设通信员速度为x,队伍的速度为y
来回共用时：a/(x-y)+a/(x-y)=2ax/(x^2-y^2)
队伍走过的路程为y*2ax/(x^2-y^2)=√3a
2(x/y)/[(x/y)^2-1]=√3
√3(x/y)^2-2(x/y)-√3=0
[√3(x/y)+1]*[(x/y)-√3]=0
解得(x/y)=√3
通信员走过的路程=x*2ax/(x^2-y^2)
=2a(x/y)^2/[(x/y)^2-1]
=2a*(√3)^2/[(√3)^2-1]
=2a*3/(3-1)
=3a
物理的队列行程问题.求高手详细解答过程.