岸成一定的角度θ,所以有
,即
。
图5
因为
,所以只有在
时,船才有可能垂直河岸渡河。
(3)若
,则不论船的航向如何,总是被水冲向下游,怎样才能使漂下的距离最短呢?
如图6所示,设船头v船与河岸成θ角。合速度v与河岸成α角。可以看出:α角越大,船漂下的距离x越短,那么,在什么条件下α角最大呢?以v水的矢尖为圆心,v船为半径画圆,当v与圆相切时,α角最大,根据![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624848370.gif)
图6
船头与河岸的夹角应为
,船沿河漂下的最短距离为:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624895577.gif)
此时渡河的最短位移:![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624926433.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624676330.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624692427.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624707277.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624723278.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624629334.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241146247544037.jpg)
图4
(2)如图5所示,渡河的最小位移即河的宽度。为了使渡河位移等于L,必须使船的合速度v的方向与河岸垂直,即使沿河岸方向的速度分量等于0。这时船头应指向河的上游,并与河岸成一定的角度θ,所以有
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624754340.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624645404.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241146247853847.jpg)
图5
因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624801320.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624598286.gif)
(3)若
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624614287.gif)
如图6所示,设船头v船与河岸成θ角。合速度v与河岸成α角。可以看出:α角越大,船漂下的距离x越短,那么,在什么条件下α角最大呢?以v水的矢尖为圆心,v船为半径画圆,当v与圆相切时,α角最大,根据
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624848370.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241146248635805.jpg)
图6
船头与河岸的夹角应为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624661402.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624895577.gif)
此时渡河的最短位移:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114624926433.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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