Two cyclists set off at the same time towards each other from points A and B and met an hour later.

Two cyclists set off at the same time towards each other from points A and B and met an hour later. Arriving at points B and A, respectively, the cyclists immediately turned back and met again. How long after the first meeting did this happen?

1. Suppose the first time cyclists met at point X, and the second time – at point Y. Then, before the first meeting, they traveled the distance together:

S1 = AX + BX = S, where

S = AB is the distance between points A and B.

2. Arriving at points B and A, respectively, each of the cyclists traveled distance S. After that, having met at point Y, the cyclists traveled additionally the distance BY and AY, respectively, and together, from the beginning of the movement:

S2 = 2S + BY + AY = 2S + S = 3S.

3. Thus, two cyclists together before the first meeting covered the distance:

S1 = vt1 = S, (1)

and before the second meeting – the distance:

S2 = vt2 = 3S, (2)

where v = v1 + v2 is the total speed of cyclists.

4. From equations (1) and (2) we obtain:

t2 / t1 = S2 / S1 = 3;
t2 = 3t1 = 3 (h);
t2 – t1 = 3 – 1 = 2 (h).
Answer: after 2 hours.