甲、乙两人分别从A、B两地同时出发，相向而行，出发时他们的速度比是3:2。他们第一次相遇后，甲的速度……
推荐回答（3个）
相遇前
甲的速度3x
乙的速度2x
相遇后
甲的速度3x(1+20%)=3.6x
乙的速度2x(1+30%)=2.6x
设第一次相遇的时间为t
(2xt/3.6x)*2.6x+7=3xt
xt=7/(3-13/9)=7*9/14=4.5KM
S=5xt1=4.5*5=20.25KM
又分析了一下这个题目，存在数学解法
无论相遇前还是相遇后，甲乙两人所用时间相同，那么各自所走路程比等于速度比，通过比例转换求。
3:2=x：3（1+20%）
x=5.4
7÷【（5.4-2.6）/（5.4+3.6）】=20.25
你猜猜看啊，由于甲和乙是同时出发的，从他们开始到相遇这段时间，时间是不是相等的，那么就可以猜出路程比为3/2了，对吧，也就是说，A到B若为s的话，那么，甲乙相遇的时候，甲走了3/5*s，乙走了2/5*s，对吧，再接下去好玩了，甲是走完全程的，（也就是说，还剩下2/5），乙走完的只是全程不到点，但是他们走的时间是一样的啊，所以，利用时间相等列下方程：（2/5s）/（3v*1.2）=（3/5s-7)/(2v*1.3),就是求s了，这种计算你自己算下好了
设甲速度3X公里每小时，乙2X公里每小时，且相遇时间为1小时。则总里程5X公里。
跟据后半程有：
2X/(1.2*3X)=(3X-7)/(1.3*2X)
解方程即可
X=4.5
5X=22.5公里
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