example
A and B start at the same time and run in a 300-meter lap. A is 6 meters per second and B is 4 meters per second.
How many laps did A run when she caught up with B for the second time?
Basic equivalence relation: catch-up time *
Speed difference = capture distance
The speed difference of this problem is: 6-4=2.
After A caught up with B for the first time, the catching distance was 300 meters around Huanbao.
After catching up for the first time, it can be regarded as starting at the same time and place, so the problem of catching up for the second time is transformed into a class to solve the problem of catching up for the first time
The time for A to catch up with B for the first time is: 300/2= 150 seconds.
A catches up with B for the first time: 6 *150 = 900 meters.
This is B run: 4* 150=600 meters.
This shows that A caught up with B at the starting point, so the second chasing problem can be simplified as the distance traveled during the first chasing multiplied by two.
A caught up with B for the second time: 900+900= 1800.
B * * * escape: 600+600= 1200.
That is, A ran 1800/300=6 laps.
B ran 1200/300=4。
circle
The problem raised by walking in the opposite direction is called the encounter problem.
At the same time, two trains departed from two stations 624.5 kilometers apart and met five hours later. It is known that the speed of passenger cars is 70 kilometers per hour, and how many kilometers per hour are trucks?
624.5÷5-70
= 124.9-70
=54.9 km
A school is moving at a speed of 4 kilometers per hour, and a classmate at the end of the team runs to the front of the team and returns to the end of the team. This classmate's speed is 12 kilometers per hour. A round trip took 14.4 minutes, please captain.
Solution: students run to the front of the team to chase the questions;
Speed difference:12-4 = 8;
Students running to the end of the line is a problem;
Sum of speeds: 4+12 =16;
Method 1: Proportional method;
The distance between catching up and meeting is equal;
∴ Speed ratio: 8:16 =1:2;
Time ratio: 2:1;
Meeting time:14.4÷ (2+1) = 4.8;
Team length: 4.8 ÷ 60×16 =1.28;
Method 2: unit 1 method;
Set the group leader as1;
Time for students to run to the front of the queue:1÷ (12-4) =1
Time for students to run to the finish line:1÷ (12+4) =1
Time per copy: 14.4 minutes (
+
)= 14.4×
Time to return to the end of the line: (14.4×)
)×
= 14.4×
Team length: (14.4×
)÷60× 16= 1.28;
A: The length of the motorcade is1.28km. ..
3) Pursuit problem:
① A speed 3, B speed 2, distance 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?
Key: A runs one lap more than B, and the time is 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.