A line with length of L = 100m and speed of V1 = 3, the correspondent arrived at the head of the line at the speed of V2 = 5m / s from the end of the line, And immediately return to the end of the team with speed v2?

If the team is regarded as static, the person will arrive at the top of the queue at the speed of v2-v1 at T1 = L / v2-v1
After returning, it is equivalent to returning at the speed of V1 + V2, T2 = L / V1 + v2,
Distance = V2 * (T1 + T2) displacement is equal to team movement distance = V1 * (T1 + T2)

A line with length L and speed V1. The correspondent arrives at the head of the line with speed V2 (V2 > V1) from the end of the line and returns to the end of the line with speed V2 immediately. Find the distance of the line during this period

In the process of the correspondent running from the end of the line to the head of the line, the speed of the correspondent relative to the team is: v = v2-v1, so the time required for the correspondent to run from the end of the line to the head of the line is: T1 = AV2 − v1. In the process of the correspondent running from the head of the line to the end of the line, the speed of the correspondent relative to the team is: V '= V1 + V2, so the correspondent runs from the head of the line to the head of the line The time required for Wu paiwei is: T2 = AV2 + V1, so the distance of the team in this period of time is: x = V1 (T1 + T2) = V1 (AV1 + V2 + AV2 − V1) = 2v1v22av22 − v21. A: the distance of the team in this period of time is 2v1v22av22 − v21

In order to convey an order, the correspondent runs from the end of the line to the head of the line, with the speed V2 = 3m / s, and then immediately returns to the end of the line with the same speed as the line
(1) How long does it take for a correspondent to leave the team and return to the end of the line
v2t-v1t=L ①
Suppose that the speed of the correspondent from the head to the tail is V2 ′, and its value is V2 ′ = V1 = 1.6m/s. Why is the speed from the head to the tail the same as that of the team?

Then immediately return to the end of the line at the same speed as the line

Why does the amount of material that loses electrons when metal reacts with acid = the amount of material that produces hydrogen atom = the amount of material that produces hydrogen?
Please prove it. It's better to have examples and analysis

The electron is conserved. In the reaction, the hydrogen element changes from the original positive 1 valence to hydrogen (0 valence), so the amount of the material that the metal reacts with the acid to lose the electron = the amount of the material that generates the hydrogen atom, and the amount of the material that generates 1mol hydrogen needs to transfer 2mol electrons, so why the amount of the material that the metal reacts with the acid to lose the electron = the amount of the material that generates the hydrogen

A line with length of L = 100m and speed of V1 = 3, the correspondent arrived at the head of the line at the speed of V2 = 5m / s from the end of the line, And immediately return to the end of the team with speed v2?