What are the formulas that can be used for the problem of encountering and chasing?

Kilometers and seconds and meters with 4 meters

Think about the process ...... yourself

I wish you to study hard and make progress every day.

1. If the speed of A is x, then the speed of B is x+4

Therefore, there is an equation 60-12 x=60+12 x+4, and the solution is x=8

Let A x B (x+4).

48 x = 72 (x+4) x = 8 A = 8 - 40 2 points = 7:40 points.

7:40 a.m.

8 = 25 150 25 = 6 6 seconds.

7-5) = 28 seconds.

Let B x meters per second, and A (x+2) meters per second.

2x=2*4

x=4 B = 4 m A = 6 m.

The questions are very simple, so let's do it myself.

I'll make up other answers next time I have time...

These questions must have been talked about by the teacher in class, because there are too many questions, netizens don't have so much time to answer them for you one by one, in this case, you have to rely on yourself. Take a good look at the example problems taught by the teacher, then compare and summarize various situations (this step is very important), and then do a few exercises to consolidate it, and you can solve this kind of problem by yourself in the future!

There are too many questions, it must be summer vacation homework.

It's summer homework, right? Happiness, more than ours!

Do your homework online too? Not really!

Go to the People's Education Edition Teacher's Book and take a look.

The six formulas for the pursuit problem are as follows:1. The speed of the encounter distance and the time of the encounter.

2. The time of the encounter, the distance of the encounter, the speed and.

3. Speed and meeting distance and meeting time.

4. The distance of the encounter = the distance taken by A + the distance taken by B.

5. A's speed = meeting distance and meeting time - B's speed.

6. A's journey = the journey of meeting - B's journey to Zen.

The problem of chasing and encountering two objects moving on the same straight line or closed figure is usually classified as a chasing problem. This kind of question is often taken in exams, and it is a large category of questions in the journey of the Raid Mountain.

Categorization of travel issues1. Encounter problems.

Multiple objects move in opposite directions, usually finding the time of encounter or the whole process.

2. The problem of flowing water boating.

The boat itself has power, and even if the water does not flow, the boat has its own speed, but in the flowing water, it is either pushed by the flowing water, or it is pushed by the flowing water, or it is pushed by the flowing water, which makes the speed of the boat change in the flowing water, while the bamboo raft and the like have no speed, its speed is the speed of the water.

3. Train travel problems.

The length of the train is actually the length of the train, which is a characteristic of the train journey.

4. Clock problems.

The clock problem can be seen as a special problem of two people chasing or meeting on a special circular track, but the two "people" here are the minute hand and the hour hand of the clock. However, in many clock problems, we often encounter various "strange clocks", or "broken clocks", whose hour and minute hands will travel different degrees per minute than regular clocks, which requires us to learn to analyze different problems independently.

1. Problem description.

1. Encounter problems.

Time, distance and speed and.

Distance and speed and time.

Speed and distance and time.

2. Catch up with the problem.

Time, distance difference, speed difference.

Distance difference Speed difference time.

Difference in speed, distance difference, time.

2. Example Questions: Shed Letters.

1. AB and 2 are 300 kilometers apart, A and B start from AB and 2 at the same time, and go in the opposite direction, A travels 20 kilometers per hour, B travels 30 kilometers per hour, how many hours later the two meet?

Analysis: Meet the problem, find the time.

300 (20 30) 6 (hours).

2. AB and B are 300 kilometers apart, A and B start from AB and 2 at the same time, and go in the same direction (B is in front of A), A travels 30 kilometers per hour, B travels 20 kilometers per hour, and A can catch up with B after a few hours?

Analyze disturbance and burial: chase the problem and find the time.

300 (30 20) 30 (hours).

Summary: The above is the simplest situation of encountering and chasing problems - a linear encounter, chasing and chasing, and in the future, it will also involve circular encounters and chasing and multiple round-trips. Only by mastering the most basic situations can we lay a solid foundation for the next step of learning.

The encounter problem is when two objects depart from two places at the same time, face to face, and after a period of time, the two objects will inevitably meet on the way.

It differs from the general travel problem in that it is not the motion of one object, so the velocity it studies includes the velocity of two objects, that is, the sum of the bending velocity.

The time from departure to meeting is the time of encounter, the distance from departure to meeting is the distance of meeting, and the distance per unit time is the sum of the speeds of two objects. Note: It must be synchronized at the same time.

Step 1: Read the question.

It depends on whether it is an encounter or a chasing problem. The encounter problem is generally from two different places in the opposite direction, while the pursuit is generally in the same direction, but the time of departure is different or there is a certain distance at the time of departure.

Step 2: Set the formula.

The formula for the encounter problem is s and = v and t meet, that is, the distance traveled by A + the distance traveled by B = the sum of the speeds of A and B The time of the encounter.

The formula for catching up with the problem is s difference = v difference t chasing, that is, the distance between A and B when they set off = the difference between the speed of A and B and the time of pursuit.

Step 3: Operation.

Note: To see the units in the question, if the unit of velocity is km h, and the unit of time is s, then the unit conversion should be performed.

It is an effective way to solve such problems by mastering the research methods and solutions of the problems of pursuit, encounter, and multi-solution, understanding the reasons for the formation of multiple solutions, analyzing the process of motion in detail, thinking and summarizing, and comparing and classifying.

1 Catch up on the problem.

2 Encounter Problems.

1) Two objects moving in the same direction: the problem of meeting is the problem of chasing.

2) Two objects moving in opposite directions: meet when the algebraic sum of the displacements that occur in each direction is equal to the distance between the two objects at the beginning.

3 The key condition in the problem of chasing, meeting or avoiding collision between two objects in the same straight line:

Its essence is to analyze and discuss whether two objects can reach the same spatial position in the same time.

2) Grasp the relationship.

1 Two relations: the time relationship and the displacement relationship.

2 A condition: that is, the velocity of the two is equal, it is often the critical condition for whether the objects can catch up, cannot catch up, or the distance between the two is the largest and smallest, and it is also the entry point for analysis and judgment.

c) Common situations.

v1 (in the back) is less than v2 (in the front).

1. A: Uniform acceleration (v1) ====>>> B: Constant speed (v2) will definitely be able to catch up.

2. A: Constant speed (v1) ====>>> B: Uniform deceleration (v2) will definitely be able to catch up.

When v1=v2, the distance between the two is the largest. (At the beginning, B with a large velocity is in front, and A at the back has a smaller velocity, and the spacing is getting larger and larger, only when the velocity of A is greater than the speed of B, the spacing can be smaller and smaller, so when the two velocities are equal, the spacing is the largest.) ）

v1 (in the back) is greater than v2 (in the front).

3. A: Constant speed (v1) ====>>> B: Uniform acceleration (v2) may not be able to catch up.

4. A: Uniform deceleration (v1) ====>>> B: Constant speed (v2) may not be able to catch up.

When a uniform deceleration object catches up with an object moving at a uniform speed in the same direction, the critical condition for catching up or not catching up is:

v chaser = v chased, at this point s = 0

i.e. v chaser v chaser will definitely be able to catch up.

v chaser v2, it will collide, if v1 = v2, it will just collide.

If t is not solved, it means that the two cannot be in the same position at the same time and cannot catch up.

If it can't catch up, when v1=v2, the distance between the two is the smallest. (At the beginning, A with a large velocity is behind, B in front of the speed is smaller, and the spacing is getting smaller and smaller, only when the speed of B is greater than the speed of A, the spacing can be larger and larger, so when the two velocities are equal, the spacing is the smallest.) ）

Note: The critical condition for an encounter (or collision) is that the velocities of two objects are exactly in phase when they are in the same position.

5. Itinerary issues.

1. Encounter problems.

Distance sum = speed and time of encounter.

2. Catch up with the problem.

Distance difference = speed difference catch-up time.

3 Flowing water boating.

Downstream speed = boat speed + water speed.

Speed against the water = speed of the boat - speed of the water.

Boat speed = (speed with the water + speed against the water) 2

Water velocity = (velocity against the water - velocity against the water) 2

4 Multiple encounters.

Linear Distance: The number of journeys traveled by A and B = the number of encounters 2-1 Circular Distance: The number of journeys traveled by A and B = the number of encounters.

Among them, the distance traveled by A = the distance traveled in a single whole course The total number of the whole journey is 5 circular runway.

6 Application of positive and negative proportional relationships in travel problems.

The distance is certain, and the speed is inversely proportional to the time.

The speed is constant, and the distance is directly proportional to the time.

Time is constant, distance is proportional to speed.

7 Catch-up problems on the clock face.

The hour and minute hands are in a straight line;

The hour and minute hands are at right angles.

8 Combine some types of score, engineering, and difference problems.

9 Itinerary issues often use the "going back in time" and "assuming" ways of thinking.

Encounter Questions:

Encounter distance = speed and meeting time.

Encounter time = encounter distance speed and.

Speed and = Distance of Encounter, Time of Encounter.

Pursuit Problem: Pursuit Distance = Speed Difference Pursuit Time.

Catch-up time = catch-up distance Speed difference.

Speed difference = chase distance catch up time.

Encounter Questions:

The distance traveled between the two cars and the time of the meeting.

The time of the encounter The distance of the encounter is the speed of the two cars.

The speed of the two vehicles and the distance they met at the time of the meeting.

Catch up on the problem. Chase distance, speed difference between two cars, chase time.

Chase time Chase distance The speed difference between the two cars.

The difference in speed between the two cars, the distance between them, and the time of their pursuit.

Encounter: t=s (v1+v2).

Catch-up: t=s (v1-v2).

Catch up on time to ask for v1>v2, otherwise you can't catch up.

First calculate the velocity equal, find the acceleration, and analyze it in detail.

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The basic quantity relationship of the itinerary problem is:

Speed Time = Distance.

Distance velocity = time.

Distance Time = Velocity.

1. Encounter problems.

The sum of the speeds Encounter time = distance between two places.

The distance between two places The sum of the velocities = the time between them.

The distance between the two places Encounter time = the sum of the speeds.

2. Catch up with the problem.

Chase distance Difference in velocity = catch up time.

The difference in speed Catch-up time = catch-up distance.

Chase distance Catch up time = difference in velocity.

Fast Slow = Difference in speed.

Chase the problem.

Velocity difference * time = distance between the two.

The distance between the two velocity difference = time.

The distance between the two time = speed difference.

Encounter Questions:Speed and * time = distance.

Distance Speed and = Time.

Distance Time = Speed and.

Encounter Questions:

The distance traveled between the two cars and the time of the meeting.

The time of the encounter The distance of the encounter is the speed of the two cars.

The speed of the two vehicles and the distance they met at the time of the meeting.

Catch up on the problem. Chase distance, speed difference between two cars, chase time.

Chase time Chase distance The speed difference between the two cars.

The difference in speed between the two cars, the distance between them, and the time of their pursuit.