题目内容
一质点由A点出发沿直线AB运动,行程的第一部分是加速度大小为a1的匀加速运动,接着做加速度大小为a2的匀减速运动,抵达B点时恰好静止.如果AB的总长度是s,试求质点走完AB所用的时间t.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115857206656.gif)
设两种运动的连接处速度为v,因为起点A和终点B速度均为零,所以两部分运动的平均速度均为
,这样全程的平均速度也是
,则全程所用时间为:
t=
=
=
,
即v=
①
由前一部分运动得v=a1t1,由后一部分运动得v=a2t2
所以t=t1+t2=
+
=
②
将①式代入②式得
t=
.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115857222223.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115857222223.gif)
t=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115857237216.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115857253371.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115857268245.gif)
即v=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115857284239.gif)
由前一部分运动得v=a1t1,由后一部分运动得v=a2t2
所以t=t1+t2=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115857300241.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115857315244.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115857331557.gif)
将①式代入②式得
t=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115857206656.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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