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« A bicycle left point A of the circular track» — 251 tasks found
(views: 605 , answers: 13 )
A cyclist left point A of the circular track, and 10 minutes later a motorcyclist followed him. 2 minutes after departure, he caught up with the cyclist for the first time, and 3 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 5 km. Give your answer in km/h.
Task B14 ()(views: 624 , answers: 11 )
A cyclist left point A of the circular track, and 20 minutes later a motorcyclist followed him. 5 minutes after departure, he caught up with the cyclist for the first time, and another 10 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 10 km. Give your answer in km/h.
The correct answer has not yet been determined
Task B14 ()(views: 691 , answers: 11 )
A cyclist left point A of the circular track, and 10 minutes later a motorcyclist followed him. 5 minutes after departure he caught up with the cyclist for the first time, and another 15 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 10 km. Give your answer in km/h.
Answer: 60
Task B14 ()(views: 612 , answers: 11 )
A cyclist left point A of the circular track, and 30 minutes later a motorcyclist followed him. 5 minutes after departure, he caught up with the cyclist for the first time, and another 47 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 47 km. Give your answer in km/h.
The correct answer has not yet been determined
Task B14 ()(views: 608 , answers: 9 )
A cyclist left point A of the circular track, and 20 minutes later a motorcyclist followed him. 5 minutes after departure, he caught up with the cyclist for the first time, and another 19 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 19 km. Give your answer in km/h.
The correct answer has not yet been determined
Task B14 ()(views: 618 , answers: 9 )
A cyclist left point A of the circular track, and 20 minutes later a motorcyclist followed him. 2 minutes after departure, he caught up with the cyclist for the first time, and another 30 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 50 km. Give your answer in km/h.
The correct answer has not yet been determined
Task B14 ()(views: 610 , answers: 9 )
A cyclist left point A of the circular track, and 30 minutes later a motorcyclist followed him. 5 minutes after departure, he caught up with the cyclist for the first time, and another 26 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 39 km. Give your answer in km/h.
The correct answer has not yet been determined
Task B14 ()(views: 622 , answers: 9 )
A cyclist left point A of the circular track, and 50 minutes later a motorcyclist followed him. 5 minutes after departure he caught up with the cyclist for the first time, and another 12 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 20 km. Give your answer in km/h.
The correct answer has not yet been determined
Task B14 (“Elementary school teacher” - Topic. Analysis of the work of primary school teachers' school education. Develop individual routes that promote the professional growth of teachers. Strengthening the educational and material base. Organizational and pedagogical activities. Continue the search for new technologies, forms and methods of teaching and education. Directions of work of primary school.
“Youth and elections” - Development of political legal consciousness among young people: Youth and elections. Development of political legal consciousness in schools and secondary specialized institutions: A set of measures to attract young people to elections. Why don't we vote? Development of political legal consciousness in preschool educational institutions:
“Afghan War 1979-1989” - The Soviet leadership brings a new president, Babrak Karmal, to power in Afghanistan. Results of the war. Soviet-Afghan War 1979-1989 On February 15, 1989, the last Soviet troops were withdrawn from Afghanistan. Reason for war. After the withdrawal of the Soviet Army from the territory of Afghanistan, the pro-Soviet regime of President Najibullah lasted another 3 years and, having lost Russian support, was overthrown in April 1992 by Mujahideen commanders.
“Signs of divisibility of natural numbers” - Relevance. Pascal's test. A sign that numbers are divisible by 6. A sign that numbers are divisible by 8. A sign that numbers are divisible by 27. A sign that numbers are divisible by 19. A sign that numbers are divisible by 13. Identify signs of divisibility. How to learn to calculate quickly and correctly. Test for divisibility of numbers by 25. Test for divisibility of numbers by 23.
“Butlerov’s Theory” - The prerequisites for the creation of the theory were: Isomerism-. The importance of the theory of the structure of organic substances. The science of the spatial structure of molecules - stereochemistry. The role of creating a theory of the chemical structure of substances. Learn the basic principles of the theory of chemical structure of A. M. Butlerov. The main provisions of the modern theory of the structure of compounds.
“Mathematics competition for schoolchildren” - Mathematical terms. The part of a line connecting two points. Students' knowledge. Contest of cheerful mathematicians. Task. A ray dividing an angle in half. The angles are all right. Time interval. Contest. The most attractive. Speed. Radius. Getting ready for winter. Jumping dragonfly. Figure. Playing with the audience. Sum of angles of a triangle.
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We continue to consider motion problems. There is a group of problems that differ from ordinary motion problems - these are problems involving circular motion (circular track, movement of clock hands). In this article we will consider such tasks. The principles of the solution are the same, the same (formula for the law of rectilinear motion). But there are small nuances in the approaches to the solution.
Let's consider the tasks:
Two motorcyclists start simultaneously in the same direction from two diametrically opposite points on a circular track, the length of which is 22 km. How many minutes will it take for the motorcyclists to meet each other for the first time if the speed of one of them is 20 km/h greater than the speed of the other?
At first glance, to some, problems involving circular motion may seem complex and somewhat confusing in comparison with ordinary tasks involving rectilinear motion. But this is only at first glance. This problem easily turns into a linear motion problem. How?
Let’s mentally turn the circular track into a straight line. There are two motorcyclists standing on it. One of them lags behind the other by 11 km, as it is stated in the condition that the length of the route is 22 kilometers.
The speed of the person lagging behind is 20 kilometers per hour higher (he is catching up with the one in front). Here is the task of linear motion.
So, we take the desired value (the time after which they become equal) to be x hours. Let’s denote the speed of the first one (located in front) as y km/h, then the speed of the second one (catching up) will be y + 20.
Let's enter the speed and time into the table.
Fill in the “distance” column:
The second travels a distance (to meet) 11 km more, which means
11/20 hours is the same as 33/60 hours. That is, 33 minutes passed before they met. You can see how to convert hours to minutes, and vice versa, in the article ““.
As you can see, the speed of the motorcyclists itself does not matter in this case.
Answer: 33
Decide for yourself:
Two motorcyclists start simultaneously in the same direction from two diametrically opposite points on a circular track, the length of which is 14 km. How many minutes will it take for the motorcyclists to meet each other for the first time if the speed of one of them is 21 km/h greater than the speed of the other?
From one point on a circular track, the length of which is 25 km, two cars started simultaneously in the same direction. The speed of the first car is 112 km/h, and 25 minutes after the start it was one lap ahead of the second car. Find the speed of the second car. Give your answer in km/h.
This task can also be interpreted, that is, it can be presented as a task on rectilinear motion. How? Just …
Two cars simultaneously start moving in the same direction. The speed of the first is 112 km/h. After 25 minutes, he is ahead of the second by 25 km (as it is said that by one lap). Find the speed of the second. In tasks involving movement, it is very important to imagine the process of this movement itself.
We will make the comparison by distance, since we know that one was 25 kilometers ahead of the other.
For x we take the desired value – the speed of the second. Travel time is 25 minutes (25/60 hours) for both.
Fill in the “distance” column:
The distance covered by the first is 25 km greater than the distance covered by the second. That is:
The speed of the second car is 52 (km/h).
Answer: 52
Decide for yourself:
From one point on a circular track, the length of which is 14 km, two cars started simultaneously in the same direction. The speed of the first car is 80 km/h, and 40 minutes after the start it was one lap ahead of the second car. Find the speed of the second car. Give your answer in km/h.
A cyclist left point A of the circular track, and 40 minutes later a motorcyclist followed him. 8 minutes after departure, he caught up with the cyclist for the first time, and another 36 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 30 km. Give your answer in km/h.
This task is relatively difficult. What is immediately worth noting? This means that a motorcyclist travels the same distance with a cyclist, catching up with him for the first time. Then he catches up with him again a second time, and the difference in distances traveled after the first meeting is 30 kilometers (the length of the circle). Thus, it will be possible to create two equations and solve their system. We are not given the speeds of the road users, so we can enter two variables. A system of two equations with two variables is solved.
So, let's convert minutes to hours, since the speed must be found in km/h.
Forty minutes is 2/3 of an hour, 8 minutes is 8/60 of an hour, 36 minutes is 36/60 of an hour.
We denote the speeds of the participants as x km/h (for the cyclist) and y km/h (for the motorcyclist).
For the first time, the motorcyclist overtook the cyclist after 8 minutes, that is, 8/60 of an hour after the start.
Up to this point, the cyclist had already been on the road for 40+8=48 minutes, that is, 48/60 hours.
Let's write this data into a table:
Both traveled the same distances, that is
The motorcyclist then caught up with the cyclist a second time. This happened 36 minutes later, that is, 36/60 hours after the first overtaking.
Let’s create a second table and fill in the “distance” column:
Since it is said that after 36 minutes the motorcyclist caught up with the cyclist again. This means that he (the motorcyclist) traveled a distance equal to 30 kilometers (one lap) plus the distance that the cyclist traveled during this time. This is the key to composing the second equation.
One lap is the length of the track, it is 30 km.
We get the second equation:
We solve the system of two equations:
So y = 6 ∙10 = 60.
That is, the speed of the motorcyclist is 60 km/h.
Answer: 60
Decide for yourself:
A cyclist left point A of the circular track, and 30 minutes later a motorcyclist followed him. 10 minutes after departure, he caught up with the cyclist for the first time, and another 30 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 30 km. Give your answer in km/h.
The next type of problems involving circular motion is, one might say, “unique.” There are tasks that are solved orally. And there are some that are extremely difficult to solve without understanding and careful reasoning. We are talking about problems about clock hands.
Here is an example of a simple task:
The clock with hands shows 11 hours 20 minutes. How many minutes will it take for the minute hand to line up with the hour hand for the first time?
The answer is obvious, in 40 minutes, when it is exactly twelve. Even if they couldn’t understand it right away, after drawing the dial(having made a sketch) on the sheet, you can easily determine the answer.
Examples of other tasks (not easy):
The clock with hands shows 6 hours 35 minutes. In how many minutes will the minute hand line up with the hour hand for the fifth time? Answer: 325
The clock with hands shows 2 o'clock exactly. In how many minutes will the minute hand line up with the hour hand for the tenth time? Answer: 600
Decide for yourself:
The clock with hands shows 8 hours 00 minutes. In how many minutes will the minute hand line up with the hour hand for the fourth time?
Are you convinced that it is very easy to get confused?
In general, I am not a supporter of giving such advice, but here it is needed, since on the Unified State Exam with such a task you can easily get confused, calculate incorrectly, or simply lose a lot of time solving it.
You can solve this problem in one minute. How? Just!
*Further information in the article is closed and available only to registered users! The registration (login) tab is located in the MAIN MENU of the site. After registration, log in to the site and refresh this page.
That's all. I wish you success!
Sincerely, Alexander.
P.S: I would be grateful if you tell me about the site on social networks.
The article discusses problems to help students: to develop skills in solving word problems in preparation for the Unified State Exam, when learning to solve problems to create a mathematical model of real situations in all parallels of primary and high school. It presents tasks: on movement in a circle; to find the length of a moving object; to find the average speed.
I. Problems involving movement in a circle.
Circular motion problems turned out to be difficult for many schoolchildren. They are solved in almost the same way as ordinary movement problems. They also use the formula. But there is a point to which we would like to pay attention.
Task 1. A cyclist left point A of the circular track, and 30 minutes later a motorcyclist followed him. 10 minutes after departure, he caught up with the cyclist for the first time, and another 30 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 30 km. Give your answer in km/h.
Solution. The speeds of the participants will be taken as X km/h and y km/h. For the first time, a motorcyclist overtook a cyclist 10 minutes later, that is, an hour after the start. Up to this point, the cyclist had been on the road for 40 minutes, that is, hours. The participants in the movement traveled the same distances, that is, y = x. Let's enter the data into the table.
Table 1
The motorcyclist then passed the cyclist a second time. This happened 30 minutes later, that is, an hour after the first overtaking. How far did they travel? A motorcyclist overtook a cyclist. This means he completed one more lap. This is the moment
which you need to pay attention to. One lap is the length of the track, it is 30 km. Let's create another table.
table 2
We get the second equation: y - x = 30. We have a system of equations: 
In the answer we indicate the speed of the motorcyclist.
Answer: 80 km/h.
Tasks (independently).
I.1.1. A cyclist left point “A” of the circular route, and 40 minutes later a motorcyclist followed him. 10 minutes after departure, he caught up with the cyclist for the first time, and another 36 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 36 km. Give your answer in km/h.
I.1. 2. A cyclist left point “A” of the circular route, and 30 minutes later a motorcyclist followed him. 8 minutes after departure, he caught up with the cyclist for the first time, and another 12 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 15 km. Give your answer in km/h.
I.1. 3. A cyclist left point “A” of the circular route, and 50 minutes later a motorcyclist followed him. 10 minutes after departure, he caught up with the cyclist for the first time, and another 18 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 15 km. Give your answer in km/h.
Two motorcyclists start simultaneously in the same direction from two diametrically opposite points on a circular track, the length of which is 20 km. How many minutes will it take for the motorcyclists to meet each other for the first time if the speed of one of them is 15 km/h greater than the speed of the other?
Solution.
Picture 1
With a simultaneous start, the motorcyclist who started from “A” traveled half a lap more than the one who started from “B”. That is, 10 km. When two motorcyclists move in the same direction, the removal velocity v = -. According to the conditions of the problem, v = 15 km/h = km/min = km/min – removal speed. We find the time after which the motorcyclists reach each other for the first time.
10:= 40(min).
Answer: 40 min.
Tasks (independently).
I.2.1. Two motorcyclists start simultaneously in the same direction from two diametrically opposite points on a circular track, the length of which is 27 km. How many minutes will it take for the motorcyclists to meet each other for the first time if the speed of one of them is 27 km/h greater than the speed of the other?
I.2.2. Two motorcyclists start simultaneously in the same direction from two diametrically opposite points on a circular track, the length of which is 6 km. How many minutes will it take for the motorcyclists to meet each other for the first time if the speed of one of them is 9 km/h greater than the speed of the other?
From one point on a circular track, the length of which is 8 km, two cars started simultaneously in the same direction. The speed of the first car is 89 km/h, and 16 minutes after the start it was one lap ahead of the second car. Find the speed of the second car. Give your answer in km/h.
Solution.
x km/h is the speed of the second car.
(89 – x) km/h – removal speed.
8 km is the length of the circular route.
The equation.
(89 – x) = 8,
89 – x = 2 15,
Answer: 59 km/h.
Tasks (independently).
I.3.1. From one point on a circular track, the length of which is 12 km, two cars started simultaneously in the same direction. The speed of the first car is 103 km/h, and 48 minutes after the start it was one lap ahead of the second car. Find the speed of the second car. Give your answer in km/h.
I.3.2. From one point on a circular track, the length of which is 6 km, two cars started simultaneously in the same direction. The speed of the first car is 114 km/h, and 9 minutes after the start it was one lap ahead of the second car. Find the speed of the second car. Give your answer in km/h.
I.3.3. From one point on a circular track, the length of which is 20 km, two cars started simultaneously in the same direction. The speed of the first car is 105 km/h, and 48 minutes after the start it was one lap ahead of the second car. Find the speed of the second car. Give your answer in km/h.
I.3.4. From one point on a circular track, the length of which is 9 km, two cars started simultaneously in the same direction. The speed of the first car is 93 km/h, and 15 minutes after the start it was one lap ahead of the second car. Find the speed of the second car. Give your answer in km/h.
The clock with hands shows 8 hours 00 minutes. In how many minutes will the minute hand line up with the hour hand for the fourth time?
Solution. We assume that we are not solving the problem experimentally.
In one hour, the minute hand travels one circle, and the hour hand travels one circle. Let their speeds be 1 (laps per hour) and 
Start - at 8.00. Let's find the time it takes for the minute hand to catch up with the hour hand for the first time.
The minute hand will move further, so we get the equation
This means that for the first time the arrows will align through
Let the arrows align for the second time after time z. The minute hand will travel a distance of 1·z, and the hour hand will travel one circle more. Let's write the equation:
Having solved it, we get that .
So, through the arrows they will align for the second time, after another - for the third time, and after another - for the fourth time.
Therefore, if the start was at 8.00, then for the fourth time the hands will align through
4h = 60 * 4 min = 240 min.
Answer: 240 minutes.
Tasks (independently).
I.4.1.The clock with hands shows 4 hours 45 minutes. In how many minutes will the minute hand line up with the hour hand for the seventh time?
I.4.2. The clock with hands shows 2 o'clock exactly. In how many minutes will the minute hand line up with the hour hand for the tenth time?
I.4.3. The clock with hands shows 8 hours 20 minutes. In how many minutes will the minute hand line up with the hour hand for the fourth time? fourth
II. Problems to find the length of a moving object.
A train, moving uniformly at a speed of 80 km/h, passes a roadside pole in 36 s. Find the length of the train in meters.
Solution. Since the speed of the train is indicated in hours, we will convert the seconds to hours.
1) 36 sec = 
2) find the length of the train in kilometers.
80· 
Answer: 800m.
Tasks (independently).
II.2. A train, moving uniformly at a speed of 60 km/h, passes a roadside pole in 69 s. Find the length of the train in meters. Answer: 1150m.
II.3. A train, moving uniformly at a speed of 60 km/h, passes a forest belt 200 m long in 1 min 21 s. Find the length of the train in meters. Answer: 1150m.
III. Medium speed problems.
On a math exam, you may encounter a problem about finding the average speed. We must remember that the average speed is not equal to the arithmetic mean of the speeds. The average speed is found using a special formula:
If there were two sections of the path, then 
.
The distance between the two villages is 18 km. A cyclist traveled from one village to another for 2 hours, and returned along the same road for 3 hours. What is the average speed of the cyclist along the entire route?
Solution:
2 hours + 3 hours = 5 hours - spent on the entire movement,
.The tourist walked at a speed of 4 km/h, then for exactly the same time at a speed of 5 km/h. What is the average speed of the tourist along the entire route?
Let the tourist walk t h at a speed of 4 km/h and t h at a speed of 5 km/h. Then in 2t hours he covered 4t + 5t = 9t (km). The average speed of a tourist is = 4.5 (km/h).
Answer: 4.5 km/h.
We note that the average speed of the tourist turned out to be equal to the arithmetic mean of the two given speeds. You can verify that if the travel time on two sections of the route is the same, then the average speed of movement is equal to the arithmetic mean of the two given speeds. To do this, let us solve the same problem in general form.
The tourist walked at a speed of km/h, then for exactly the same time at a speed of km/h. What is the average speed of the tourist along the entire route?
Let the tourist walk t h at a speed of km/h and t h at a speed of km/h. Then in 2t hours he traveled t + t = t (km). The average speed of a tourist is
= (km/h).The car covered some distance uphill at a speed of 42 km/h, and downhill at a speed of 56 km/h.
.
The average speed of movement is 2 s: (km/h).
Answer: 48 km/h.
The car covered some distance uphill at a speed of km/h, and down the mountain at a speed of km/h.
What is the average speed of the car along the entire route?
Let the length of the path section be s km. Then the car traveled 2 s km in both directions, spending the entire journey 
.
The average speed of movement is 2 s: 
(km/h).
Answer: km/h.
Consider a problem in which the average speed is given, and one of the speeds needs to be determined. Application of the equation will be required.
The cyclist was traveling uphill at a speed of 10 km/h, and down the mountain at some other constant speed. As he calculated, the average speed was 12 km/h.
.III.2. Half the time spent on the road, the car was traveling at a speed of 60 km/h, and the second half of the time at a speed of 46 km/h. Find the average speed of the car along the entire journey.
III.3. On the way from one village to another, the car walked for some time at a speed of 60 km/h, then for exactly the same time at a speed of 40 km/h, then for exactly the same time at a speed equal to the average speed on the first two sections of the route . What is the average speed of travel along the entire route from one village to another?
III.4. A cyclist travels from home to work at an average speed of 10 km/h, and back at an average speed of 15 km/h, since the road goes slightly downhill. Find the average speed of the cyclist all the way from home to work and back.
III.5. A car traveled from point A to point B empty at a constant speed, and returned along the same road with a load at a speed of 60 km/h. At what speed was he driving empty if the average speed was 70 km/h?
III.6. The car drove for the first 100 km at a speed of 50 km/h, for the next 120 km at a speed of 90 km/h, and then for 120 km at a speed of 100 km/h. Find the average speed of the car along the entire journey.
III.7. The car drove for the first 100 km at a speed of 50 km/h, the next 140 km at a speed of 80 km/h, and then 150 km at a speed of 120 km/h. Find the average speed of the car along the entire journey.
III.8. The car drove for the first 150 km at a speed of 50 km/h, for the next 130 km at a speed of 60 km/h, and then for 120 km at a speed of 80 km/h. Find the average speed of the car along the entire journey.
III. 9. The car drove for the first 140 km at a speed of 70 km/h, the next 120 km at a speed of 80 km/h, and then 180 km at a speed of 120 km/h. Find the average speed of the car along the entire journey.
Lesson type: repeating and generalizing lesson.
Lesson objectives:
- educational – repeat methods for solving various types of word problems involving movement
- developing – develop students’ speech through enriching and complicating its vocabulary, develop students’ thinking through the ability to analyze, generalize and systematize material
- educational – formation of a humane attitude among students towards participants in the educational process
Lesson equipment:
- interactive board;
- envelopes with assignments, thematic control cards, consultant cards.
Lesson structure.
Main stages of the lesson |
Tasks to be solved at this stage |
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| Organizational moment, introductory part |
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| Preparing students for active work (repetition) |
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| Stage of generalization and systematization of the studied material (work in groups) |
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| Checking the work, making adjustments (if necessary) |
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| Summing up the lesson. Homework analysis |
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Forms of organization of students’ cognitive activity:
- frontal form of cognitive activity - at stages II, IY, Y.
- group form of cognitive activity - at stage III.
Teaching methods: verbal, visual, practical, explanatory - illustrative, reproductive, partially - search, analytical, comparative, generalizing, traductive.
During the classes
I. Organizational moment, introductory part.
The teacher announces the topic of the lesson, the objectives of the lesson and the main points of the lesson. Checks the class's readiness for work.
II. Preparing students for active work (repetition)
Answer the questions.
- What kind of motion is called uniform (movement at a constant speed).
- What is the formula for the path with uniform motion ( S = Vt).
- From this formula, express the speed and time.
- Specify units of measurement.
- Conversion of speed units
III. Stage of generalization and systematization of the studied material (work in groups)
The whole class is divided into groups (5-6 people per group). It is advisable to have students of different skill levels in the same group. Among them, a group leader (the strongest student) is appointed, who will lead the work of the group.
All groups receive envelopes with assignments (they are the same for all groups), consultant cards (for weak students) and thematic control sheets. In the thematic control sheets, the group leader gives grades to each student in the group for each task and notes the difficulties that the students encountered when completing specific tasks.
Card with tasks for each group.
№ 5.
No. 7. The motor boat traveled 112 km upstream of the river and returned to the point of departure, spending 6 hours less on the return journey. Find the speed of the current if the speed of the boat in still water is 11 km/h. Give your answer in km/h.
No. 8. The motor ship travels along the river to its destination 513 km and, after stopping, returns to the point of departure. Find the speed of the ship in still water if the current speed is 4 km/h, the stay lasts 8 hours, and the ship returns to the point of departure 54 hours after departure. Give your answer in km/h.
No. 9. From pier A to pier B, the distance between which is 168 km, the first motor ship set off at a constant speed, and 2 hours after that, the second one set off after it, at a speed 2 km/h higher. Find the speed of the first ship if both ships arrived at point B at the same time. Give your answer in km/h.
Sample thematic control card.
| Class ________ Student's full name___________________________________ | ||
Job No. |
Comment |
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Cards consultants.
| Card No. 1 (consultant) |
| 1. Driving on a straight road |
| When solving problems involving uniform motion, two situations often occur. If the initial distance between objects is S, and the velocities of the objects are V1 and V2, then: a) when objects move towards each other, the time after which they will meet is equal to . b) when objects move in one direction, the time after which the first object will catch up with the second is equal to , ( V 2 > V 1) |
Example 1. The train, having traveled 450 km, was stopped due to snow drift. Half an hour later the path was cleared, and the driver, increasing the speed of the train by 15 km/h, brought it to the station without delay. Find the initial speed of the train if the distance traveled by it to the stop was 75% of the entire distance.
X= -75 does not fit the conditions of the problem, where x > 0. Answer: the initial speed of the train is 60 km/h. |
| Card No. 2 (consultant) |
2. Driving on a closed road |
| If the length of a closed road is S, and the speeds of objects V 1 and V 2, then: a) when objects move in different directions, the time between their meetings is calculated by the formula; |
| Example 2. At a competition on a circuit track, one skier completes a lap 2 minutes faster than the other and an hour later beats him by exactly a lap. How long does it take each skier to complete the circle? Let S m – length of the ring route and x m/min and y m/min – speeds of the first and second skiers, respectively ( x> y) . Then S/x min and S/y min – the time it takes the first and second skiers to complete the lap, respectively. From the first condition we obtain the equation. Since the speed of removal of the first skier from the second skier is ( x- y) m/min, then from the second condition we have the equation . Let's solve the system of equations.
Let's make a replacement S/x= a And S/y= b, then the system of equations will take the form:
60(a+ 2) – 60a = a(a+ 2)a 2 + 2a- 120 = 0. The quadratic equation has one positive root a = 10 then b = 12. This means that the first skier completes the circle in 10 minutes, and the second skier in 12 minutes. Answer: 10 min; 12 min. |
| Card No. 3 (consultant) |
3. Movement along the river |
| If an object moves with the flow of a river, then its speed is equal to Vflow. =Vob. +
Vcurrent If an object moves against the flow of a river, then its speed is equal to Vagainst the current = V inc. – Vcurrent. The object’s own speed (speed in still water) is equal to The speed of the river flow is The speed of the raft is equal to the speed of the river flow. |
| Example 3. The boat went 50 km downstream of the river, and then traveled 36 km in the opposite direction, which took it 30 minutes longer than along the river. What is the boat's own speed if the speed of the river is 4 km/h? Let the boat's own speed be X km/h, then its speed along the river is ( x+ 4) km/h, and against the flow of the river ( x- 4) km/h. The time it takes for the boat to move along the river flow is hours, and against the river flow is hours. Since 30 minutes = 1/2 hour, then according to the conditions of the problem we will create the equation =. Multiply both sides of the equation by 2( x+ 4)(x- 4) >0 . We get 72( x+ 4) -100(x- 4) = (x+ 4)(x- 4) x 2 + 28x- 704 = 0 x 1 =16, x 2 = - 44 (excluded, since x> 0). So, the boat's own speed is 16 km/h. Answer: 16 km/h. |
IV. Problem solving analysis stage.
Problems that caused difficulty for students are analyzed.
No. 1. From two cities, the distance between which is 480 km, two cars simultaneously drove towards each other. After how many hours will the cars meet if their speeds are 75 km/h and 85 km/h?
- 75 + 85 = 160 (km/h) – approach speed.
- 480: 160 = 3 (h).
Answer: the cars will meet in 3 hours.
No. 2. From cities A and B, the distance between which is 330 km, two cars simultaneously left towards each other and met after 3 hours at a distance of 180 km from city B. Find the speed of the car that left city A. Give the answer in km/ h.
- (330 – 180) : 3 = 50 (km/h)
Answer: the speed of a car leaving city A is 50 km/h.
No. 3. A motorist and a cyclist left at the same time from point A to point B, the distance between which is 50 km. It is known that a motorist travels 65 km more per hour than a cyclist. Determine the speed of the cyclist if it is known that he arrived at point B 4 hours 20 minutes later than the motorist. Give your answer in km/h.
Let's make a table.
Let's create an equation, taking into account that 4 hours 20 minutes =
,Obviously, x = -75 does not fit the conditions of the problem.
Answer: The cyclist's speed is 10 km/h.
No. 4. Two motorcyclists start simultaneously in the same direction from two diametrically opposite points on a circular track, the length of which is 14 km. How many minutes will it take for the motorcyclists to meet each other for the first time if the speed of one of them is 21 km/h greater than the speed of the other?
Let's make a table.
Let's create an equation.
, where 1/3 hour = 20 minutes.Answer: in 20 minutes the motorcyclists will pass each other for the first time.
No. 5. From one point on a circular track, the length of which is 12 km, two cars started simultaneously in the same direction. The speed of the first car is 101 km/h, and 20 minutes after the start it was one lap ahead of the second car. Find the speed of the second car. Give your answer in km/h.
Let's make a table.
Let's create an equation.
Answer: the speed of the second car is 65 km/h.
No. 6. A cyclist left point A of the circular track, and 40 minutes later a motorcyclist followed him. 8 minutes after departure, he caught up with the cyclist for the first time, and another 36 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the route is 30 km. Give your answer in km/h.
Let's make a table.
Movement before the first meeting |
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cyclist |
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