如图1所示,某人从路灯C的下方P点处,以速度v向左匀速直线行走到Q点,人头M的影子由A点运动到B点.问人头M的影子怎样运动,速度是多少?1解析设路灯高为H,人高为h;t=0时刻,人站在P点,人头M的影子在A点;经时间t,人位于Q点,人头的影子在B点.因人作匀速直线运动,则在t时间内里人行走的距离PQ=vt.头M的影子移动的距离为AB,设AB=x, As shown in Figure 1, someone from the street below the C at point P, the speed v left straight to a straight line walking point Q, the head of M’s shadow from point A to point B. Asked the head of M’s shadow how to exercise, speed 1 is the resolution of street light height H, height h; t = 0 moment, the person standing P, M head shadow at point A; For a straight line due to human motion, then in t time walking distance PQ = vt. Head M shadow moving distance AB, set AB = x,