滑块以某一初速度冲上斜面做匀减速直线运动,到达斜面顶端时的速

滑块以某一初速度冲上斜面做匀减速直线运动,到达斜面顶端时的速度恰为零.已知滑块通过斜面中点时的速度为v,则滑块在前一半路程中的平均速度大小为

A. v                                          B.( +1)v

C.v                                                    D.

答案

解析:本题考查的是灵活应用匀变速直线运动的规律的能力,根据题意画出运动过程示意图,如图,设斜面总长为2s,在斜面底端时,物体的速度为v0,加速度的大小为a,则根据匀变速运动速度位移关系vt2-v02=2as,得

AC的过程-v02=-2a·2s

BC的过程-v2=-2a·s

所以v02=2v2,v0=

v

再根据

=,得前一半路程中

==v.

答案:A


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