题目内容
在“研究匀变速直线运动” 的实验中,小车拖着纸带的运动情况如图所示,图中A、B、C、D、E为相邻的记数点,相邻的记数点的时间间隔是0.10s,标出的数据单位是cm,则打点计时器在打A点时小车的瞬时速度
=________m/s,小车运动的加速度是
=_________
。![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241558452527653.jpg)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155844908351.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155845049283.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824155845174783.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241558452527653.jpg)
0.52(3分) 2.4(3分)
试题分析:A点为OB段的中间时刻,由中间时刻的瞬时速度等于平均速度可得
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241558452671352.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241558453921618.png)
点评:本题难度较小,对于匀变速直线运动某一时刻的速度大小可由“中间时刻的瞬时速度等于平均速度”求得,加速度大小可由逐差法求得
![](http://thumb2018.1010pic.com/images/loading.gif)
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