高一的一道物理题，急急
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\u5219\u52a0\u901f\u9636\u6bb5\u7ecf\u8fc7\u8def\u7a0b\u4e3as=\uff081/2\uff09gt² \u6700\u5927\u901f\u5ea6\u4e3av=gt
\u51cf\u901f\u9636\u6bb5\u52a0\u901f\u5ea6a=\uff08975N-650N\uff09/65kg=5m/s²
\u51cf\u901f\u9636\u6bb5\u7528\u65f6t`=\uff08v-6m/s\uff09/\uff085m/s²\uff09
\u51cf\u901f\u9636\u6bb5\u8def\u7a0bs`=vt`+\uff081/2)at`² s+s`=11.4
\u4ee3\u5165\u6570\u636e\u53ef\u7b97\u51fa\uff1at=3.6s
v=gt=36m/s t\u603b=t+t`=3.6+6=9.6s
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通过前一半位移的时间T1=x/15
设通过后一半位移的时间为T2
x=12(T2)/2+6(T2)/2=9T2
T2=x/9
全程平均速度为
V=2x/(T1+T2)=11.25km/h
整个位移为S
一半位移的时间为T＝S／2×15＝S／30
对余下的一半位移有S／2＝12×T`／2＋6×T`/2＝9T`，所以有S＝18T`，T`为余下的时间
整个过程的平均速度V＝S/(T+T`)=S/(S/30+S/18)消去S可以求出V
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