A few seventh grade math problems, solving, urgent! 1. Party A and Party B set out from place a to place B at the speed of 4km per hour at the same time. After 2.5 hours, Party A returned to place a to get the documents. He rushed back to place a at the speed of 6km per hour. After getting the documents, he went back to catch up with Party B at the same speed. As a result, Party A and Party B arrived at the same time, The speed of the ship in still water is 10km / h, and the speed of the current is 2km / h. how many kilometers must the ship go to return to the original Wharf in 8 hours

A few seventh grade math problems, solving, urgent! 1. Party A and Party B set out from place a to place B at the speed of 4km per hour at the same time. After 2.5 hours, Party A returned to place a to get the documents. He rushed back to place a at the speed of 6km per hour. After getting the documents, he went back to catch up with Party B at the same speed. As a result, Party A and Party B arrived at the same time, The speed of the ship in still water is 10km / h, and the speed of the current is 2km / h. how many kilometers must the ship go to return to the original Wharf in 8 hours

Party A and Party B set out from place a to place B at the speed of 4km per hour at the same time. After 2.5 hours, Party A returned to place a to pick up the documents. He rushed back to place a at the speed of 6km per hour. After getting the documents, he went back to catch up with Party B at the same speed. As a result, Party A and Party B arrived at the same time. It is known that Party A was delayed for 15 minutes in the office when picking up the documents, so we can find the distance between a and B
Let the distance between a and B be s km
(S/4)=(2.5/4)+[(S+2.5)/6]+(15/60),
The solution is s = 15.5km, that is, the distance between a and B is 15.5km
A ship is going down the river from the wharf and up the river. It intends to return to the original wharf within 8 hours. All the time, the speed of the ship in still water is 10 km / h, and the speed of the current is 2 km / h. ask how many kilometers the ship has to walk at most to return to the original wharf within 8 hours
The downstream velocity is 10 + 2 = 12 km / h, and the upstream velocity is 10-2 = 8 km / h,
The velocity ratio of downstream to upstream is 12:8 = 3:2,
Because if the distance is equal, the time is inversely proportional to the speed,
Therefore, the sailing time ratio of downstream and upstream is 2:3,
It is known that the sailing time of downstream and upstream is no more than 8 hours,
It can be concluded that the sailing time along the current shall not exceed 8 × [2 / (2 + 3)] = 3.2 hours,
Therefore, the ship must return if it can walk 12 × 3.2 = 38.4km at most