Assuming that the initial velocity of A is V A and the initial velocity of B is V B, they meet at C after time t 1 at the time of departure, and A catches up with B after time t2, then the following equation can be obtained.

(V A +v B) t 1=a

V b t 1=b

V B t2=a-b-b/2=a-3b/2

2v A t2=2b+v B t2=2b+a-3b/2=a+b/2

According to the first two equations of the above four equations, we can get:1+VA/VB = A/B.

According to the last two equations of the above four equations, we can get: 2v A /v B =(a+b/2)/(a-3b/2).

In the above two formulas, V A /v B is eliminated by substitution, and x = A/B:x = 1/2, 5/2 is obtained.

Where x= 1/2 does not meet the requirement that a is greater than or equal to 2b, so a/b=5/2, that is, a/(2b)=5/4.