First of all, Party B leaves two hours earlier than Party A, which means that Party B leaves two hours earlier than Party A. If the speed of Party B is m, Party B leaves two hours (S 1 = 2m). Secondly, at this time, A starts, and the speed is 2.5 times that of B, that is, 2.5M；; Third, if A wants to catch up with B, it means that the distance of A is equal to the distance of B; Suppose it takes time t for A to catch up with B, then the distance traveled by A is =2.5Mt, and the distance traveled by B after time t is S2=Mt, then the total distance traveled by B is S(B)= S(B)= S 1+S2 = 2M+Mt, because A and B travel in the same path, so S (A) =S (B).

Let b's speed be m, and Serie A's speed be 2.5M

B starts at 7 o'clock, and A starts at 9 o'clock, so B goes for two hours first. Assuming that the distance traveled is S 1, there are:

S 1=2M

If it takes time for A to catch up with B, then the distance between A and B is equal, that is, S (A) =S (B).

S(a)= 2.5 metric tons

Let S2 be the distance traveled by B in t time, then S2 = mt

The total distance traveled by b is s (b) = 2m+mt.

So: 2m+mt = 2.5mt

The solution is t=4/3.

Generally speaking, the process of mathematics pursuit is to consider equal distance.