一道物理题
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参考一物体在斜面顶端由静止开始匀加速下滑，最初3秒内通过的位移为4.5M，最后3秒内通过的为10.5M，求斜面长
悬赏分：5 - 解决时间：2010-8-24 17:33
提问者： dykusa - 三级
最佳答案
检举
静止开始匀加速下滑，最初3秒内通过的位移为4.5M
加速度a=2s1/t1^2=2*4.5/3^2=1m/s^2
最后t=3秒内通过的为s2=10.5M
最后3秒的平均速度v=10.5/3=3.5m/s
v=(v1+v2)/2
v1=v2-at
所以v=(v2-at+v2)/2=v2-1/2at
v2=v+1/2at=3.5+1/2*1*3=5m/s
所以斜面长s=v2^2/(2a)=5^2/(2*1)=12.5m
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