16公里算远的距离吗?

设乙出发后,X小时能追上甲
16+4X=12X
X=2(小时)

甲乙两人分别在相距50公里的地方同时相向出发,甲每小时走16公里,乙每小时走18公里,多少分钟后,两人相距25公里?（写出两种情况）
(50+25) ÷( 18+16) = 75 ÷ 34 = 34分之 75 小时
(50-25) ÷( 18+16) = 25 ÷ 34 = 34分之 25 小时

设：乙出发后X小时追上甲
16+4X=12
解出X=2
所以,乙出发2小时后追上甲
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设时间为t,则：
(12t)^2 + (10-16t)^2=6^2,144t^2+256t^2-320t+100-36=0,25t^2-20t+4=0,(5t-2)^2=0,5t-2=0,t=0.4
即最快0.4小时后,二人距离6km

设x小时后能追上甲
16+4x=12x
8x=16
x=2
2小时后能追上甲

设行驶时间为x 两小时前两船的距离 为根号下（16-8x）²+（6x）² 对时间求导就是了
两小时以后 两船的距离为 根号下【8（4-x）】²+（6x）² 对时间求导 然后代入时间就是变化速度了
具体自己算

两次提速后,两车的速度和是一样的,因此相遇需要的时间也是一样的,也就是两次提速后和提速前相比减少的时间是一样,但是甲乙缩短的距离不一样,甲提速的时候,乙少走了16km,乙提速的时候,甲少走了12km,
因此乙速/甲速=16/12=4/3
设甲速是x,提速前相遇需要时间t,那么乙速就是4x/3,提速前速度和是7x/3
两次提速后的速度和都是7x/3+5,提速后相遇需要时间是(7x/3)/(7x/3=5)=t*7x/(7x+15),
因此提速后时间减少了t-t*7x/(7x+15)=15t/(7x+15)
x*15t/(7x+15)=12
=>t=(28x+60)/(5x)=(28+60/x)/5
设x=30,那么t=6,
甲速=30,乙速=30*4/4=40,总程=70*6=420
第一次相遇,甲走180,乙走240
甲提速后是420/75*35=196,196-180=16
乙提速后是420/75*45=252,252-240=12
符合要求
设x=60,那么t=29/5,
甲速=60,乙速=80,总程=28*29=812
提速前甲=60*29/5=348,乙=812-348=464
甲提速后812/145*65=364,364-348=16
乙提速后812/145*85=476,476-464=12
也符合要求.
实际上所有满足上面的t和x关系的都符合要求,因此题目有问题.漏或错了条件.

确定甲乙两船的距离S与时间t的函数 然后对时间t求导即可
步骤如下：
设甲初始位置坐标（0 0）则乙初始位置坐标为（0 16）
那么在经过时间t后,甲船距离原点 6t 乙船距离原点 16-8t
则甲乙两船在经过时间t后的距离S=根号下 (6t)^2+(16-8t)^2
s对t求导 然后将t=1代入得结果.

设x小时乙能追上甲
4x+16=12x
x=2

根号就是平方根的意思
因为16*16=256
所以16=根号256
256除以16应该=16,除的话……1都没有吧

（16+4）X10÷（16-4）=50/3（小时）

1,航程=（16+4）*12=240
2时间=航程/速度=240/（16-4）=20

轮船在静水中的速度x公里/小时,水的流速y公里/小时
x+y=20
x-y=16
x=18,y=2
轮船在静水中的速度18公里/小时,水的流速2公里/小时

设行驶时间为x 两小时前两船的距离 为根号下（16-8x）²+（6x）² 对时间求导就是了
两小时以后 两船的距离为 根号下【8（4-x）】²+（6x）² 对时间求导 然后代入时间就是变化速度了
具体自己算