Topic: Speed of approach and speed of removal.
Target: introduce new concepts of “approach speed and removal speed”, develop the ability to solve motion problems.
Org moment.
Open the notebooks Number. Classwork.
On the tables there is a green blue pen, a simple pencil, a ruler, a felt-tip pen
Write down the formula and solution.
Find the formula used to solve it.
The cyclist moved at a speed of 100 m/min, how much distance did he cover in 3 minutes?
Write down the formula and solution.
In 20 minutes the boy covered 800 meters on a skateboard. How fast was he moving?
Write down the formula and solution.
Find the formula used to solve it.
Tourists on a hike move at a speed of 5 km/hour. How long will it take them to cover 25 km?
Formulation of the problem.
– Listen to the problem: two ships set off simultaneously to meet each other. The speed of one is 70 km/h, the speed of the other is 80 km/h. 10 hours later they met. What is the distance between ports?
– What does “simultaneously” mean?
- Let's simulate the problem.(There is a visual display at the board)
– How many kilometers did the first ship approach the meeting place in an hour? Second?
Children solve a problem, student at the board. We are checking the solution.
70 * 10 = 700 km distance traveled by 1 ship;
80 * 10 = 800 km distance covered by 1 ship;
700 + 800 = 1500 km distance between two ports.
– There is a second way to solve this problem.
The topic of our lesson today is APPROACH SPEED AND REMOVAL SPEED.
Let's formulate the objectives of the lesson
– What goal will we set for the next stage of the lesson?(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 velocities of the moving objects)
– Let's try to derive formulas for the speed of approach. Let us remember what letters indicate speed and how approach occurs.
– Compare 2 drawings. What did you notice? What is the difference? Are the speed types the same?
– What do you think, in which drawing will we talk about the speed of approach, and where – about the speed of removal?
Explanation of the concepts of “approach speed” and “removal speed”.
Go to Slide 4 “1) Oncoming traffic.”
– Look at the screen.
– What can you say about the movement of Malvina and Buratino?
-What movement is this?
– At what point were Malvina and Buratino after 1 minute, after 2 minutes, after 3 minutes? Let's fill out the table.
– How much does the distance between them decrease every minute?
– At what point and after how many minutes did the meeting take place?
- Let's draw a conclusion.
Go to Slide 5 “2) Movement in opposite directions.”
– Look at the screen.
– What can you say about the movement of Signor Tomato and Cipollino?
-What movement is this? Let's fill out the table.
– From what points did their movement begin? Let's fill out the table.
– At what point were Signor Tomato and Cipollino after 1 minute, after 2 minutes, after 3 minutes? Let's fill out the table.
– What happens to the distance between objects?
– How much does the distance between them increase every minute?
– Will there be a meeting?
- Let's draw a conclusion.
Take some leaves. Write me the formula for the speed of approach and the formula for the speed of removal
Check on slide
– Consider the problem diagrams, determine what speed of movement we are talking about (approaching or moving away), connect with a suitable expression and calculate it.
Solution to the problem next slide
Lesson summary.
– Our lesson has come to an end. What did you learn about in class today? What is important to know to determine the speed of approaching or moving away? What did you especially like or remember?
Students check the assignment using Slides 12–13.e
Let's consider problems in which we are talking about movement in one direction. In such problems, two objects move in the same direction at different speeds, moving away from each other or approaching each other.
Speed of approach problems
The speed at which objects move closer to each other is called approach speed.
To find the speed of approach of two objects that are moving in the same direction, you need to subtract the smaller from the larger speed.
Task 1. A car left the city at a speed of 40 km/h. After 4 hours, a second car drove out after him at a speed of 60 km/h. How many hours will it take for the second car to catch up with the first?
Solution: Since at the time the second car left the city, the first one had already been on the road for 4 hours, during this time it managed to move away from the city by:
40 4 = 160 (km)
The second car is moving faster than the first, which means that every hour the distance between the cars will decrease by the difference in their speeds:
60 - 40 = 20 (km/h) is closing speed cars
By dividing the distance between the cars by the speed of their approach, you can find out how many hours later they will meet:
160: 20 = 8 (h)
1) 40 · 4 = 160 (km) - distance between cars
2) 60 - 40 = 20 (km/h) - approaching speed of cars
3) 160: 20 = 8 (h)
Answer: The second car will catch up with the first in 8 hours.
Task 2. From two villages that are 5 km apart, two pedestrians left at the same time in the same direction. The speed of the pedestrian walking in front is 4 km/h, and the speed of the pedestrian walking behind is 5 km/h. How many hours after leaving will the second pedestrian catch up with the first?
Solution: Since the second pedestrian moves faster than the first, the distance between them will decrease every hour. This means you can determine the speed of approach of pedestrians:
5 - 4 = 1 (km/h)
Both pedestrians left at the same time, which means the distance between them is equal to the distance between the villages (5 km). By dividing the distance between pedestrians by the speed of their approach, we find out how long it will take for the second pedestrian to catch up with the first:
The solution to the problem by actions can be written as follows:
1) 5 - 4 = 1 (km/h) - this is the speed of approaching pedestrians
2) 5: 1 = 5 (h)
Answer: After 5 hours, the second pedestrian will catch up with the first.
Removal speed task
The speed at which objects move away from each other is called removal rate.
To find the speed of removal of two objects that are moving in the same direction, you need to subtract the smaller from the larger speed.
Task 2. Two cars left at the same time from the same point in the same direction. The speed of the first car is 80 km/h, and the speed of the second is 40 km/h.
1) What is the speed of removal between cars?
2) What will be the distance between the cars after 3 hours?
3) After how many hours will the distance between them be 200 km?
Solution: First, we find out the speed at which the cars are moving away from each other; to do this, we subtract the smaller one from the higher speed:
80 - 40 = 40 (km/h)
Every hour, cars move 40 km away from each other. Now you can find out how many kilometers there will be between them in 3 hours; to do this, multiply the removal rate by 3:
40 3 = 120 (km)
To find out how many hours later the distance between cars will be 200 km, you need to divide the distance by the speed of removal:
200: 40 = 5 (h)
Answer:
1) The speed of removal between cars is 40 km/h.
2) After 3 hours there will be 120 km between the cars.
3) After 5 hours there will be a distance of 200 km between the cars.
Back forward
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Lesson objectives:
Didactic:
- introduce the concepts of “approach speed” and “removal speed” and the ability to check the correctness of calculations;
- consolidate the ability to read and build movement patterns;
- develop and consolidate the ability to solve motion problems, the ability to compose inverse problems;
- consolidate computational skills in addition, subtraction, multiplication and division of numbers, as well as skills in computational operations with fractions;
Educational:
- development of creative abilities, memory, ability to think logically;
- development of mathematical literate speech;
Educational: nurturing interest in mathematics;
Equipment: Textbook L.G. Peterson “Mathematics 4th grade, part 2”, test cards. Computer, projector, interactive whiteboard. Illustrative material (presentation in MS PowerPoint format)<Презентация.ppt>.
During the classes
Organizing time.
- Hello, guys, sit down! Check if you have everything ready for the lesson.
- Let's remember the landing rules.
– Write down the number.
Purpose of the lesson (Setting a learning task).
– Please remember how many objects can simultaneously move along a number beam? Where can objects begin to move? In what directions can objects move? How fast can objects move?
– Today we will find out what “approach speed” and “removal speed” are, what you need to know to determine what speed it is, how to find the speed of approach or removal.
– Let’s write down the topic of the lesson “Speed of approach and speed of removal.”
Mathematical dictation.
- The minuend is 130, the subtrahend is 111. Find the difference.
- Dividend 480, divisor 40. Find the quotient.
- How much more is 200 than 184?
- What is 2/3 of 27?
- How many times is 320 greater than 20?
- What number was tripled to get 57?
- Divide the sum of 95 and 105 by 10.
- 2/5 of the number is 12. Find the whole number.
Individual tasks.
Performed on the board by 2 students during a mathematical dictation.
Exercise 1.
| S | V | t | Formula | |
| I | ? km | 45 km/h | 7 hours | |
| II | 180 m | ? m/min | 5 minutes | |
| III | 960 m | 16 m/s | ? With | |
| IV | ? km | 60 km/h | 60 min |
Task 2.
Draw the movement of points on a coordinate ray and write down the formula for the movement of points:
- The movement of point A begins from the point with coordinate (6) in the right direction at a speed of 3 unit segments per hour. The movement of point B begins from the point with coordinate (14) in the left direction at a speed of 1 unit segment per hour. What are the coordinates of these points after 1 hour, 2 hours?
- The movement of point A begins from the point with coordinate (6) in the left direction at a speed of 3 unit segments per hour. The movement of point B begins from the point with coordinate (14) in the right direction at a speed of 1 unit segment per hour. What are the coordinates of these points after 1 hour, 2 hours?
Checking mathematical dictation and individual assignments.
Checking the mathematical dictation.
– A word is encrypted in the answers to the mathematical dictation. To decipher it, the alphabet of the Russian language will help us.
– Each answer corresponds to the serial number of a letter in the alphabet. Write the letters on a line.
Go to Slide 2 “Mathematical dictation”.
– What did you do? Let's check.
For each click on Slide 2, one column of the table is filled.
– Whoever gets the word “speed” gives himself a 5.
– What 2 groups can the numbers of the mathematical dictation be divided into?
- to even/odd
- round/non-round;
– What is “movement speed”?
Checking task 1.
| S | V | t | Formula | |
| I | 315 km | 45 km/h | 7 hours | S=V*t |
| II | 180 m | 36 m/min | 5 minutes | V=S:t |
| III | 960 m | 16 m/s | 6 s | t=S:V |
| IV | 60 km | 60 km/h | 60 min | S=V*t |
– How to find the distance, knowing the speed and time of the object?
– How to find the speed, knowing the distance and time of the object?
– How to find time, knowing the distance and speed of an object?
Checking task 2.
– Compare 2 drawings. What did you notice? What is the difference? Are the speed types the same?
– What do you think, in which drawing will we talk about the speed of approach, and where – about the speed of removal?
Exercise for the eyes.
Explanation of the concepts of “approach speed” and “removal speed”.
Working with exercise 1 of lesson 24 (Slides 3–6). As the explanation progresses, students are asked questions about what they see on the screen and after their answers, the student fills out the table on the board, the rest in the textbooks, then the teacher moves on to the next step of the animation.
Go to Slide 3 “1) Oncoming traffic.”
– Look at the screen.
– What can you say about the movement of Malvina and Buratino?
-What movement is this?
– At what point were Malvina and Buratino after 1 minute, after 2 minutes, after 3 minutes? Let's fill out the table.
- Let's draw a conclusion.
Go to Slide 4 “2) Movement in opposite directions.”
– Look at the screen.
– What can you say about the movement of Signor Tomato and Cipollino?
-What movement is this? Let's fill out the table.
– From what points did their movement begin? Let's fill out the table.
– At what point were Signor Tomato and Cipollino after 1 minute, after 2 minutes, after 3 minutes? Let's fill out the table.
– What happens to the distance between objects?
– Will the meeting take place?
- Let's draw a conclusion.
Go to Slide 5 “3) Moving in pursuit.”
– Look at the screen.
– What can you say about the movement of Crocodile Gena and Cheburashka?
-What movement is this?
– From what points did their movement begin? Let's fill out the table.
– At what point were Crocodile Gena and Cheburashka after 1 minute, after 2 minutes, after 3 minutes? Let's fill out the table.
– How much does the distance between them decrease every minute?
– At what point and after how many minutes did the meeting take place?
- Let's draw a conclusion.
Go to Slide 6 “4) Movement with a lag.”
– Look at the screen
– What can you say about the movement of Donut and Dunno?
-What movement is this?
– From what points did their movement begin?
- At what point were Donut and Dunno after 1 minute, after 2 minutes, after 3 minutes? Let's fill out the table.
– What happens to the distance between objects? Why?
– How much does the distance between them increase every minute?
– Will the meeting take place?
- Let's draw a conclusion.
– What is “closing speed”? ( This is the distance by which objects approach each other per unit time.)
– What is “removal speed”? ( This is the distance that objects move away per unit time.)
Drawing up a reference diagram.
Go to Slide 7 “Basic diagram”.
– We will draw up reference diagrams for all types of movement.
Physical education minute.
We came to the forest meadow,
Lifting your legs higher
Through bushes and hummocks,
Through branches and stumps.
Who walked so high -
Didn't trip, didn't fall.
Solving problems with comments.
To consolidate knowledge, students understand and solve problems for all types of movement.
– Let’s solve several problems and determine what speed we are talking about: approaching or moving away? What is it equal to? And the heroes of the fairy tale “The Golden Key” will help us with this.
Working with Slides 8–11. Students determine from the Slide which reference diagram the problem belongs to and suggest a way to solve it.
Working with the class:
- Go to Slide 8 “Task of moving in opposite directions.”
Individually:
- The robbers are chasing Pinocchio, who is running away from them at a speed of 19 units/min. How does the distance between Pinocchio and the robbers change if they run at a speed of 23 units/min.
- Compose an inverse problem to the 1st problem.
- Change the condition of the 2nd problem so that it is solved “-”.
- Change the condition of the 4th problem so that it is solved “+”.
Independent problem solving (test).
To test knowledge and skills on this topic, students received test cards with the task “Establish a correspondence between the problem diagram and its solution (options 1 and 2).”
– Consider the diagrams of the problems, determine what speed of movement we are talking about (approaching or moving away), connect with a suitable expression and calculate it.
Mutual verification of problem solutions.
Students check the assignment using Slides 12–13.
Lesson summary.
– Our lesson has come to an end. What did you learn about in class today? What is important to know to determine the speed of approaching or moving away? What did you especially like or remember?
Homework.
Examples, task
Giving marks and encouraging students.
Throughout the lesson, students' work and answers were assessed verbally and with reward medals.
List of used sources and literature.
- Textbook L.G. Peterson“Mathematics 4th grade, part 2.”
- Pictures from the personal website of Nikolai Kozlov http://nkozlov.ru/library/s318/d3458/
How to find closing speed*? and got the best answer
Answer from Star Lord[newbie]
If objects move in the same direction, then subtract.
If towards each other or in different directions, then fold them.
Answer from Irisha ***[newbie]
+
Answer from shpg okay[newbie]
-
Answer from Egor Bagrov[active]
X+Z=Y (X-speed, Z-speed2,Y-response)
Answer from Huck Finn[guru]
Theory:
All problems related to movement are solved using one formula. Here it is: S=Vt. S is the distance, V is the speed of movement, and t is the time. This formula is the key to solving all these problems, and everything else is written in the text of the problem; the main thing is to read and understand the problem carefully. The second important point is the reduction of all data in the problem of quantities to common units of measurement. That is, if time is given in hours, then the distance should be measured in kilometers, if in seconds, then the distance in meters, respectively.
Problem solving:
So, let's look at three main examples of solving motion problems.
Two objects left one after another.
Let's assume that you are given the following task: the first car left the city at a speed of 60 km/h, half an hour later the second car left at a speed of 90 km/h. After how many kilometers will the second car catch up with the first? To solve such a problem, we have a formula: t = S /(v1 - v2). Since we know the time, but not the distance, we transform it S = t(v1 - v2). Substitute the numbers: S = 0.5 (30 min.) (90-60), S=15 km. That is, both cars will meet after 15 km.
Two objects left in the opposite direction.
If you are given a problem in which two objects set off towards each other, and you need to find out when they will meet, then you need to apply the following formula: t = S /(v1 + v2). For example, from points A and B, between which there are 43 km , a car was traveling at a speed of 80 km/h, and a bus was traveling from point B to A at a speed of 60 km/h. How long will it take for them to meet? Solution: 43/(80+60)=0.30 hours.
Two objects left at the same time in the same direction.
Given a task: a pedestrian moved from point A to point B, moving at a speed of 5 km/h, and a cyclist also left at a speed of 15 km/h. How many times faster will a cyclist get from point A to point B if it is known that the distance between these points is 10 km? First you need to find the time it takes the pedestrian to cover this distance. We rework the formula S=Vt, we get t =S/V. Substitute the numbers 10/5=2. that is, the pedestrian will spend 2 hours on the road. Now we calculate the time for the cyclist. t =S/V or 10/15=0.7 hours (42 minutes). The third action is very simple, we must find the difference in time between a pedestrian and a person on a bicycle. 2/0.7=2.8. The answer is: a cyclist will get to point B 2.8 times faster than a pedestrian, i.e. almost three times faster.
So, let's say our bodies are moving in the same direction. How many cases do you think there could be for such a condition? That's right, two.
Why does this happen? 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.
Fourth task
Kolya goes to work by car at a speed of km/h. Colleague Kolya Vova is driving at a speed of km/h. Kolya lives kilometers away from Vova.
How long will it take for Vova to catch up with 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 an hour or in minutes.
Let's compare our solutions...
The drawing looks like this:
Similar to yours? Well done!
Since the problem asks how long after the guys met, and they 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). When composing the equations, let's take time for.
So, Vova made his way to the meeting place. Kolya made his way to the meeting place. It's clear. Now let's look at the axis of movement.
Let's start with the path that Kolya took. Its path () is shown in the figure as a segment. 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 is equal to the path that Kolya took.
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 from this example you understand how important the role is played Well done drawing!
And we smoothly move on, or rather, we have already moved on to the next point of our algorithm - bringing all quantities to the same dimension.
The rule of three “Rs” - dimension, reasonableness, calculation.
Dimension.
Problems do not always give the same dimension for each participant in the movement (as was the case in our easy problems).
For example, you can find problems where it is said that bodies moved for a certain number of minutes, and their speed of movement is indicated in km/h.
We can’t just take and substitute the values into the formula - the answer will be incorrect. Even in terms of units of measurement, our answer “fails” the reasonableness test. Compare:
Do you see? When multiplying correctly, we also reduce the units of measurement, and, accordingly, we obtain a reasonable and correct result.
What happens if we don’t convert to one measurement system? The answer has a strange dimension and the result is % incorrect.
So, just in case, let me remind you of the meanings of the basic units 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 = hours = minutes = seconds
Advice: When converting units of measurement related to time (minutes into hours, hours into seconds, etc.), imagine a clock dial in your head. The naked eye can see that the minutes are a quarter of the dial, i.e. hours, minutes is a third of the dial, i.e. an hour, and a minute is an hour.
And now a very simple task:
Masha rode her 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 minutes from an hour (mentally imagined a clock dial, and said that minutes is a quarter of an hour), respectively - min = hours.
Reasonableness.
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 can’t be negative, right? So, rationality, that’s what it’s about)
Calculation.
See if your solution “passes” the dimensions and reasonableness, and only then check the calculations. It is logical - if there is an inconsistency with dimension and rationality, then it is easier to cross out everything and start looking for logical and mathematical errors.
“Love of tables” or “when drawing is not enough”
Movement problems are not always as simple as we solved before. Very often, in order to solve a problem correctly, you need not just draw a competent picture, but also make a table with all the conditions given to us.
First task
A cyclist and a motorcyclist left at the same time from point to point, the distance between them being kilometers. It is known that a motorcyclist travels more kilometers per hour than a cyclist.
Determine the speed of the cyclist if it is known that he arrived at the point minutes later than the motorcyclist.
This is the task. Pull yourself together and read it several times. Have you read it? Start drawing - a straight line, a point, a point, two arrows...
In general, draw, and now we’ll compare what you got.
It's a bit empty, isn't it? Let's draw a table.
As you remember, all movement tasks consist of the following components: speed, time and path. It is these columns that any table in such problems will consist of.
True, we will add one more column - Name, about whom we write information - a motorcyclist and a cyclist.
Also indicate in the header dimension, in which you will enter the values there. You remember how important this is, right?
Did you get a table like this?
Now let's analyze everything we have and at the same time enter the data into the table and figure.
The first thing we have is the path that the cyclist and motorcyclist took. It is the same and equal to km. Let's bring it in!
Let's take the speed of the cyclist as, then the speed of the motorcyclist will be...
If with such a variable the solution to the problem does not work, it’s okay, we’ll take another one until we reach the winning one. This happens, the main thing is not to be nervous!
The table has changed. We only have one column left unfilled - time. How to find time when there is a path and speed?
That's right, divide the distance by the speed. Enter this into the table.
Now our table is filled in, now we can enter the data into the drawing.
What can we reflect on it?
Well done. Speed of movement of motorcyclist and cyclist.
Let's re-read the problem again, look at the picture and the completed table.
What data is not reflected in the table or figure?
Right. The time the motorcyclist arrived before the cyclist. We know that the time difference is minutes.
What should we do next? That's right, convert the time given to us from minutes to hours, because the speed is given to us in km/h.
The magic of formulas: drawing up and solving equations - manipulations leading to the only correct answer.
So, as you may have guessed, now we will make up the equation.
Drawing up the equation:
Look at your table, at the last condition that is not included in it and think, the relationship between what and what can we put into the equation?
Right. We can create an equation based on the time difference!
Logical? The cyclist rode more; if we subtract the motorcyclist’s time from his time, we will get the difference given to us.
This equation is rational. If you don’t know what this is, read the topic “”.
We bring the terms to a common denominator:
Let's open the brackets and present similar terms: Phew! Got it? Try your hand at the following problem.
Solution of the equation:
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. Let's decide!
We received two possible answers. Let's see what we got for? That's right, the speed of the cyclist.
Let us remember the “3P” rule, more specifically “reasonableness”. Do you know what I mean? Exactly! Speed cannot be negative, so our answer is km/h.
Second task
Two cyclists set out on a -kilometer ride at the same time. The first one drove at a speed that was one 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 second to the finish line. Give your answer in km/h.
Let me remind you of the solution algorithm:
- Read the problem a couple of times and understand all the details. Got it?
- Start drawing a picture - in which direction are they moving? how far did they travel? Did you draw it?
- Check that all your quantities are of the same dimension and begin to briefly write out the conditions of the problem, making a table (do you remember what graphs are there?).
- While you are writing all this, think about what to take for? Have you chosen? Write it down in the table! Well, now it’s simple: we make up an equation and solve. Yes, and finally - remember the “3Rs”!
- I've done everything? Well done! I found out that the speed of the cyclist is km/h.
-"What color is your car?" - "She's beautiful!" Correct answers to the questions asked
Let's continue our conversation. So what is the speed of the first cyclist? km/h? I really hope you’re not nodding yes now!
Read the question carefully: “What is the speed of first cyclist?
Do you understand what I mean?
Exactly! Received is not always the answer to the question posed!
Read the questions carefully - perhaps after finding them you will need to perform some more manipulations, for example, add km/h, as in our task.
One more 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.
Watch the dimensions not only during the solution itself, but also when writing down the answers.
Circular movement problems
Bodies in problems can move not necessarily straight, but also in a circle, for example, cyclists can ride along a circular track. Let's look at this problem.
Task No. 1
A cyclist left a point on the circular route. Minutes later, he had not yet returned to the point and the motorcyclist left the point after him. Minutes after leaving, 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 route is km. Give your answer in km/h.
Solution to problem No. 1
Try to draw a picture for this problem and fill out a table for it. Here's what I got:
Between meetings, the cyclist traveled a distance, and the motorcyclist - .
But at the same time, the motorcyclist drove exactly one lap more, as can be seen from the figure:
I hope you understand that they didn't actually drive in a spiral - the spiral just schematically shows that they drive in a circle, passing the same points on the route several times.
Got it? Try to solve the following problems yourself:
Tasks for independent work:
- Two motorcycles start at the same time in one direction of the two dia-metral-but-pro-ti-on- false points of a circular route, the length of which is equal to km. After how many minutes do the cycles become equal for the first time, if the speed of one of them is km/h higher than the speed of the other? ho-ho?
- From one point on a circular highway, the length of which is equal to km, at one time there are two motorcyclists in the same direction. The speed of the first motorcycle is equal to km/h, and minutes after the start it was ahead of the second motorcycle by one lap. Find the speed of the second motorcycle. Give your answer in km/h.
Solutions to problems for independent work:
- Let km/h be the speed of the first motor cycle, then the speed of the second motor cycle is equal to km/h. Let the cycles be equal for the first time in a few hours. In order for the cycles to be equal, the faster one must overcome them from the beginning distance equal to the length of the route.
We get that the time is hours = minutes.
- Let the speed of the second motorcycle be equal to km/h. In an hour, the first motorcycle traveled more kilometers than the second, so we get the equation:
The speed of the second motorcyclist is km/h.
Current problems
Now that you are excellent at solving problems “on land,” let’s move into the water and look at the scary problems associated with the current.
Imagine that you have a raft and you lower it into the lake. What's happening to him? Right. It stands because a lake, a pond, a puddle, after all, is still water.
The current speed in the lake is .
The raft will only move if you start rowing yourself. The speed it acquires will be the raft's own speed. It doesn’t matter where you swim - left, right, the raft will move at the speed with which you row. It's clear? It's logical.
Now imagine that you are lowering a raft onto the river, you turn away to take the rope..., you turn around, and it... floats away...
This happens because the river has a current speed, which carries your raft in the direction of the current.
Its speed is zero (you are standing in shock on the shore and not rowing) - it moves at the speed of the current.
Got it?
Then answer this question: “At what speed will the raft float down the river if you sit and row?” Thinking about it?
There are two possible options here.
Option 1 - you go with the flow.
And then you swim at your own speed + the speed of the current. The flow seems to help you move forward.
2nd option - t 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 at which you move is equal to your own speed - the speed of the current.
Let's say you need to swim a kilometer. When will you cover this distance faster? When will you go with the flow or against it?
Let's solve the problem and check.
Let's add to our path data on the speed of the current - km/h and the raft's own speed - km/h. How much time will you spend moving with and against the current?
Of course, you coped with this task without difficulty! It takes an hour with the current, and an hour against the current!
This is the whole essence of the tasks at movement with the current.
Let's complicate the task a little.
Task No. 1
The boat with the motor took an hour to travel from point to point, and an hour to return.
Find the speed of the current if the speed of the boat in still water is km/h
Solution to problem No. 1
Let us denote the distance between points as, and the speed of the current as.
| Path S | Speed v, km/h |
Time t, hours |
|
| A -> B (upstream) | 3 | ||
| B -> A (downstream) | 2 |
We see that the boat takes the same path, respectively:
What did we charge for?
Current speed. Then this will be the answer :)
The speed of the current is km/h.
Task No. 2
The kayak left from point to point located km from. After staying at point for an hour, the kayak went back and returned to point c.
Determine (in km/h) the kayak's own speed if it is known that the speed of the river is km/h.
Solution to problem No. 2
So let's get started. Read the problem several times and make a drawing. I think you can easily solve this on your own.
Are all quantities expressed in the same form? No. Our rest time is indicated in both hours and minutes.
Let's convert this into hours:
hour minutes = h.
Now all quantities are expressed in one form. Let's start filling out the table and finding what we'll take for.
Let be the kayak's own speed. Then, the speed of the kayak downstream is equal and against the current is equal.
Let's write down this data, as well as the path (as you understand, it is the same) and time, expressed in terms of path and speed, in a table:
| Path S | Speed v, km/h |
Time t, hours |
|
| Against the stream | 26 | ||
| With the flow | 26 |
Let's calculate how much time the kayak spent on its journey:
Did she swim for all the hours? Let's reread the task.
No, not all. She had an hour of rest, so from hours we subtract the rest time, which we have already converted into hours:
h the kayak really floated.
Let's bring all the terms to a common denominator:
Let's open the brackets and present similar terms. Next, we solve the resulting quadratic equation.
I think you can handle this on your own too. What answer did you get? I have km/h.
Let's sum it up
ADVANCED LEVEL
Movement tasks. Examples
Let's consider examples with solutionsfor each type of task.
Moving with the flow
Some of the simplest tasks are river navigation problems. Their whole essence is as follows:
- if we move with the flow, the speed of the current is added to our speed;
- if we move against the current, the speed of the current is subtracted from our speed.
Example #1:
The boat sailed from point A to point B in hours and back again in hours. Find the speed of the current if the speed of the boat in still water is km/h.
Solution #1:
Let us denote the distance between points as AB, and the speed of the current as.
Let's put all the data from the condition into the table:
| Path S | Speed v, km/h |
Time t, hours | |
| A -> B (upstream) | AB | 50-x | 5 |
| B -> A (downstream) | AB | 50+x | 3 |
For each row of this table you need to write the formula:
In fact, you don't have to write equations for each row of the table. We see that the distance traveled by the boat back and forth is the same.
This means that we can equate the distance. To do this, we use immediately formula for distance:
Often you have to use formula for time:
Example #2:
A boat travels a distance of kilometers against the current an hour longer than with the current. Find the speed of the boat in still water if the speed of the current is km/h.
Solution #2:
Let's try to create an equation right away. The time upstream is an hour longer than the time upstream.
It is written like this:
Now, instead of each time, let’s substitute the formula:
We have received an ordinary rational equation, let’s solve it:
Obviously, speed cannot be a negative number, so the answer is km/h.
Relative motion
If some bodies are moving relative to each other, it is often useful to calculate their relative speed. It is equal to:
- the sum of velocities if bodies move towards each other;
- speed differences if bodies move in the same direction.
Example No. 1
Two cars left points A and B simultaneously towards each other at speeds km/h and km/h. In how many minutes will they meet? If the distance between points is km?
I solution method:
Relative speed of cars km/h. This means that if we are sitting in the first car, it seems motionless to us, but the second car is approaching us at a speed of km/h. Since the distance between the cars is initially km, the time it will take for the second car to pass the first:
Method II:
The time from the start of movement to the meeting of the cars is obviously the same. Let's designate it. Then the first car drove the path, and the second - .
In total they covered all the kilometers. Means,
Other movement tasks
Example #1:
A car left point A to point B. At the same time, another car left with him, which drove exactly half the way at a speed of km/h less than the first, and drove the second half of the way at a speed of km/h.
As a result, the cars arrived at point B at the same time.
Find the speed of the first car if it is known that it is greater than km/h.
Solution #1:
To the left of the equal sign we write down the time of the first car, and to the right - of the second:
Let's simplify the expression on the right side:
Let's divide each term by AB:
The result is an ordinary rational equation. Having solved it, we get two roots:
Of these, only one is larger.
Answer: km/h.
Example No. 2
A cyclist left point A of the circular route. Minutes later, he had not yet returned to point A, and a motorcyclist followed him from point A. Minutes after leaving, 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 route is km. Give your answer in km/h.
Solution:
Here we will equate the distance.
Let the speed of the cyclist be, and the speed of the motorcyclist - . Until the moment of the first meeting, the cyclist had been on the road for minutes, and the motorcyclist had been on the road for .
At the same time, they traveled equal distances:
Between meetings, the cyclist traveled a distance, and the motorcyclist - . But at the same time, the motorcyclist drove exactly one lap more, as can be seen from the figure:
I hope you understand that they didn’t actually drive in a spiral; the spiral just schematically shows that they drive in a circle, passing the same points on the route several times.
We solve the resulting equations in the system:
SUMMARY AND BASIC FORMULAS
1. Basic formula
2. Relative motion
- This is the sum of speeds if the bodies move towards each other;
- difference in speed if bodies move in the same direction.
3. Moving with the flow:
- If we move with the current, the speed of the current is added to our speed;
- if we move against the current, the speed of the current is subtracted from the speed.
We helped you deal with movement problems...
Now it's your turn...
If you carefully read the text and solved all the examples yourself, we are willing to bet that you understood everything.
And this is already half the way.
Write below in the comments, have you figured out the movement problems?
Which ones cause the most difficulties?
Do you understand that tasks for “work” are almost the same thing?
Write to us and good luck on your exams!
