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Thoughts on solving a sixth-grade math travel problem
Draw a picture, calculate the position of point A in each distance according to the proportion, and let AB distance be s,

The speed ratio between Party A and Party B is 72: 48 = 3: 2.

In other words, when A ran back and forth for three groups, B ran back and forth for two groups. At this time, both A and B returned to the starting point and started the cycle.

Draw the cycle as shown above in proportion, and take the distance of B running AB as the unit time.

Use an hourglass model? It is known that the second frontal encounter is (2/5) AB distance from point A..

Note: hourglass shape, with 2-4/3=2/3 at the top? , the bottom is 2- 1= 1, and the ratio is 2:3? [The second purple circle indicates this point]

A cycle, only four units, A catches up with B from behind. At this time, both of them are at point A, and the second time A catches up with B is also at point A (the time is 8 units).

So the distance between the two points mentioned in the question is (2/5) AB distance = 80m.

AB distance = 80 ÷ (2/5) = 200m.