题目内容
如图2-11(原图2-16)所示,物体自O点由静止开始做匀加速直线运动,A、B、C、D为其轨道上的四点,测得AB=2m,BC=3m,CD=4m.且物体通过AB、BC、CD所用的时间相等,求OA间的距离.[3]
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114413995754.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114413995754.gif)
1.125m
作匀变速直线运动的物体,中间时刻的瞬时速度等于这一段的平均速度,设物体通过AB、BC、CD各段所用的时间均为t0,则
, ①
, ②
又
, ③
, ④
由于AB=2m,BC=3m,CD=4m.解①、②、③、④式可得:
,
,
=1.125m,
所以OA间的距离为1.125m。
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114414026594.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114414042604.gif)
又
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114414120459.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114414151486.gif)
由于AB=2m,BC=3m,CD=4m.解①、②、③、④式可得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114414166447.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114414182414.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114414213493.gif)
所以OA间的距离为1.125m。
![](http://thumb2018.1010pic.com/images/loading.gif)
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