一物体正在做匀变速直线运动，在第一秒内和第三秒内通过的路程分别为2m和4m，求3s内的平均速度？
\u4e00\u7269\u4f53\u6b63\u5728\u505a\u5300\u53d8\u901f\u76f4\u7ebf\u8fd0\u52a8\uff0c\u5728\u7b2c\u4e00\u79d2\u5185\u548c\u7b2c\u4e09\u79d2\u5185\u901a\u8fc7\u7684\u8def\u7a0b\u5206\u522b\u4e3a2m\u548c4m\uff0c\u7b2c\u4e00\u79d2\u7684\u5e73\u5747\u901f\u5ea6\u662fV0.5=2m/s
\u7b2c\u4e09\u79d2\u7684\u5e73\u5747\u901f\u5ea6\u662fV2.5=4m/s
\u6240\u4ee5\u52a0\u901f\u5ea6\u4e3a1m/s^2
\u6240\u4ee5V2=3.5m/s
3\u79d2\u5185\u7684\u5e73\u5747\u901f\u5ea6\u4e3aV1.5=3m/s
\u56e0\u4e3a\u662f\u5300\u53d8\u901f\u76f4\u7ebf\u8fd0\u52a8\uff0c\u6240\u4ee5\u52a0\u901f\u5ea6\u6052\u5b9a
\u8bbe\u7b2c\u4e00\u79d2\u521d\u901f\u5ea6\u4e3aV0\uff0c\u52a0\u901f\u5ea6\u4e3aa\uff0c\u5219:
\u7b2c\u4e00\u79d2\u672b\uff08\u5373\u7b2c\u4e8c\u79d2\u521d\uff09\u901f\u5ea6\u4e3aV0+a\uff0c
\u7b2c\u4e8c\u79d2\u672b\uff08\u5373\u7b2c\u4e09\u79d2\u521d\uff09\u901f\u5ea6\u4e3aV0+2a\uff0c
\u7b2c\u4e09\u79d2\u672b\u901f\u5ea6\u4e3aV0+3a
\u7531\u4ee5\u4e0a\u53ef\u77e5\uff0c
\u7b2c\u4e00\u79d2\u5185\u7684\u5e73\u5747\u901f\u5ea6\u4e3a\uff08V0+V0+a\uff09/2\uff0c
\u7b2c\u4e09\u79d2\u5185\u7684\u5e73\u5747\u901f\u5ea6\u4e3a\uff08V0+2a+V0+3a\uff09/2\uff0c
\u7531\u8ddd\u79bb = \u5e73\u5747\u901f\u5ea6 * \u65f6\u95f4 \u53ef\u5f97\uff1a
\uff082*V0+a\uff09/2 * 1(S) = 2(m)
\uff082*V0+5a\uff09/2 * 1(s) = 4(m)
\u5f97 a = 1\uff0cV0 = 1.5
\u5219\u7b2c2\u79d2\u672b\u7684\u901f\u5ea6\u4e3a V0+2a = 1.5 + 2 = 3.5 (m/s)
X3-X1=2aT^2
4-2=a*1^2
a=2m/s^2
S=Vot+at^2/2
2=Vo*1+2*1^2/2
Vo=1m/s
V3=Vo+2*3
=7m/s
三秒内平均速度=（Vo+V3)/2=(1+7)/2=4m/s
由：s=v0t+at^2/2，可得：s1=v0*1+0.5a，
s3=3v0+9a/2-2v0-4a*2=v0*1+5a/2
则：s3-s1=2=2a，解得：a=2 m/s^2，v0=1 m/s
则：v3=3a+v0=7 m/s
v平=(v0+v3)/2=(1+7)/2=4m/s
应该是匀加速运动吧，第一秒速度2m/s,第三秒速度4m/s，第二秒应该经过3m，平均速度应该是3m/s。
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绛旓細绗竴绉掔殑骞冲潎閫熷害鏄疺0.5=2m/s 绗笁绉掔殑骞冲潎閫熷害鏄疺2.5=4m/s 鎵浠ュ姞閫熷害涓1m/s^2 鎵浠2=3.5m/s 3绉掑唴鐨勫钩鍧囬熷害涓篤1.5=3m/s
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