Downstream stroke = (ship speed+water speed) × downstream time

Upstream stroke = (ship speed-water speed) × upstream time

Downstream speed = ship speed+current speed

Current speed = ship speed-water speed

Still water velocity = (downstream velocity+upstream velocity) ÷2

Water velocity: (downstream velocity-upstream velocity) ÷2

Meeting questions (straight line)

The formula of opposite movement: meeting time = distance ÷ speed and (speed of A × time+speed of B × time = distance).

Reciprocal formula: opposite distance = speed and× time (speed of A× time+speed of B× time = opposite distance)

Encounter a problem (ring)

Distance of a+distance of b = circumference of the ring.

Meet many times

Linear distance: the total distance between line A and line B * * * = number of encounters ×2- 1.

Circular distance: the total distance between line A and line B * * * = the number of encounters.

Where the distance of the * * * line = the distance traveled in a complete trip * * * the number of complete trips.

Catch up with the problem

The formula for driving in the same direction: (slow comes first, fast comes later) catch-up time = catch-up distance ÷ speed difference.

If you are on a circular runway: (fast in front, slow in the back) catching distance = speed difference × time catching distance ÷ time = speed difference.

Distance of a+distance of b = total distance.

Suppose that the speed of A is X km/h, the speed of B is Y km/h, and A starts from A and B. When they meet for the first time, the distance from A is 4 km, that is, A has walked for (4/X) hours, and the distance from B to B is.

Y× (4/x) 〉 km, so that we can know that the whole route is S = 4+Y× (4/x) 〉, so that we can also know that this topic is about the distance from B when we meet for the first time, and compare this distance with the distance of 3 km from B when we meet for the second time. Therefore, for the convenience of later explanation, this distance [y× (4/x)] is expressed by j.

After meeting for the first time, A needs to walk 3+[y× (4/x)] to see B for the second time, while in A, B needs to walk 4+s-3 to see A .. So two people can write an equation at the same time, as follows:

{3+〔Y×(4/X)〕}/X=(4+S-3)/Y

(where s is the whole distance, which has been given above. I won't write it all here for the convenience of writing, but it's best to write it all when doing the problem, or I won't understand it. )

By sorting out the above formula, we can get,

4Y^2-XY-5X^2=0

Break this formula down into

(Y+X)(4Y-5X)=0

You can get the relationship between x and y, Y=-X or

Y=5X/4

Because the speed of two people can't be negative, the first relationship is rejected, and the second relationship is available.

Therefore, if we bring this relationship into the distance formula J, we can get that J=(5X/4 )×4/X=5.

Therefore, we know that when A and B meet for the first time, the distance from B is 5 kilometers, and when they meet for the second time, the distance from B is 3 kilometers, so the distance between the two meeting places is 2 kilometers.