a、b两物体从同一位置沿同一直线运动，它们的速度图象如图所示，求在60s末,a在b前方还是后方，相距多远？ 
500m.
题目分析：由图象可知，a在前60s的位移为sa=
+
=
(40+10)×20 m +(60－20)×40 m =2100m
b在前60s的位移为sb=
=×80×(60-20) m ="1600m"
所以a在b前方，相距为2100m－1600m=500m.
点评：理解速度图象的斜率、“面积”等来分析物体的运动情况，是应具备的能力
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