题目内容
甲、乙两个同学在直跑道上进行4×100m接力(如图所示),他们在奔跑时有相同的最大速度,乙从静止开始全力奔跑需跑出25m才能达到最大速度,这一过程可看作匀加速直线运动.现在甲持棒以最大速度向乙奔来,乙在接力区伺机全力奔出.若要求乙接棒时奔跑的速度达到最大速度的80%,则:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250017228444445.jpg)
(1)乙在接力区须奔出多少距离?
(2)乙应在距离甲多远时起跑?
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250017228444445.jpg)
(1)乙在接力区须奔出多少距离?
(2)乙应在距离甲多远时起跑?
(1)16m (2)24m
试题分析:(1)乙从静止开始全力奔跑,做初速度为零的匀加速运动,需跑出
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001722859535.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001722891367.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001722906689.png)
若要求乙接棒时奔跑速度达到最大速度的80%,则
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250017229221026.png)
联立以上各式解得:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001722937571.png)
(2)在接棒过程中,乙做初速度为零的匀加速直线运动,设乙在距甲
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001722953324.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250017229691368.png)
根据运动学公式有:
乙的位移:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/201408250017229841025.png)
甲的位移:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001723000557.png)
由几何关系知:
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001723203512.png)
以上各式联立解得 :
![](http://thumb.1010pic.com/pic2/upload/papers/20140825/20140825001723218598.png)
所以乙应在距离甲24m时起跑.
![](http://thumb2018.1010pic.com/images/loading.gif)
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