题目内容
(1)甲何时追上乙?甲追上乙时甲的速度多大?此时甲离出发点多远?
(2)在追赶过程中,甲、乙之间何时有最大距离?这个距离为多大?
(1)40s 20m/s 400m(2)10s末甲、乙之间距离最大 225m
(1)设如图所示,甲经过时间t追上乙,则有:
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241155297573506.jpg)
s甲=
(1分)
s乙= v0t (1分)
s甲=s0+s乙 (1分)
代入数值解得: t = 40s和t = -20s(舍去) (1分)
此时甲的速度: v甲= at =0.5×40m/s ="20m/s " (1分)
甲离出发点的位移:s甲=
=
×0.5×402m =" 400m " (1分)
(2)在追赶过程中,当甲的速度小于乙的速度时,甲、乙之间的距离仍在继续增大;当甲的速度大于乙的速度时,甲、乙之间的距离便不断减小;当v甲=v乙时,甲、乙之间的距离达到最大值。 (1分)
由:v甲=at= v乙
得:t=
s="10s " 即10s末甲、乙之间距离最大。 (1分)
且:Smax=s0+v乙t-
="200m" +5×10m-
×0.5×102 m =" 225m " (2分)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241155297573506.jpg)
s甲=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115529788403.gif)
s乙= v0t (1分)
s甲=s0+s乙 (1分)
代入数值解得: t = 40s和t = -20s(舍去) (1分)
此时甲的速度: v甲= at =0.5×40m/s ="20m/s " (1分)
甲离出发点的位移:s甲=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115529788403.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115529804226.gif)
(2)在追赶过程中,当甲的速度小于乙的速度时,甲、乙之间的距离仍在继续增大;当甲的速度大于乙的速度时,甲、乙之间的距离便不断减小;当v甲=v乙时,甲、乙之间的距离达到最大值。 (1分)
由:v甲=at= v乙
得:t=
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115529835473.gif)
且:Smax=s0+v乙t-
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115529788403.gif)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824115529804226.gif)
![](http://thumb2018.1010pic.com/images/loading.gif)
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