设甲车速度V甲,乙车速度V乙,第一次相遇时间t1,第二次相遇时间t2,全程s。
第一次相遇
V甲*t1=60····························(1)
V乙*t1=s-60···························(2)
(1)除以(2)得:
V甲/V乙=60/(s-60)··················(3)
第二次相遇
V甲*t2=2s-40·························(4)
V乙*t2=s+40··························(5)
(4)除以(5)得:
V甲/V乙=(2s-40)/(s+40)········(6)
由(3)、(6)可得
60/(s-60)=(2s-40)/(s+40)
解得s=110
追问太难了啊,四年级哪看得懂