Fundamentals of the basics

Without which it is impossible to solve problems on the movement

For successful solution movement tasks you need to keep one simple formula in mind all the time:

To make it easier to remember this formula, think about how you would answer the following question:

“How many kilometers can I ride a bicycle in an hour, moving at a speed of km / h?”

You, without hesitation, will answer - km. Here you go. Congratulations! You always knew this formula well, you just could not formulate it.

From our formula it is easy to express all its components:

Formula for speed:

Formula for time:

The following pyramid helps a lot of people to remember the formula:

Got it? Now let's look at the detailed problem solving algorithmon the move. It consists of large steps.

Let us examine in more detail some of the features and subtleties that arise when solving movement tasks.

A little about attentiveness in reading tasks

Read the problem several times. Become aware of it so that you can understand absolutely everything.

For example, there are often difficulties with the concept of " own speed boats/boats, etc. Think what that might mean? That's right, the speed of the boat in standing water, for example, in a pond when it is not affected by the current velocity. By the way, in tasks they sometimes write “find the speed of a boat in still water”. Now you know that the boat's own speed and the speed of the boat in still water are the same, so don't get lost if you come across both of these definitions.

Peculiarities of painting in motion tasks - who goes where, who came to whom, and where they all met)

Make a drawing and write everything on it along the way. known quantities(well, or under it, if you don’t know how to display them schematically). The drawing should clearly reflect the whole meaning of the task. The drawing should be made in such a way that the dynamics of movement is visible on it - directions of movement, meetings, turns, turns. A high-quality drawing allows you to understand the problem without looking at its text. He is your main clue for further formulating the equation.

Consider possible types movements of two bodies:

Movement towards each other.

• If bodies move towards each other, then their speed of approach is equal to the sum of their speeds:

Do not trust? Let's see in practice.

Suppose two cars left a point and a point towards each other. The speed of one car is km/h and the speed of a car is km/h. They met within hours. What is the distance between points and?

1 solution:

1. (km) - the distance traveled by the first car
2. (km) - the distance traveled by the second car
3. (km) - the distance that both cars traveled, that is, the distance between points and.

Solution 2:

1. (km) - the distance that cars have traveled, that is, the distance between points and.

Both solutions are correct. It’s just that the second one is more rational, so memorize the formula (it’s absolutely logical, right?), And to assimilate, solve the following problem:

Misha and Vasya rode a bicycle towards each other. Misha's speed is km/h, Vasya's speed is km/h. The guys met in an hour. What path did they take together?

I decided? It turned out that the speed of approach is equal to km/h, and the path is equal to km. Now let's figure out how time is calculated in such a case.

• If the initial distance between the bodies is equal, then the time after which they will meet is calculated by the formula:

Based on the previous formula, this is quite logical, however, let's try to check it in practice. So, the task is - From point and point, cars move towards each other with speeds of km / h and km / h. Distance between points - km. How long will it take for the cars to meet?

1 solution:

Let be the time that the cars travel, then the path of the first car is , and the path of the second car is . Their sum and will be equal to the distance between the points and - .

Let's solve the equation:

(h) - the time after which the cars met.

Solution 2:

1. (km/h) - vehicle approach speed
2. (h) - the time that the cars were on the road.

Movement in opposite sides.

• If bodies move away from each other, then their removal speed is equal to the sum of their speeds:

Try to solve the problem yourself and prove the correctness of this formula as in the previous case. And here is the problem: cars left Moscow in opposite directions. The speed of one car is km/h, the speed of the other is km/h. How far apart will the cars be after one hour?

I decided? Solving in the first way, it turned out that the path taken by the first car is km, and the second is km. Accordingly, the distance between the cars is km. Solving in the second way, it turns out that the speed of removal is equal to km / h, and the path is equal to km / h h \u003d km.

Now let's see how the time is calculated in such a case.

• The time spent by the bodies on the way, when moving away from each other, is equal to the distance traveled (that is, if there was initially a certain distance between the bodies, then it should be subtracted from the total distance) to the distance divided by the sum of the speeds of the bodies:

As you can see, a formula similar to the one we derived with movement of bodies towards each other. Do you think that this cannot be? Check it out in practice!

Suppose two cars are moving in opposite directions with speeds and km/h. At a stop, the distance between them was km. How long did the cars take?

Try to solve this problem in the two ways that were described in going to a meeting. I decided? Is the formula correct? Let's compare the answers: the equation obtained by solving option 1 -; when solving option 2 - removal speed - km / h, travel time - hours.

But what if the bodies are initially at a certain distance from each other? It looks something like this:

How to solve such problems then? Very simple. When making a decision, we need to take into account.

• If there is any initial distance between the bodies, then the path formula is as follows:

Is it logical? Express from this formula the time of the meeting of two bodies, and then compare what we got.

Did you manage? Then we solve the problem for this formula.

From different points city ​​N in the directions opposite to each other, two motorcyclists left. The initial distance between them was km. The speed of the first motorcyclist was km/h; the speed of the second is km/h. After what time will the distance between them be equal to km?

What answer did you get? I got an hour. Let's check everything in detail. The path that the motorcyclists actually traveled is km km km. The speed of their removal from each other is km/h. Divide km by km/h and get hours - the time that motorcyclists spent on the road.

Movement in one direction.

So let's say our bodies move in the same direction. How many cases do you think there might be for such a condition? That's right, two.

Why is it so? I am sure that after all the examples you will easily figure out how to derive these formulas.

Got it? Well done! It's time to solve the problem. Kolya goes to work by car at a speed of km/h. Colleague Kolya Vova travels at a speed of km/h. Kolya lives at a distance of km from Vova. How long will it take Vova to overtake Kolya if they left the house at the same time?

Did you count? Let's compare the answers - it turned out that Vova will catch up with Kolya in hours or minutes.

Let's compare our solutions.

The drawing looks like this:

Similar to yours? Well done!

Since the problem asks how long the guys met and left at the same time, the time they traveled will be the same, as well as the meeting place (in the figure it is indicated by a dot). Making equations, take the time for.

So, Vova made his way to the meeting place. Kolya made his way to the meeting place. This is clear. Now we deal with the axis of movement.

Let's start with the path that Kolya did. Its path () is shown as a segment in the figure. And what does Vova's path () consist of? That's right, from the sum of the segments and, where is the initial distance between the guys, and equal to path done by Kolya.

Based on these conclusions, we obtain the equation:

Got it? If not, just read this equation again and look at the points marked on the axis. Drawing helps, doesn't it?

hours or minutes minutes.

I hope this example made you understand how important role a well-composed drawing plays, and we smoothly move on, or rather, have already moved on to next paragraph of our algorithm - bringing all quantities to the same dimension.

The rule of three "R" - dimension, reasonableness, calculation.

Dimension.

The tasks do not always give the same dimension for each participant in the movement (as was the case in our easy tasks). For example, you can find tasks where it is said that the bodies moved for a certain number of minutes, and the speed of their movement is indicated in km / h. We can't just take and substitute the values ​​in the formula - the answer will be wrong. Even in terms of units of measurement, our answer “will not pass” the test for reasonableness. Compare:

See? With proper multiplication, we also reduce the units of measurement, and, accordingly, we get a reasonable and correct result. And what happens if we do not translate into one system of measurement? The answer has a strange dimension and % is an incorrect result.

So, just in case, I will remind you of the meaning basic units measurements of length and time.

• Length units:

centimeter = millimeters

decimeter = centimeters = millimeters

meter = decimeters = centimeters = millimeters

kilometer = meters

• Time units:

minute = seconds

hour = minutes = seconds

day = hour = minute = second

Advice: When converting units of measurement related to time (minutes to hours, hours to seconds, etc.), imagine a clock face in your head. It can be seen with the naked eye that minutes is a quarter of the dial, i.e. hours, minutes is a third of the dial, i.e. hours, and a minute is an hour.

And now a very simple task:

Masha rode a bicycle from home to the village at a speed of km/h for minutes. What is the distance between the car house and the village?

Did you count? The correct answer is km.

minutes is an hour, and another minute from an hour (mentally imagined the clock face, and said that minutes is a quarter of an hour), respectively - min h.

Intelligence.

Do you understand that the speed of a car cannot be km/h, unless, of course, we are talking about a sports car? And even more so, it cannot be negative, right? So, reasonableness, that's about it)

Calculation.

See if your solution "passes" the dimension and reasonableness, and only then check the calculations. It is logical - if there is an inconsistency with dimension and reasonableness, then it is easier to cross out everything and start looking for logical and mathematical errors.

"Love for tables" or "when drawing is not enough"

Far from always, the tasks for movement are as simple as we solved before. Very often, in order to correctly solve the problem, you need not only to draw a competent drawing, but also to compile a table with all the conditions given to us.

From point to point, the distance between which is km, a cyclist and a motorcyclist left at the same time. It is known that a motorcyclist travels more miles per hour than a cyclist. Determine the speed of the cyclist if it is known that he arrived at the point a minute later than the motorcyclist.

Here is such a task. Pull yourself together and read it several times. Read? Start drawing - a straight line, a point, a point, two arrows ... In general, draw, and now let's compare what you got.

Kind of empty, right? We draw a table. As you remember, all movement tasks consist of components: speed, time and path. It is from these graphs that any table in such problems will consist. True, we will add one more column - the name, about whom we write information - a motorcyclist and a cyclist. Also in the header, indicate the dimension in which you will enter the values ​​\u200b\u200bthere. You remember how important this is, right?

Do you have a table like this?

Now let's analyze everything that we have, and in parallel enter the data into a table and into a figure.

The first thing we have is the path that the cyclist and motorcyclist have traveled. It is the same and equal to km. We bring in!

We argue further. We know that a motorcyclist travels more km/h than a cyclist, and in the problem we need to find the speed of the cyclist... Let's take the speed of the cyclist for, then the speed of the motorcyclist will be... If with such variable solution task will not work - it's okay, let's take another one until we reach the victorious one. This happens, the main thing is not to be nervous!

The table has changed. We have left not filled only one column - time. How to find the time when there is a path and speed? That's right, divide the path by the speed. Enter it in the table.

So our table has been filled, now you can enter data into the figure. What can we reflect on it? Well done. The speed of movement of a motorcyclist and a cyclist.

Reread the problem again, look at the figure and the completed table. What data is not shown in the table or in the figure? Right. The time by which the motorcyclist arrived earlier than the cyclist. We know that the time difference is minutes. What should we do next step? That's right, translate the time given to us from minutes to hours, because the speed is given to us in km / h.

Formula magic: writing and solving equations - manipulations leading to the only correct answer

So, as you may have guessed, now we will make an equation. Look at your table, at the last condition that was not included in it, and think about the relationship between what and what can we put into the equation? Correctly. We can make an equation based on the time difference!

Is it logical? The cyclist rode more, if we subtract the time of the motorcyclist from his time, we will just get the difference given to us.

This equation is rational. If you don't know what it is, read the topic "".

We bring the terms to a common denominator:

Let's open the brackets and give like terms:

From this equation we get the following:

Let's open the brackets and move everything to the left side of the equation:

Voila! We have a simple quadratic equation. We decide!

We received two responses. Look what we got for? That's right, the speed of the cyclist. We recall the rule "3P", more specifically "reasonableness". Do you understand what I mean? Exactly! Speed ​​cannot be negative, so our answer is km/h.

Phew! Got it? Try your hand at the next task.

Two cyclists set out on a 1-kilometer run at the same time. The first one was driving at a speed that was 1 km/h faster than the second one, and arrived at the finish line hours earlier than the second one. Find the speed of the cyclist who came to the finish line second. Give your answer in km/h.

I remind you: read the problem a couple of times - learn all the details. Got it? Start drawing the drawing - in which direction are they moving? how far did they travel? Did you draw? Check if all the quantities you have are of the same dimension and start writing out the condition of the problem briefly, making up a table (do you remember what columns are there?). While writing all this, think about what to take for? Chose? Record in the table! Well, now it’s simple: we make an equation and solve it. Yes, and finally - remember the "3P"!

I've done everything? Well done! It turned out that the speed of the cyclist is km / h.

-"What color is your car?" - "She's beautiful!" Correct answers to the questions

Let's continue our conversation. So what is the speed of the first cyclist? km/h? I really hope you're not nodding in the affirmative right now! Read the question carefully: "What is the speed of first cyclist? Got what I mean? Exactly! Received is not always the answer to the question! Read the questions thoughtfully - perhaps, after finding it, you will need to perform some more manipulations, for example, add km / h, as in our task.

Another point - often in tasks everything is indicated in hours, and the answer is asked to be expressed in minutes, or all the data is given in km, and the answer is asked to be written in meters. Look at the dimension not only during the solution itself, but also when writing down the answers.

Tasks for moving in a circle

The bodies in the tasks may not necessarily move in a straight line, but also in a circle, for example, cyclists can ride along a circular track. Let's take a look at this problem.

From paragraph circular track the cyclist left. In minutes he had not yet returned to the checkpoint, and a motorcyclist followed him from the checkpoint. Minutes after departure, he caught up with the cyclist for the first time, and minutes after that he caught up with him for the second time. Find the speed of the cyclist if the length of the track is km. Give your answer in km/h.

Try to draw a picture for this problem and fill in the table for it. Here's what happened to me:

Let the speed of the cyclist be, and the speed of the motorcyclist -. Until the moment of the first meeting, the cyclist was on the road for minutes, and the motorcyclist -. In doing so, they traveled equal distances:

Between meetings, the cyclist traveled the distance, and the motorcyclist -. But at the same time, the motorcyclist drove exactly one lap more, this can be seen from the figure:

I hope you understand that they didn't actually go in a spiral - the spiral just schematically shows that they go in a circle, passing the same points of the track several times.

We solve the resulting equations in the system:

Got it? Try to solve independently the following tasks:

1. Two mo-to-tsik-li-hundreds start-to-tu-yut one-but-time-men-but in one-right-le-ni from two dia-met-ral-but pro-ty-in-po- false points of a circular route, the length of a swarm is equal to km. After how many minutes, mo-the-cycle-lists are equal for the first time, if the speed of one of them is by km / h more than the speed of the other hoo?
2. From one point of the cur-go-howl of the highway, the length of some swarm is equal to km, at the same time, in one right-le-ni, there are two motorcyclists. The speed of the first motorcycle is km / h, and minutes after the start, he was ahead of the second motorcycle by one lap. Find the speed of the second motorcycle. Give your answer in km/h.

Answers:

1. Let km / h be the speed of the first mo-to-cycle-li-hundred, then the speed of the second mo-to-cycle-li-hundred is km / h. Let the first time mo-the-cycle-lists be equal in hours. In order for mo-the-cycle-li-stas to be equal, the faster one must overcome them from the beginning distance, equal in lo-vi-not to the length of the route.

   We get that the time is equal to hours = minutes.

2. Let the speed of the second motorcycle be km/h. In an hour, the first motorcycle traveled a kilometer more than the second swarm, respectively, we get the equation:

   The speed of the second motorcyclist is km/h.

Tasks for the course

Now that you're good at problem solving on land, let's move on to the water and look at the terrifying current problems.

Imagine that you have a raft and you lower it into a lake. What is happening to him? Correctly. It stands because a lake, a pond, a puddle, after all, is stagnant water. The current velocity in the lake is. The raft will only move if you start rowing yourself. The speed he gains will be own speed of the raft. No matter where you swim - left, right, the raft will move at the same speed with which you row. This is clear? It's logical.

Now imagine that you are lowering the raft onto the river, turn away to take the rope... turn around, and he... swam away... This happens because the river has a current speed that carries your raft in the direction of the current. At the same time, its speed is equal to zero (you are standing in shock on the shore and not rowing) - it moves with the speed of the current.

Got it? Then answer this question - "How fast will the raft float on the river if you sit and row?" Thinking?

There are two possible cases here:

1 case- you go with the flow, and then you go with your own speed + the speed of the current. The current seems to help you move forward.

2nd case You are swimming against the current. Hard? That's right, because the current is trying to "throw" you back. You are making more and more efforts to swim at least meters, respectively, the speed with which you move is equal to your own speed - the speed of the current.

Let's say you need to swim a mile. When will you cover this distance faster? When will you move with the flow or against?

Let's solve the problem and check. Let's add to our path data on the speed of the current - km/h and on the own speed of the raft - km/h. How much time will you spend moving with and against the current?

Of course, you easily coped with this task! Downstream - an hour, and against the current as much as an hour! This is the whole essence of the tasks on flow with the flow.

Let's complicate the task a little. A boat with a motor sailed from point to point in an hour, and back in an hour. Find the speed of the current if the speed of the boat in still water is km/h

Let's denote the distance between the points as, and the speed of the current as.

We will enter all the data from the condition in the table:

	Path S	speed v, km/h	time t, hours
A -> B (upstream)			4
B -> A (downstream)			2

We see that the boat makes the same path, respectively:

What did we charge for? Flow speed. Then this will be the answer :) The current speed is km / h.

The kayak went from point to point, located km away. After staying at point for an hour, the kayak set off and returned to point c. Determine (in km/h) the own speed of the kayak if it is known that the speed of the river is km/h.

So let's get started. Read the problem several times and draw a picture. I think you can easily solve this on your own. Are all quantities expressed in the same form? No. Rest time is indicated in both hours and minutes. Let's convert this to hours.

hour minutes = h.

Now all quantities are expressed in one form. Let's start filling out the table and looking for what we'll take for.

Let be the own speed of the kayak. Then, the speed of the kayak downstream is equal, and against the current is equal.

Let's write this data, as well as the path (as you understand, it is the same) and the time expressed in terms of path and speed, in a table:

	Path S	speed v, km/h	time t, hours
Against the stream	30
With the flow	30

Let's calculate how much time the kayak spent on its trip:

Did she swim all hours? Rereading the task. No, not all. She had a rest of an hour of minutes, respectively, from the hours we subtract the rest time, which we have already translated into hours:

H - the kayak really swam.

In motion problems, three interrelated quantities are considered:

S - distance (traveled path),

t - travel time and

V - speed - the distance traveled per unit of time.

• Distance is the product of speed and time of travel;

S = V t

• Velocity is the distance a body travels per unit of time;
• Velocity is the quotient of distance divided by travel time;

V=S/t

• Time is the quotient of distance divided by speed.

t=S/V

What are the possible situations?

Situation one

Two objects start moving towards each other at the same time.

Oncoming traffic.

Situation two

Two objects move simultaneously in opposite directions.

Movement in opposite directions from the same point

Situation three

Two objects move simultaneously in the same direction.

When solving these problems, it is necessary to use the concepts of “rapprochement speed” and “removal speed”.

TASK 1

AT this moment the distance between two taxi drivers is 345 km. At what distance will the taxi drivers be in two hours if the speed of one is 72 km/h and the other is 68 km/h, and they drive towards each other at the same time?

First way to solve
1) 72 + 68 \u003d 140 (km / h) - approach speed taxi drivers.
2) 140 * 2 \u003d 280 (km) - this is the distance taxi drivers will approach each other in 2 hours.
3) 345 - 280 = 145 (km) - taxi drivers will be at this distance in 2 hours.
Answer: 145 km.

The second way to solve

1) 72 * 2 \u003d 144 (km) - this distance will be covered by one taxi driver in 2 hours.

2) 68 * 2 \u003d 136 (km) - this distance will be covered by another taxi driver in 2 hours.

3) 144+ 136 \u003d 280 (km) - this is the distance taxi drivers will approach each other in 2 hours.

4) 345 - 280 = 145 (km) - taxi drivers will be at this distance in 2 hours.

Answer: 145 km.

TASK 2

The distance between cities A and B is 720 km. A fast train leaves A for B at a speed of 80 km/h. After 2 hours, I went out to meet him from B to A

a passenger train with a speed of 60 km/h. In how many hours after the departure of the passenger train will these trains meet?

Solution

1) 80 * 2 = 160 (km) - a fast train passed in 2 hours.

2) 720-160=560(km) - trains have to pass.

3) 80 + 60 = 140 (km / h) - the speed of approach of 2 trains.

4) 560:140=4(h) - there was a passenger train on the way.

Answer: 4 hours.

TASK 3

Two buses departed from two points towards each other at the same time. The speed of one bus is 45 km/h and the speed of the other bus is 72 km/h. The first bus before the meeting drove 135 km.
Find the distance between points.

Solution

First way to solve
2) 72 * 3 = 216 (km) - the second bus passed before the meeting.
3) 135 + 216 = 351 (km) – distance between points.
Answer: 351 km.

The second way to solve

1) 135: 45 = 3 (hours) - the buses were driving to the meeting.

2) 45 +72 = 117 (km/h). - approach speed buses.

3) 117 * 3 = 351 (km) – distance between points.

Answer: 351 km.

What is convergence speed?

TASK 4

The car and the bus left two cities located at a distance of 740 km towards each other with speeds of 70 km/h and 50 km/h. How far apart will the cars be after 5 hours?

Solution

First way to solve

1) 50 * 5 = 250 (km) - a car will pass before the meeting.

2) 70 * 5 = 350 (km) - the bus will pass before the meeting.

3) 250 + 350 = 600 (km) - at this distance they will approach each other.

4) 740 -600 = 140 (km) - this distance will be between them in 5 hours.

Answer: 140 km.

The second way to solve

1) 50 + 70 = 120 (km/h) – approach speed buses and cars.

2) 120 * 5 = 600 (km) - this is the distance they will approach each other.

3) 740 - 600 = 140 (km) - this distance will be between them in 5 hours.

Answer: 140 km.

TASK 5

Two racing cars drove towards each other. The distance between them was 660 km. One was traveling at a speed of 100 km/h and the other 120 km/h. How soon will they meet?

Solution

1) 100+120=220(km/h)- approach speed machines.

2) 660:220=3(h) - after this time the racing cars will meet.

Answer: after 3 hours.

TASK 6

Two tigers ran out of the same den at the same time in opposite directions. The speed of one tiger is 48 km / h, and the other is 54 km / h. What will be the distance between the tigers after 3 hours?

Solution

First way to solve

1) 48 * 2 = 96 (km) - one tiger will run in 2 hours.

2) 54 * 2 = 108 (km) - another tiger will run in 2 hours.

3) 96 + 108 = 204 (km) - will be between the tigers in 2 hours.

Answer: 204 km.

The second way to solve

1) 48 + 54 \u003d 102 (km / h) - the speed of removal of tigers.

2) 102 * 2 = 204 (km) - will be between the tigers in 2 hours.

Answer: 204 km.

TASK 7

Maxim and Sasha left the school at a speed of 50 m/min. Roma followed them after 6 minutes at a speed of 80 m/min. In how many minutes will Roma catch up with Maxim and Sasha?

Solution

1) 80 - 50 = 30 (km/h) – approach speed boys.

2) 50 * 6 = 300 (km) - this was the distance between the boys before leaving Roma's school.

3) 300 : 30 = 10 (min.) - after such a time, Roma will catch up with his friends.

Answer: after 10 minutes.

RESULTS

1) When solving problems on the motion of two objects, the concepts « approach speed" and " removal rate ».

2) When solving problems on oncoming traffic and moving in opposite directions approach speed and removal rate are addition the speed of moving objects.

3) When solving problems for movement in one direction approach speed and removal rate are found by subtracting the speeds of moving objects.

Lesson type: lesson on introducing new knowledge

Goals:

• learn to read and write information presented in the form of various mathematical models;
• consolidate the ability to solve problems for movement based on formulas;
• improve skills oral account develop auditory and visual attention, memory, logical thinking, mathematical speech;
  consolidate knowledge of the relationship between the studied units of measurement;
• get acquainted with the new concept of "rapprochement speed";
• continue to learn to check correct and evaluate the results of their work.

Equipment: presentation, cards.

DURING THE CLASSES

1. Psychological attitude students

2. Actualization of students' knowledge

3. Working with cards

All students have cards.

- Let's start with a warm-up:

SPEED		DISTANCE

• A cyclist was moving at a speed of 100 m/min, how far did he cover in 3 minutes?
• In 20 minutes on a skateboard, the boy overcame 800 meters. How fast was he moving?
• Tourists on a hike move at a speed of 5 km / h, how long will it take them to overcome 25 km?
• Write a problem for a classmate.

Self-compilation of the conditions of the problem, a classmate reports the answer. Examination

- Guys, what do you think is the topic of our today's lesson? (Problem solving)

What quantities will we work with in today's lesson? (speed, time, distance)
What is our goal at this stage of the lesson? (To consolidate problem solving skills, remember the relationship of quantities)

- Let's look at our ladder of success, and everyone will determine for themselves what step you are on in mastering this topic. ( Attachment 1 ) Draw your little man on the corresponding step.

4. Group work

Each group receives an A3 sheet and a task (musical intro)(Attachment 1 )

a) - Tell us about these values ​​​​according to the plan

1. Definition
2. Formula
3. Units of measurement

(One representative comes to the board)

b) Make up a problem on the picture

– Listen to the condition of the problem: two ships set off at the same time to meet each other. The speed of one is 70 km/h, the speed of the other is 80 km/h. After 10 hours they met. What is the distance between ports?
What does "at the same time" mean?
Let's simulate the problem. (At the board is a visual display)
- How many km per hour did the first ship approach the meeting point? Second?

Children solve the problem, the student at the blackboard. We check the solution.

70*10 = 700 km distance traveled by 1 vessel;
80 * 10 = 800 km distance traveled by 1 vessel;
700 + 800 = 1500 km distance between two ports.

- If someone decided the second way, we invite you to the board, please explain.

What did you learn in the first act?

If no one decides:

Let's get back to our modeling. By how many kilometers per hour did the distance between the ships decrease? (The ships left at the same time, which means that every hour the distance between them decreases by the sum of two speeds.)

This sum of speeds is called approach speed

70 + 80 =150 (km/h)

– Knowing the speed and time, what can we find? (Distance)

What are our goals for the next stage of the lesson? (To get acquainted with a new concept, using a new concept, derive a formula. Understand that with the joint, simultaneous movement of two objects towards each other, for each unit of time, the distance is reduced by the sum of the speeds of moving objects)

- Let's try to deduce the convergence rate formulas. Let's remember what letters indicate the speed, how the approach occurs.

in) Multi-level task

- Make up problems according to the schemes, choose and solve.

Children solve problems.

- Who chose 1 task? Why How did you solve 1 problem, did you use new knowledge?
- Who chose the 2nd task?

Checking the solution (combining groups)

6. Summing up

- Take a card, complete the tasks.

Children work on cards.

- Put all the cards on which you worked today in a notebook, hand in notebooks at a break.

– Back to our ladder of success, have you advanced on the steps of knowledge?
- How did you draw the second man? Why?
- What did you learn in the lesson?
– Who needs to practice solving problems of a new type?

7. Homework: page 91 №5 read the tasks, choose for homework, the one that is more interesting to you.

Lesson grades.

TASKS FOR ONCOMING TRAFFIC

Most simple tasks on oncoming traffic begin to decide already in the 4th grade. The solution of such problems is usually carried out in 2 - 3 steps. In all tasks for oncoming traffic, such a concept is used as approach speed, i.e. the total speed of two bodies with which they move towards each other. The approach speed is a key value in solving problems for oncoming traffic.

The main formula for solving problems for oncoming traffic is the same formula, where the distance is expressed in terms of speed and time:

S = v t

A feature of the application of this formula is that the speed of approach of two bodies is taken as the speed, i.e. the sum of their speeds. This is the speed of oncoming traffic, which we talked about. Thus, the formula for solving problems for oncoming traffic can be written as follows:

S = v (approach) t

v (approach) = v 1 + v 2

where v 1 is the speed of the 1st body, v 2 is the speed of the 2nd body.

Examples of tasks for oncoming traffic:

1) From two piers, the distance between which is 90 km, two motor ships simultaneously left towards each other. The first ship was moving at a speed of 20 km/h, the second at a speed of 25 km/h. How many hours later did they meet?

2) Two swallows fly at a speed of 23 m/s. In how many seconds will they meet if the distance between them is 920 m?

3) Two trains left two cities at the same time towards each other. One train was traveling at a speed of 63 km/h. What was the speed of the second train if the distance between cities is 564 km? The trains met after 4 hours.

4) From two berths, the distance between which is 90 km, two boats simultaneously left towards each other. The first went at a speed of 8 km/h, the second - at a speed of 10 km/h. How many hours later did the boats meet?

5) A cyclist and a motorcyclist left the village and the city at the same time towards each other. The cyclist was traveling at a speed of 16 km/h and the motorcyclist was traveling at a speed of 54 km/h. The cyclist traveled 48 km before meeting. How far did the motorcyclist travel before meeting?

6) Two boys simultaneously ran towards each other along a sports track, the length of which is 200 m. They met after 20 s. The first one ran at a speed of 5 m/s. How fast was the second boy running?

7) Two stations left at the same time freight trains and met 5 hours later. One train traveled 29 km per hour, and the other 35 km. What is the distance between these stations?

8) 2 buses left two cities at the same time towards each other. The speed of the first bus is 25 km/h, the speed of the second is 50 km/h. The first bus passed 100 km before the meeting. How many kilometers did the second bus travel before the meeting?

9) The distance between the two cities is 81 km. Two cyclists left at the same time towards each other. One cyclist travels 3 km more per hour than another. At what distance from the cities did they meet if the meeting took place 3 hours after the departure?

10) Two riders left at the same time towards each other from two points, the distance between which is 100 km. The riders met after 4 hours. Find the speed of the first rider if the speed of the second is 13 km/h.

11) A boat and a boat departed from two piers at the same time towards each other. Before the meeting, the boat traveled 48 km, and the boat - 24 km. Boat speed - 8 km / h. Find the speed of the boat.

12) Two boats simultaneously departed from two piers towards each other, which met after 3 hours. The speed of one boat is 15 km / h, the speed of the second boat is 18 km / h. Find the distance between the piers.

13) Two motorcyclists left two cities at the same time towards each other. One motorcyclist was moving at a speed of 80 km/h. He traveled 320 km before meeting. How far did the second motorcyclist travel before the meeting if he was moving at a speed of 65 km/h?

14) A boat and a boat departed from two piers at the same time towards each other and met after 3 hours. The speed of the boat is 15 km / h, the speed of the boat is 4 times more. Find the distance between the piers.

15) Two planes simultaneously took off from two airfields towards each other and met after 3 hours. The speed of one plane is 600 km per hour, and the second plane is 900 km / h. Find the distance between airfields.

16) From two cities, the distance between which is 840 km, 2 trains left at the same time towards each other. The speed of the first train is 100 km/h, the second - 10 km/h more. In how many hours will the trains meet?

17) A boat and a boat departed from two piers at the same time towards each other. They met after 5 hours. The speed of the boat is 12 km/h, and the speed of the boat is 5 times greater. Find the distance between the piers.

18) A steamboat sailed from one pier at 11 o'clock in the morning, passing at 15 km / h, and from another pier towards it at 3 o'clock the next morning another steamer departed, passing at 17 km / h. In how many hours after the departure of the second steamer will they meet if there are 380 km between the piers?

19) Two tourists, the distance between which is 140 km, left towards each other one after the other after 3 hours. How many hours after the departure of the first one will they meet if the first traveled 10 km/h and the second 12 km/h?

20) A motor ship and a boat left the two piers towards each other at the same time. The ship was moving at a speed of 33 km / h, and the boat - 25 km / h. After 3 hours they met. What is the distance between the piers?

21) From two villages at the same time, a girl came out towards each other, who moved at a speed of 3 km / h, and a boy, who moved 2 times faster than the girl. The meeting took place 4 hours later. What is the distance between villages?

22) Two trains go towards each other from two stations, the distance between which is 385 km. The first one left earlier by 2 hours and moves at a speed of 53 km/h. 3 hours after the second train left, they met. What is the speed of the second train?

23) From two cities, the distance between which is 484 km, two trains left at the same time towards each other. The speed of one train is 45 km/h. Determine the speed of the other train if the trains meet after 4 hours.

24) Passenger and freight trains set off from two cities at the same time towards each other. They met 12 hours later. What is the distance between cities if it is known that the speed of a passenger train is 75 km/h, and that of a freight train is 35 km/h?

25) Two trains left two cities at the same time towards each other. One was walking at a speed of 42 km / h, and the other - 52 km / h. After 6 hours the trains met. Find the distance between cities.

26) The distance along the river between the two cities is 275 km. A steamboat and a barge left these cities at the same time towards each other. The ship was moving at a speed of 28 km/h. Find the speed of the barge if it is known that it met the steamer 5 hours after leaving.

27) From two cities, the distance between which is 1380 km, two trains left at the same time towards each other and met after 10 hours. The speed of one of them is 75 km/h. Find the speed of the other train.

28) The distance between the villages is 48 km. After how many hours will two pedestrians meet, who went out at the same time towards each other, if the speed of one is 3 km/h and the other is 5 km/h?

29) From the village to the city 340 km. A motorcyclist traveled from a village to a city at a speed of 42 km/h. After 2 hours, a cyclist rode towards him at a speed of 22 km / h. In how many hours will they meet?

30) Two motorcyclists left two cities at the same time towards each other and met after 10 minutes. The speed of one of them is 920 m/min, and the other one is 970 m/min. Find the distance between cities.

31) Two trains left one city to another at the same time towards each other and met after 9 hours. The speed of one train is 48 km/h, and the speed of the other is 5 km/h more than the other. Find the distance between cities.

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First, let's recall the formulas that are used to solve such problems: S = υ t, υ = S: t, t = S: υ
where S is the distance, υ is the speed of movement, t is the time of movement.

When two objects move uniformly at different speeds, the distance between them either increases or decreases for each unit of time.

Approach speed is the distance that objects approach each other per unit of time.
Removal speed is the distance that objects are removed per unit of time.

Approach movement oncoming traffic and pursuit. move to remove can be divided into two types: movement in opposite directions and lagging behind.

The difficulty for some students is to correctly put "+" or "-" between speeds when finding the speed of approach of objects or the speed of receding.

Consider a table.

It can be seen from it that when objects move in opposite directions them speeds add up. When moving in one direction - subtracted.

Examples of problem solving.

Task number 1. Two cars are moving towards each other with speeds of 60 km/h and 80 km/h. Determine the speed at which the cars are approaching.
υ 1 = 60 km/h
υ 2 = 80 km/h
Find υ sat
Solution.
υ sat \u003d υ 1 + υ 2- closing speed in different directions )
υ sat \u003d 60 + 80 \u003d 140 (km / h)
Answer: the approach speed is 140 km/h.

Task number 2. Two cars left the same point in opposite directions at speeds of 60 km/h and 80 km/h. Determine the rate at which machines are removed.
υ 1 = 60 km/h
υ 2 = 80 km/h
Find υ beats
Solution.
υ beats = υ 1 + υ 2- removal rate (the “+” sign, since it is clear from the condition that the cars are moving in different directions)
υ beats = 80 + 60 = 140 (km/h)
Answer: the removal speed is 140 km/h.

Task number 3. From one point in one direction, first a car left at a speed of 60 km/h, and then a motorcycle at a speed of 80 km/h. Determine the speed at which the cars are approaching.
(We see that here is the case of movement in pursuit, so we find the speed of approach)
υ av = 60 km/h
υ mot = 80 km/h
Find υ sat
Solution.
υ sat \u003d υ 1 - υ 2- closing speed (the “–” sign, since it is clear from the condition that the cars are moving in one direction)
υ sat \u003d 80 - 60 \u003d 20 (km / h)
Answer: The approach speed is 20 km/h.

That is, the name of the speed - approach or removal - does not affect the sign between the speeds. Only direction matters.

Let's consider other tasks.

Task number 4. Two pedestrians left the same point in opposite directions. The speed of one of them is 5 km / h, the other - 4 km / h. How far apart will they be after 3 hours?
υ 1 = 5 km/h
υ 2 = 4 km/h
t = 3 h
Find S
Solution.
in different directions)
υ beats = 5 + 4 = 9 (km/h)

S = υ beat t
S = 9 3 = 27 (km)
Answer: after 3 hours the distance will be 27 km.

Task number 5. Two cyclists simultaneously started towards each other from two points, the distance between which is 36 km. The speed of the first is 10 km/h, the second is 8 km/h. In how many hours will they meet?
S = 36 km
υ 1 = 10 km/h
υ 2 = 8 km/h
Find t
Solution.
υ sat \u003d υ 1 + υ 2 - speed of approach (the “+” sign, since it is clear from the condition that the cars are moving in different directions)
υ sat = 10 + 8 = 18 (km/h)
(meeting time can be calculated using the formula)
t = S: υ Sat
t = 36: 18 = 2 (h)
Answer: See you in 2 hours.

Task number 6. Two trains left the same station in opposite directions. Their speeds are 60 km/h and 70 km/h. In how many hours will the distance between them be 260 km?
υ 1 = 60 km/h
υ 2 = 70 km/h
S = 260 km
Find t
Solution .
1 way
υ beats \u003d υ 1 + υ 2 - removal rate (sign “+” since it is clear from the condition that pedestrians are moving in different directions)
υ beats = 60 + 70 = 130 (km/h)
(The distance traveled is found by the formula)
S = υ beat t ⇒ t= S: υ beats
t = 260: 130 = 2 (h)
Answer: after 2 hours the distance between them will be 260 km.
2 way
Let's make an explanatory drawing:

It can be seen from the figure that
1) after a given time, the distance between the trains will be equal to the sum of the distances traveled by each of the trains:
S = S 1 + S 2;
2) each of the trains traveled the same time (from the condition of the problem), which means that
S 1 \u003d υ 1 t-distance traveled by 1 train
S 2 \u003d υ 2 t- distance traveled by train 2
Then,
S= S1 + S2= υ 1 t + υ 2 t = t (υ 1 + υ 2)= t υ beats
t = S: (υ 1 + υ 2)- the time for which both trains will travel 260 km
t \u003d 260: (70 + 60) \u003d 2 (h)
Answer: The distance between trains will be 260 km in 2 hours.

1. Two pedestrians simultaneously came out towards each other from two points, the distance between which is 18 km. The speed of one of them is 5 km / h, the other - 4 km / h. In how many hours will they meet? (2 h)
2. Two trains left the same station in opposite directions. Their speeds are 10 km/h and 20 km/h. In how many hours will the distance between them be 60 km? (2 h)
3. From two villages, the distance between which is 28 km, two pedestrians came out towards each other at the same time. The speed of the first is 4 km/h, the speed of the second is 5 km/h. How many kilometers per hour do pedestrians approach each other? How far apart will they be after 3 hours? (9 km, 27 km)
4. The distance between the two cities is 900 km. Two trains left these cities towards each other with speeds of 60 km/h and 80 km/h. How far apart were the trains 1 hour before the meeting? Is there an extra condition in the task? (140 km, yes)
5. A cyclist and a motorcyclist left the same point in the same direction at the same time. The speed of a motorcyclist is 40 km/h and that of a cyclist is 12 km/h. What is the speed of their removal from each other? In how many hours will the distance between them be 56 km? (28 km/h, 2 h)
6. From two points 30 km apart, two motorcyclists left at the same time in the same direction. The speed of the first is 40 km/h, the second is 50 km/h. In how many hours will the second overtake the first?
7. The distance between cities A and B is 720 km. A fast train leaves A for B at a speed of 80 km/h. After 2 hours, a passenger train left B to A towards him at a speed of 60 km/h. In how many hours will they meet?
8. A pedestrian left the village at a speed of 4 km/h. After 3 hours, a cyclist followed him at a speed of 10 km / h. How many hours does it take the cyclist to overtake the pedestrian?
9. The distance from the city to the village is 45 km. A pedestrian leaves the village for the city at a speed of 5 km/h. An hour later, a cyclist rode towards him from the city to the village at a speed of 15 km/h. Which of them will be closer to the village at the time of the meeting?
10. Old task. A young man went from Moscow to Vologda. He walked 40 miles a day. A day later, another young man was sent after him, passing 45 versts a day. In how many days will the second overtake the first?
11. Old problem. The dog saw a hare in 150 fathoms, which runs 500 fathoms in 2 minutes, and the dog in 5 minutes - 1300 fathoms. The question is, at what time will the dog overtake the hare?
12. Old problem. Two trains left Moscow for Tver at the same time. The first passed at 39 versts and arrived in Tver two hours earlier than the second, which passed at 26 versts. How many miles from Moscow to Tver?