题目内容
分析:
a≥![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114210287494.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114210287494.gif)
后车刹车做匀减速运动,当后车运动到与前车车尾即将相遇时,如后车车速已降到等于甚至小于前车车速,则两车就不会相撞,故取s后=s+s前和v后≤v前求解
解法一:取取上述分析过程的临界状态,则有
v1t-
a0t2=s+v2t
v1-a0t = v2
a0 =![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114210287494.gif)
所以当a≥
时,两车便不会相撞。
解法一:取取上述分析过程的临界状态,则有
v1t-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114210318225.gif)
v1-a0t = v2
a0 =
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114210287494.gif)
所以当a≥
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824114210287494.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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