Catch-up distance ÷ speed difference = catch-up time (catch-up in the same direction)
Speed difference = catching distance/catching time
Distance a traveled-distance b traveled = distance to catch up with the time difference. Meeting: the distance, speed and time of meeting.
Speed sum × meeting time = meeting distance
Meet distance ÷ meet time = speed and.
Distance traveled by A+distance traveled by B = total distance Example: A and B set off at the same time and ran around the 300-meter circular runway. A runs 6 meters per second, and B runs 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 = catch-up distance
The speed difference of this problem is: 6-4=2 (m/s).
After A caught up with B for the first time, the catching distance was 300 meters around the circular runway.
After the first catch-up, the two can be regarded as starting at the same time, so the problem of the second catch-up is transformed into a problem similar to solving the first catch-up.
The first time for A to catch up with B is: 300÷2= 150 (seconds).
A overtook B for the first time, running: 6× 150=900 (meters).
This shows that A caught up with B at the starting point, so the second chase problem can be simplified as the distance run multiplied by two in the first chase.
A catches up with B for the second time: 900+900= 1800 (meters).
Then a run 1800÷300=6 laps.
The conventional method to solve the catch-up problem is to list the equations according to the displacement equation. The displacement formula of uniform linear motion is a quadratic equation, so the quadratic trinomial (y=ax2+bx+c) and discriminant (△=b? -4ac).
In addition, when two (or several) objects are moving, one of them is often taken as the reference, that is, only the other (or several) objects are moving to make them "stationary". This simplifies the research process, so the problem of traceability is often solved by changing the reference method. At this time, the initial velocity and acceleration of other objects relative to the reference object should be determined before the motion of other objects can be determined.
In the final analysis, the more practical equation can solve all problems. I don't think arithmetic is a good way to solve problems. We should learn to solve them with equations.
1. A, B and C are located on the same straight line, and the distance from mile to mile to a and Cl is equal. Party A and Party B set out from A and C respectively at the same time. Party A meets Party B at a distance of-0/00 meters from Mile Mile/Kloc. After meeting, they move on. After arriving at Station C, Party A immediately returned and passed by 300 meters. What's the distance between station A and station C?
2. On the expressway, a car with a length of 4m and a speed of 1 10km/h is ready to overtake a truck with a length of 12m and a speed of100 km/h. How many hours does it take for the car to catch up with the truck completely?
Xiao Wang and Xiao Li go to the park from school at the same time. Xiao Wang drives at a speed of 10 km per hour. Xiao Li has something to leave late. In order to arrive with Xiao Wang at the same time, Xiao Li traveled at the speed of 12km per hour, but when Xiao Wang traveled 2/3 of the distance, his speed slowed down by 2km per hour. As a result, Xiao Li caught up with him 2 kilometers away from the park and asked about the distance from school to the park and how Li was late.
4. Party A, Party B and Party C travel at the speed of 30m, 40m and 50m per minute respectively. A and B set out in the same direction at the same time, and C chased A and B from B at the same time. It took more than ten minutes to catch up with B after catching A. How many meters is there between A and B?
On a street, a cyclist and a pedestrian walk in the same direction, and the speed of cyclist is three times that of pedestrian. Every 10 minute, a bus passes pedestrians. If the bus leaves from the departure station at the same time every 20 minutes, how many times will it be assigned?
6. The normal running speed of the train from Station A to bilibili is 60 km/h. When a train leaves Station A 8 minutes late, the driver will increase the speed to 80 km/h and arrive at the station 2 minutes early.