A and B set out from a and B at the same time and went opposite each other. B's speed was 23 times that of A. after they met, they continued to move forward. A arrived at B, and B returned immediately when they arrived at A. It is known that the distance between the place where they met for the second time and the place where they met for the first time is 3000 meters. What is the distance between a and B?

A and B set out from a and B at the same time and went opposite each other. B's speed was 23 times that of A. after they met, they continued to move forward. A arrived at B, and B returned immediately when they arrived at A. It is known that the distance between the place where they met for the second time and the place where they met for the first time is 3000 meters. What is the distance between a and B?

2 + 3 = 53000 ^ [35 × 3-1 - (1-35)], = 3000 ^ [95-1-25], = 3000 ^ [45 − 25], = 3000 ^ [25, = 7500 (m), answer: the distance between a and B is 7500 M

A and B set out from a and B at the same time, facing each other. B's speed is 2 / 3 of a's. after meeting, they continue to move forward. A goes to B. B goes to a
Return immediately. It is known that the location of their second meeting is 20 meters away from the location of their first meeting. How many meters is the distance between AB and ab?

Speed ratio: A: B = 3:2
5 copies
To meet for the first time a line 3, distance a place 3
To the second encounter, a line 9, a distance of 1
The place of the second meeting is 2 minutes away from the place of the first meeting
Each: 20 △ 2 = 10 (m)
The distance between a and B is: 10x5 = 50 (m)

Party A and Party B start from a and B at the same time and walk towards each other. The speed of Party B is 3 / 2 of that of Party A. after they meet, they continue to move forward and Party A arrives at B
Then: B returns immediately after arriving at A. It is known that the location of the second meeting is 3000 meters away from the location of the first meeting. Find the distance between a and B

The first meeting point is C, the second meeting point is D, ab = s, a -- C -- D -- B
∵ v b = (3 / 2) V A, the distance b took at the first meeting:
BC = [S / (VA + VB)] * VB = s * [(3 / 2) / (1 + 3 / 2)] = 3S / 5
From the beginning to the second meeting, they walked three times AB distance, 3S,
And the distance s a = AB + BD = [3S / (V A + v b)] * V A = 3S / (1 + 3 / 2) = 6S / 5, BD = 6S / 5-s = s / 5;
Then CD = bc-bd = 3S / 5-s / 5 = 2S / 5 = 3000m, х s = 7500m, that is ab = 7500m

A and B set out from a and B at the same time and went opposite each other. B's speed was 23 times that of A. after they met, they continued to move forward. A arrived at B, and B returned immediately when they arrived at A. It is known that the distance between the place where they met for the second time and the place where they met for the first time is 3000 meters. What is the distance between a and B?

2 + 3 = 53000 ^ [35 × 3-1 - (1-35)], = 3000 ^ [95-1-25], = 3000 ^ [45 − 25], = 3000 ^ [25, = 7500 (m), answer: the distance between a and B is 7500 M