Meeting distance = speed × meeting time

Meeting time = meeting distance/speed and

Speed Sum = Meeting Distance/Meeting Time

Catch up with the problem

Catch-up distance = speed difference× catch-up time

Catch-up time = catch-up distance ÷ speed difference

Speed difference = catching distance ÷ catching time

A team is 450 meters long and advances at a speed of 90 meters per minute. Someone took something from the end of the line to the head of the line and immediately returned to the end of the line at a speed of 3 meters per second. How long does it take to go back and forth?

Comments: This problem is actually divided into two processes: ① The process from the tail to the head is a catching-up process, which is equivalent to the last person catching up with the previous person; ② The process from beginning to end is a meeting process, which is equivalent to meeting people from beginning to end.

In the process of catching up, if the catching-up time is x seconds and the speed of the leading team is 90 m/min = 1.5 m/s, then the driving distance of the leading team is1.5 m; If the pursuer's speed is 3m/s, the distance traveled by the pursuer is 3x meters. From the reciprocal relationship in the pursuit problem, "the distance between the pursuer and the pursued = the original distance apart", there are:

3x- 1.5x=450 ∴x=300

In the process of meeting, let the meeting time be y seconds, and the speed of the team and the returnees has not changed, then the distance traveled by the followers is 1.5y meters, and the distance traveled by the returnees is 3y meters. According to the equality relation in the encounter problem, "the distance traveled by A+the distance traveled by B = the total distance" is 3Y+ 1.5Y = 450 ∴.

Therefore, the round-trip time is x+y=300+ 100=400 (seconds).