A cyclist left point a of the circular track. A cyclist (cm) left point A of the circular track. How to solve? I. Circular motion problems
Sections: Mathematics
The article discusses tasks to help students: to develop skills in solving word problems in preparation for the Unified State Exam, when learning to solve problems on composing mathematical model 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.
Figure 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.
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 down the mountain 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, the next 130 km at a speed of 60 km/h, and then 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.
Posted on 03/23/2018
A cyclist left point A of the circular route.
After 30 minutes, he had not yet returned to point A and a motorcyclist followed him from point A. 10 minutes after departure he caught up with the cyclist for the first time,
and 30 minutes later I 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
math problem
education
answer
comment
add to favorites
Svetl-ana02-02
23 hours ago
If I understood the condition correctly, the motorcyclist left half an hour after the cyclist started. In this case the solution looks like this.
A cyclist covers the same distance in 40 minutes, and a motorcyclist in 10 minutes; therefore, the speed of a motorcyclist is four times the speed of a cyclist.
Let's say a cyclist moves at a speed of x km/h, then the speed of the motorcyclist is 4x km/h. Before the second meeting, (1/2 + 1/2 + 1/6) = 7/6 hours will pass from the moment the cyclist starts and (1/2 + 1/6) = 4/6 hours from the moment the motorcyclist starts. By the time of the second meeting, the cyclist will have covered (7x/6) km, and the motorcyclist will have covered (16x/6) km, having overtaken the cyclist by one lap, i.e. having traveled 30 km more. We get the equation.
16x/6 - 7x/6 = 30, from where
So, the cyclist was traveling at a speed of 20 km/h, which means the motorcyclist was traveling at a speed of (4*20) = 80 km/h.
Answer. The speed of the motorcyclist is 80 km/h.
comment
add to favorites
thank
Vdtes-t
22 hours ago
If the solution is in km/h, then the time must be expressed in hours.
Let's denote
v speed of the cyclist
m motorcyclist speed
After ½ hour, a motorcyclist followed the cyclist from point A. ⅙ hour after departure he caught up with the cyclist for the first time
We write down the path traveled before the first meeting in the form of an equation:
and another ½ hour after that, the motorcyclist caught up with him for the second time.
We write down the path traveled to the second meeting in the form of an equation:
We solve a system of two equations:
We simplify the first equation (multiplying both sides by 6):
Substitute m into the second equation:
The cyclist's speed is 20 km/h
Determining the speed of a motorcyclist
Answer: the speed of the motorcyclist is 80 km/h
“Lesson Tangent to a circle” - Prove that line AC is tangent to a given circle. Problem 1. Given: env.(O;OM), MR – tangent, angle KMR=45?. Calculate the length of BC if OD=3cm. General lesson. Draw a tangent to the given circle. Topic: “circle”. Solution: Problem solving. Practical work. Make notations and notes.
“Tangent to a circle” - Property of a tangent. Let d be the distance from the center O to the straight line KM. The segments AK and AM are called tangent segments drawn from A. Tangent to a circle. Then. A tangent to a circle is perpendicular to the radius drawn to the point of tangency. Proof. Let us prove that if AK and AM are tangent segments, then AK = AM, ?OAK = ? OAM.
“Circumference and Circle” - Calculate. Find the circumference. Find the radius of the circle. Find the area of the shaded figure. Circle. Circular sector. Draw a circle with center K and radius 2 cm. Complete the statement. Independent work. Circumference. Circle. Area of a circle. Calculate the length of the equator. Game.
“Equation of a circle” - Construct a circle in your notebook, given by equations: Center of the circle O(0;0), (x – 0)2 + (y – 0)2 = R 2, x2 + y2 = R 2? equation of a circle with center at the origin. . O (0;0) – center, R = 4, then x2 + y2 = 42; x2 + y2 = 16. Find the coordinates of the center and the radius if AB is the diameter of the given circle.
“Circle length 6th grade” - Lesson motto: History of numbers?. The diameter of the diesel locomotive wheel is 180 cm. Lambert found for? the first twenty-seven suitable fractions. Math lesson in 6th grade Math teacher: Nikonorova Lyubov Arkadyevna. Lesson plan. Competition "Mosaic of Presentations". But you can find an infinite sequence of suitable fractions.
“Elementary school teacher” - Topic. Analysis of the work of teachers' school education primary classes. Develop individual routes that promote professional growth teachers. Strengthening the educational and material base. Organizational and pedagogical activities. Continue the search for new technologies, forms and methods of teaching and education. Areas of work 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. Reason for war. After withdrawal 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 structure theory organic matter. The science of the spatial structure of molecules - stereochemistry. The Role of Theory Creation chemical structure substances. Learn the basic principles of the theory of chemical structure of A. M. Butlerov. Basic position modern theory 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. A period of time. Contest. The most attractive. Speed. Radius. Getting ready for winter. Jumping dragonfly. Figure. Playing with the audience. Sum of angles of a triangle.
There are a total of 23,687 presentations in the topic
Problem 1. Two cars left point A for point B at the same time.
The first one drove all the way at a constant speed.
The second one drove the first half of the way at a speed
lower speed of the first by 14 km/h,
and the second half of the journey at a speed of 105 km/h,
and therefore arrived in B at the same time as the first car.
Find the speed of the first car,
if it is known that it is more than 50 km/h.
Solution: Let's take the entire distance as 1.
Let's take the speed of the first car to be x.
Then, the time it took the first car to travel the entire distance is
equals 1/x.
The second car speed for the first half of the journey, i.e. 1/2,
was 14 km/h less than the speed of the first car, x-14.
The time taken by the second car is 1/2: (x-14) = 1/2(x-14).
The second half of the journey, i.e. 1/2, the car passed
at a speed of 105 km/h.
The time he spent is 1/2: 105 = 1/2*105 = 1/210.
The times of the first and second are equal to each other.
Let's make an equation:
1/x = 1/2(x-14) + 1/210
We find the common denominator - 210x(x-14)
210(x-14) = 105x + x(x-14)
210x - 2940 = 105x + x² - 14x
x² - 119x + 2940 = 0
Solving this quadratic equation through the discriminant, we find the roots:
x1 = 84
x2 = 35. The second root does not fit the conditions of the problem.
Answer: the speed of the first car is 84 km/h.
Problem 2. From point A of a circular route, the length of which is 30 km,
Two motorists started at the same time in the same direction.
The speed of the first is 92 km/h, and the speed of the second is 77 km/h.
In how many minutes will the first motorist
will be ahead of the second 1 lap?
Solution: This task, despite the fact that it is given in 11th grade,
can be solved at the primary school level.
Let's ask just four questions and get four answers.
1. How many kilometers will the first motorist travel in 1 hour?
92 km.
2. How many kilometers will the second motorist travel in 1 hour?
77 km.
3. How many kilometers will the first motorist be ahead of the second after 1 hour?
92 - 77 = 15 km.
4. How many hours will it take for the first motorist to be 30 km ahead of the second?
30:15 = 2 hours = 120 minutes.
Answer: in 120 minutes.
Problem 3. From point A to point B, the distance between them is 60 km,
a motorist and a cyclist left at the same time.
It is known that every hour a motorist passes
90 km more than a cyclist.
Determine the speed of the cyclist if it is known that he arrived at point B 5 hours 24 minutes later than the motorist.
Solution: In order to correctly solve any problem assigned to us,
you need to stick to a certain plan.
And the most important thing is that we need to understand what we want from this.
That is, what equation do we want to arrive at under the conditions that are given.
We will compare everyone's time with each other.
A car travels 90 km per hour more than a cyclist.
This means the speed of the car is greater than the speed
cyclist at 90 km/h.
Taking the speed of the cyclist as x km/hour,
we get the speed of the car x + 90 km/h.
The travel time for a cyclist is 60/x.
Car travel time is 60/(x+90).
5 hours 24 minutes is 5 24/60 hours = 5 2/5 = 27/5 hours
Let's make an equation:
60/x = 60/(x+90) + 27/5 Reduce the numerator of each fraction by 3
20/x = 20/(x+90) + 9/5 Common denominator 5x(x+90)
20*5(x+90) = 20*5x + 9x(x+90)
100x + 9000 = 100x + 9x² + 810x
9x² + 810x – 9000 = 0
x² + 90x – 1000 = 0
Solving this equation through the discriminant or Vieta’s theorem, we get:
x1 = - 100 Doesn't fit the purpose of the problem.
x2 = 10
Answer: The cyclist's speed is 10 km/h.
Problem 4. A cyclist rode 40 km from a city to a village.
On the way back he drove at the same speed
but after 2 hours of driving I stopped for 20 minutes.
After stopping, he increased the speed by 4 km/h
and therefore spent the same amount of time on the way back from the village to the city as on the way from the city to the village.
Find the cyclist's initial speed.
Solution: we solve this problem in relation to the time spent
first to the village and then back.
A cyclist was traveling from town to village at the same speed x km/hour.
In doing so, he spent 40 hours.
In 2 hours he traveled 2 km back.
He has 40 km left to travel - 2 km that he has covered
at a speed of x + 4 km/h.
At the same time, the time he spent on the way back
consists of three terms.
2 hours; 20 minutes = 1/3 hour; (40 - 2x)/(x + 4) hours.
Let's make an equation:
40/x = 2 + 1/3 + (40 - 2x)/(x + 4)
40/x = 7/3 + (40 - 2x)/(x + 4) Common denominator 3x(x + 4)
40*3(x + 4) = 7x(x + 4) + 3x(40 - 2x)
120x + 480 = 7x² + 28x + 120x - 6x²
x² + 28x – 480 = 0 Solving this equation through the discriminant or Vieta’s theorem, we get:
x1 = 12
x2 = - 40 Doesn't fit the conditions of the problem.
Answer: The initial speed of the cyclist is 12 km/h.
Problem 5. Two cars left the same point at the same time in the same direction.
The speed of the first is 50 km/h, the second is 40 km/h.
Half an hour later, a third car left the same point in the same direction,
which overtook the first car 1.5 hours later,
than the second car.
Find the speed of the third car.
Solution: In half an hour the first car will travel 25 km, and the second 20 km.
Those. the initial distance between the first and third car is 25 km,
and between the second and third - 20 km.
When one car catches up with another, they speeds are subtracted.
If we take the speed of the third car to be x km/h,
then it turns out that he caught up with the second car after 20/(x-40) hours.
Then he will catch up with the first car in 25/(x - 50) hours.
Let's make an equation:
25/(x - 50) = 20/(x - 40) + 3/2 Common denominator 2(x - 50)(x - 40)
25*2(x - 40) = 20*2(x - 50) + 3(x - 50)(x - 40)
50x - 2000 = 40x - 2000 + 3x² - 270x + 6000
3x² - 280x + 6000 = 0 Solving this equation through the discriminant, we get
x1 = 60
x2 = 100/3
Answer: the speed of the third car is 60 km/h.
