General principle: distance = time * speed

Solution: Draw a picture.

Equivalence relation: equal time or equal distance.

1) in the opposite direction.

A speed of 3 Li, B speed of 2 Li, distance of 10 Li, walking in opposite directions at the same time, when do you meet?

Meeting time x, 2X+3X= 10, [or X(2+3)= 10]

Change:

① Lead or lag: A speed 3, B speed 2, distance 30. When A is in front of 1, they go in the opposite direction. When is the meeting time?

A speed is 3, B speed is 2, and the distance is 30. B walks 5 miles first and goes in the opposite direction. When is the meeting time?

② Find the distance: A speed of 3, B speed of 2, walk in opposite directions, meet at 2 o'clock, and find the total distance?

③ Find the speed: A speed is 3,30 minutes apart, while walking in the opposite direction, meet at 6 o'clock, and find the speed of B?

1) (this kind of problem is rare)

① The speed of A is 3, and the speed of B is 2, while walking in the opposite direction. What's the distance in five hours?

(2) The speed is 3, and it runs in the opposite direction of the same place at the same time. After 5 hours, how to find the B speed at a distance of 35?

③ The speed of A is 3, and the speed of B is 2, with a distance of 10 mile, and they run in opposite directions at the same time. After a period of time, the distance is 35. What time is it?

(4) A speed is 3, B speed is 2, and the same place runs in the opposite direction. A starts at 2 o'clock before B's departure, and after a period of time, the distance is 4 1. What time is it?

Go in the opposite direction (face to face), go back to back (back to back),

The problem is different, but the fact is the same.

3) Pursuit problem:

① The speed of A is 3, the speed of B is 2 and the distance is 5. Meanwhile, when can A catch up with B?

Neither face to face nor back to back, all facing the same direction,

A distance = distance +B distance.

Time X:3X=2X+5.

Or: (3-2)X=5,

(Different from previous thinking: A takes 3-2 more steps than B at a time, with a distance of 5. How long will it take to complete the extra distance? )

(2) Change problem, ring problem:

Circle 20, speed A 3, speed B 2, race in the same direction at the same time. When did A and B meet for the second time? How many laps did you run in one run?

The key: A runs one lap more than B, multiplied by X, then (3 -2)X=20.

X=20, A runs 20*3/20=3 laps.

This kind of circular problem is more common in competition questions and thinking questions. This is very beneficial.

3) sailing against the current:

Key: downstream speed = ship speed+current speed.

Current speed = ship speed-current speed

Ship speed refers to the speed when the water is still (still water).

① Ship speed 3, water speed 1, distance 60. What is the sailing time? Countercurrent sailing time?

60/4= 15,60/2=30.

② Water speed 1, distance 60, and round trip time 35. What's the speed of the boat?

Ship speed x, 60/(x+1)+60/(x-1) = 35.

(3) the speed difference is 60, and the time for a round trip is 35. What's the water speed?

Water velocity x, 60/(3+X)+60/(3-X)=35.

It should be said that these are the most basic travel problems. To solve the travel problem, these methods must be familiar.

I wonder if it will help you?