题目内容
做匀变速直线运动的物体初速度为12 m/s,在第6s内的位移比第5s内的位移多4m。关于物体运动情况的说法,下列正确的是 ( )
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130507585168.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130507585168.png)
A.物体的加速度为4 m/s2 |
B.物体5s末的速度是36m/s。 |
C.物体5、6两秒内的位移是72m。 |
D.物体从14m的![]() ![]() |
A![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130507585168.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130507585168.png)
根据匀变速直线运动相邻相同时间内位移差为常数:
,解得
m/s2所以A对;
根据
,5秒末速度为
,B错;
因为
∴5、6两秒内的位移:
m=64m,C错;发生14m的位移所用时间t:14=12t+0.5×4×t
,解得t=1s;
发生32m的位移所用时间t’:32=12t+0.5×4×t
,解得t=2s,所以从A到B所用时间为1s.
故选AD.
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130507975551.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241305080531011.png)
根据
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130508115559.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130508224887.png)
因为
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130508255767.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/201408241305082871086.png)
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130508318242.png)
发生32m的位移所用时间t’:32=12t+0.5×4×t
![](http://thumb.1010pic.com/pic2/upload/papers/20140824/20140824130508318242.png)
故选AD.
![](http://thumb2018.1010pic.com/images/loading.gif)
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