设甲乙两地相距X km,A车开始的速度为V km/h,B车速度为V‘ km/h,则相遇后A车的速度为(V+20) km/h,根据题意列方程组:6(V+V')= X5(V+20+V')= X两式变形得:V+V'= X/6V+V'= X/5 - 20即X/6 = X/5 - 20解得X=600 km
(va+vb)*6=s(va+20+vb)*5=s6va+6vb=5va+100+5vb=sva+vb=100s=100*6=600