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……

2024-11-24 06:26:55
推荐回答(3个)
回答1:

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

回答2:

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

回答3:

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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