匀变速直线运动推论即在纸带处理中的应用
\u5300\u53d8\u901f\u76f4\u7ebf\u8fd0\u52a8\u7684\u63a8\u8bbav\u5e73\u5747=(vo+vt)/2
2as=vt²+vo²
\u0394s=aT²
\uff081\uff09A\u3001\u6839\u636e\u5300\u53d8\u901f\u76f4\u7ebf\u8fd0\u52a8\u7684\u63a8\u8bba\u5f97\uff1a\u76f8\u90bb\u7684\u76f8\u7b49\u7684\u65f6\u95f4\u95f4\u9694\u4f4d\u79fb\u4e4b\u5dee\u76f8\u7b49\uff0e\u5373\u6839\u636e\u25b3x=aT2\u5f97\uff0cs2-s1=aT2\uff0c\u6545A\u6b63\u786e\uff0es4-s1=3aT2\uff0c\u6545B\u6b63\u786e\uff0eC\u3001\u6839\u636e\u5300\u53d8\u901f\u76f4\u7ebf\u8fd0\u52a8\u4e2d\u65f6\u95f4\u4e2d\u70b9\u7684\u901f\u5ea6\u7b49\u4e8e\u8be5\u8fc7\u7a0b\u4e2d\u7684\u5e73\u5747\u901f\u5ea6\u5f97\uff0c\u6253\u70b92\u65f6\u7269\u4f53\u7684\u901f\u5ea6\u4e3av2=s3+s22T\uff0e\u6545C\u9519\u8bef\uff0eD\u3001\u8ba1\u6570\u70b90\u70b9\u4e0d\u4e00\u5b9a\u662f\u6253\u51fa\u7684\u7b2c\u4e00\u4e2a\u70b9\uff0c\u6240\u4ee50\u70b9\u7684\u901f\u5ea6\u4e0d\u4e00\u5b9a\u7b49\u4e8e\u96f6\uff0c\u6240\u4ee5s1=v0t+12aT2\uff0c\u6545D\u9519\u8bef\uff0e\u6545\u9009AB\uff0e\uff082\uff09\u2460\u6839\u636e\u5300\u53d8\u901f\u76f4\u7ebf\u8fd0\u52a8\u7684\u7279\u70b9\uff08\u76f8\u90bb\u7684\u65f6\u95f4\u95f4\u9694\u4f4d\u79fb\u4e4b\u5dee\u76f8\u7b49\uff09\u5f97\u51fa\uff1ax45-x34=x34-x23=x23-x12=x12-x01\u654545\u4e24\u70b9\u95f4\u8ddd\u7b49\u4e8e54mm\uff0c\u6240\u4ee5\u5c5e\u4e8e\u7eb8\u5e26A\u7684\u662fC\u56fe\uff0e\u6545\u9009C\uff0e\u2461\u6839\u636e\u8fd0\u52a8\u5b66\u516c\u5f0f\u25b3x=at2\u5f97\uff1aa=\u25b3xt2=0.036?0.030(0.1)2=0.60m/s2\uff0e\u6545\u7b54\u6848\u4e3a\uff1a\uff081\uff09AB \uff082\uff09C\uff0c0.60
设三个时间t1、t2、t3，其中t2-t1=T，t3-t2=T
t1与t2间隔距离为x1，t3与t2间隔为x2
由x=v0t+(1/2)（at^2)可得：
x1=v0T+(1/2)a(t2^2-t1^2)
x2=v0T+(1/2)a(t3^2-t2^2)
x2-x1=(1/2)a[t3^2-t2^2-(t2^2-t1^2)]
t3^2-t2^2-(t2^2-t1^2)]=
T(t3+t2)-T（t1+t2）
=2T^2
所以得特x=x2-x1=aT^2
设某一时刻t位移为x1.经过时间T后位移为x2.在经过时间T后位移为x3
x1=vt+at^2/2
x2=v(t+T)+a(t+T)^2/2
x3=v(t+2T)+a(t+2T)^2/2
所以由t的任意性可得到△x＝(x3-x2)-(x2-x1)=aT^2
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