Analytical problem for movement. Tasks for the circular motion of Point a of the circular route, the length of which

From point A of the circular track, the length of which is 75 km, two cars started simultaneously in the same direction. The speed of the first car is 89 km/h, the speed of the second car is 59 km/h. In how many minutes after the start will the first car be ahead of the second by exactly one lap?

The solution of the problem

This lesson shows how, using the physical formula for determining the time in uniform motion: , make a proportion to determine the time when one car overtakes another in a circle. When solving the problem, a clear sequence of actions is indicated for solving such problems: we introduce a specific designation for what we want to find, we write down the time it takes for one and the second car to overcome a certain number of laps, given that this time is the same value - we equate the resulting equalities . The solution is finding an unknown quantity in a linear equation. To get the results, be sure to remember to substitute the number of laps obtained in the formula for determining the time.

The solution of this problem is recommended for students of the 7th grade when studying the topic “Mathematical language. Mathematical model "(Linear equation with one variable"). When preparing for the OGE, the lesson is recommended when repeating the topic “Mathematical language. Mathematical model".

Posted on 03/23/2018

A cyclist left point A of the circular track.

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 caught up with him a second time.

Find the speed of the motorcyclist if the length of the track is 30 km.

Give your answer in km/h

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Svetl-ana02-02

23 hours ago

If I understood the condition correctly, the motorcyclist left half an hour after the beginning of the start of the cyclist. In this case, the solution looks like this.

A cyclist travels the same distance in 40 minutes, and a motorcyclist in 10 minutes, so the speed of the motorcyclist is four times that of the cyclist.

Suppose a cyclist is moving at a speed of x km/h, then the speed of a motorcyclist is 4x km/h. Before the second meeting, (1/2 + 1/2 + 1/6) = 7/6 hours from the start of the cyclist and (1/2 + 1/6) = 4/6 hours from the start of the motorcyclist will pass. By the time of the second meeting, the cyclist will have traveled (7x/6) km, and the motorcyclist - (16x/6) km, overtaking the cyclist by one lap, i.e. driving 30 km more. We get an equation.

16x/6 - 7x/6 = 30, whence

So, the cyclist was traveling at a speed of 20 km/h, which means that the motorcyclist was traveling at a speed of (4*20) = 80 km/h.

Answer. The speed of the motorcyclist is 80 km/h.

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Vdtes-t

22 hours ago

If the solution is in km/h, then the time must be expressed in hours.

Denote

v speed of the cyclist

m motorcyclist speed

After ½ hour from point A, a motorcyclist followed the cyclist. ⅙ hours after departure, he caught up with the cyclist for the first time

we write in the form of an equation the path traveled before the first meeting:

and another ½ hour after that, the motorcyclist caught up with him for the second time.

we write in the form of an equation the path traveled to the second meeting:

We solve a system of two equations:

v/2+v/6=m/6

m/2=30+v/2

Simplify the first equation (by multiplying both sides by 6):

Substitute m into the second equation:

the speed of the cyclist is 20 km/h

Determine the speed of the motorcyclist

Answer: The speed of the motorcyclist is 80 km/h

Sections: Maths

Type of lesson: iterative-generalizing lesson.

Lesson Objectives:

educational

developing

educational

Lesson equipment:

interactive board;
envelopes with tasks, thematic control cards, consultant cards.

Lesson structure.

	The main stages of the lesson	Tasks to be solved at this stage
	Organizational moment, introductory part	creating a welcoming atmosphere in the classroom set students up for productive work identify missing check students readiness for the lesson
	Preparing students for active work (review)	check students' knowledge on the topic: "Solving text problems of various types for movement" implementation of the development of speech and thinking of responding students development of analytical and critical thinking of students through commenting on the answers of classmates organize the learning activities of the whole class during the response of the students called to the board
	The stage of generalization and systematization of the studied material (work in groups)	to test students' ability to solve problems of various types of movement, to form students' knowledge reflected in the form of ideas and theories, the transition from private ideas to broader generalizations to carry out the formation of moral relations of students to participants in the educational process (during group work)
	Checking the performance of work, adjusting (if necessary)	check the execution of data for groups of tasks (their correctness) continue to form students' ability to analyze, highlight the main thing, build analogies, generalize and systematize develop the ability to negotiate
	Summing up the lesson. Parsing homework	inform students about homework, explain the methodology for its implementation motivate the need and obligation to do homework sum up the lesson

Forms of organization of cognitive activity of students:

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 readiness of the class to work.

II. Preparing students for active work (review)

Answer the questions.

What kind of movement is called uniform (movement at a constant speed).
What is the path formula for uniform motion ( S=Vt).
From this formula, express the speed and time.
Specify units of measure.

Conversion of speed units

III. The stage of generalization and systematization of the studied material (work in groups)

The whole class is divided into groups (5-6 people in a group). It is desirable that in the same group there are students of different levels of training. 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 sheets of thematic control, the group leader assigns marks to each student of the group for each task and notes the difficulties that students have in completing specific tasks.

Card with tasks for each group.

№ 5.

No. 7. The motorboat passed 112 km against the current of the river and returned to the point of departure, having spent 6 hours less on the way back. 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 passes along the river to the destination 513 km and after parking returns to the point of departure. Find the speed of the ship in still water, if the speed of the current is 4 km/h, the stay lasts 8 hours, and the ship returns to the point of departure 54 hours after leaving it. Give your answer in km/h.

Sample of thematic control card.

Class ________ Full name of the student ___________________________________

job number		Comment

Consultant cards.

Card number 1 (consultant)

1. Driving on a straight road

When solving problems of uniform motion, two situations often occur.

If the initial distance between the objects is equal to S, and the speeds 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 a snow drift. Half an hour later the path was cleared, and the driver, having increased 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 total distance.

Find the whole path: 450: 0.75 = 600 (km)
Let's find the length of the second section: 600 - 450 = 150 (km)
Let's make and solve the equation:

X= -75 is not suitable for the condition of the problem, where x > 0.

Answer: The initial speed of the train is 60 km/h.

Card number 2 (consultant)

2. Driving on a closed road

If the length of the 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 ;
b) when objects move in one direction, the time between their meetings is calculated by the formula

Example 2 At competitions on the ring track, one skier completes the circle 2 minutes faster than the other and after an hour has bypassed him exactly on the circle. How long does it take each skier to complete the lap?

Let be S m is the length of the ring road and x m/min and y m/min are the speeds of the first and second skiers, respectively ( x > y) .

Then S/x min and S/y min - the time for which the first and second skiers pass the circle, 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:

. Multiply both sides of the equation by 60 a(a + 2) > 0.

60(a + 2) – 60a = a(a + 2)a 2 + 2a- 120 = 0. The quadratic equation has one positive root a = 10 then b= 12. So the first skier completes the lap in 10 minutes, and the second skier in 12 minutes.

Answer: 10 min; 12 min.

Card number 3 (consultant)

3. Movement on the river

If an object moves along the river, then its speed is equal to Vstream. =Voct. + Vtech.

If an object is moving against the current of the river, then its speed is Vagainst the current =V oct. – Vtech. The object’s own speed (speed in still water) is equal to

The speed of the river is

The speed of the raft is equal to the speed of the river.

Example 3 The boat went downstream for 50 km and then traveled 36 km in the opposite direction, which took him 30 minutes longer than downstream. What is the speed of the boat 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 current of the river ( x- 4) km/h. The time of the boat's movement along the river is equal to hours, and against the flow of the river, hours. Since 30 minutes = 1/2 hour, then, according to the condition of the problem, we will compose 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 \u003d 0 x 1 \u003d 16, x 2 \u003d - 44 (we exclude, since x> 0).

So, the own speed of the boat is 16 km/h.

Answer: 16 km/h.

IV. Problem solving stage.

Problems that caused difficulties for students are analyzed.

No. 1. From two cities, the distance between which is equal to 480 km, two cars simultaneously left towards each other. In how many hours will the cars meet if their speeds are 75 km/h and 85 km/h?

75 + 85 = 160 (km/h) – closing speed.
480: 160 = 3 (h).

Answer: the cars will meet in 3 hours.

No. 2. From cities A and B, the distance between them is 330 km, two cars left towards each other at the same time 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 your 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. From point A to point B, the distance between which is 50 km, a motorist and a cyclist left at the same time. 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 make an equation, given that 4 hours 20 minutes =

It is obvious that x = -75 does not fit the condition of the problem.

Answer: The speed of the cyclist is 10 km/h.

No. 4. Two motorcyclists start simultaneously in one direction from two diametrically opposite points of a circular track, the length of which is 14 km. In how many minutes will the motorcyclists catch up for the first time if the speed of one of them is 21 km/h more than the speed of the other?

Let's make a table.

Let's make an equation.

where 1/3 hour = 20 minutes.

Answer: After 20 minutes, the motorcyclists will line up for the first time.

No. 5. From one point of the 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 make 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 after 40 minutes a motorcyclist followed him. 8 minutes after departure, he caught up with the cyclist for the first time, and 36 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the track is 30 km. Give your answer in km/h.

Let's make a table.

Movement to the first meeting

cyclist

No. 9. From pier A to pier B, the distance between which is 168 km, the first ship set off at a constant speed, and 2 hours after that, the second one set off after it, at a speed of 2 km / h more. Find the speed of the first ship if both ships arrive at point B at the same time. Give your answer in km/h.

Let's make a table, based on their conditions, that the speed of the first ship is x km / h.

Let's make an equation:

Multiplying both sides of the equation by x

Answer: the speed of the first ship is equal to the river 12 km/h

V. Summing up the lesson.

During the summing up of the lesson, once again, students should pay attention to the principles of solving problems on movement. When giving homework, give an explanation of the most difficult tasks.

Literature.

1) Article : Mathematics of the Unified State Examination 2014 (a system of tasks from an open bank of tasks) Koryanov A.G., Nadezhkina N.V. - published on the website

Sections: Maths

The article discusses tasks to help students: to develop the skills of solving text problems in preparation for the Unified State Examination, when learning to solve problems for compiling a mathematical model of real situations in all parallels of the primary and high schools. It presents tasks: for movement in a circle; to find the length of a moving object; to find the average speed.

I. Problems for motion in a circle.

Circumferential tasks proved to be difficult for many students. They are solved in almost the same way as ordinary problems for movement. They also use the formula . But there is a point to which we pay attention.

Task 1. A cyclist left point A of the circular track, and after 30 minutes a motorcyclist followed him. 10 minutes after departure, he caught up with the cyclist for the first time, and 30 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the track is 30 km. Give your answer in km/h.

Decision. The speeds of the participants will be taken as X km/h and y km/h. For the first time, the motorcyclist overtook the cyclist 10 minutes later, that is, one hour after the start. Up to this point, the cyclist has been on the road for 40 minutes, that is, hours. The participants in the movement have traveled the same distance, that is, y = x. Let's put the data in the table.

Table 1

The motorcyclist then overtook the cyclist a second time. This happened 30 minutes later, that is, one hour after the first overtaking. What distances did they travel? The motorcyclist overtook the cyclist. And that means he drove one lap more. That's the moment

to which you need to pay attention. One circle is the length of the track, It is equal to 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 track, and after 40 minutes a motorcyclist followed him. 10 minutes after departure, he caught up with the cyclist for the first time, and 36 minutes after that he caught up with him for the second time. Find the speed of the motorcyclist if the length of the track is 36 km. Give your answer in km/h.

I.1. 2. A cyclist left point “A” of the circular track, and after 30 minutes a motorcyclist followed him. 8 minutes after departure, he caught up with the cyclist for the first time, and 12 minutes after that, he caught up with him for the second time. Find the speed of the motorcyclist if the length of the track is 15 km. Give your answer in km/h.

I.1. 3. A cyclist left point “A” of the circular track, and after 50 minutes a motorcyclist followed him. 10 minutes after departure, he caught up with the cyclist for the first time, and 18 minutes after that, he caught up with him for the second time. Find the speed of the motorcyclist if the length of the track is 15 km. Give your answer in km/h.

Two motorcyclists start simultaneously in the same direction from two diametrically opposite points of a circular track, the length of which is 20 km. In how many minutes will the motorcyclists catch up for the first time if the speed of one of them is 15 km/h more than the speed of the other?

Decision.

Picture 1

With a simultaneous start, the rider who started from “A” drove half a lap more, who started from “B”. That is 10 km. When two motorcyclists move in the same direction, the removal speed is v = -. According to the condition of the problem, v= 15 km/h = km/min = km/min is the removal speed. We find the time after which the motorcyclists catch up 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 of a circular track, the length of which is 27 km. In how many minutes will the motorcyclists catch up for the first time if the speed of one of them is 27 km/h more than the speed of the other?

I.2.2. Two motorcyclists start simultaneously in the same direction from two diametrically opposite points of a circular track, the length of which is 6 km. In how many minutes will the motorcyclists catch up for the first time if the speed of one of them is 9 km/h more than the speed of the other?

From one point of the 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.

Decision.

x km/h is the speed of the second car.

(89 - x) km / h - removal speed.

8 km - the length of the circular track.

The equation.

(89 - x) = 8,

89 - x \u003d 2 15,

Answer: 59 km/h

Tasks (independently).

I.3.1. From one point of the 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 of the 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 of the 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 of the 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 8:00. After how many minutes will the minute hand align with the hour hand for the fourth time?

Decision. We assume that we do not solve the problem experimentally.

In one hour, the minute hand goes one circle, and the hour part of the circle. Let their speeds be 1 (laps per hour) and Start - at 8.00. Find the time it takes for the minute hand to overtake the hour hand for the first time.

The minute hand will go further, so we get the equation

So, for the first time, the arrows will line up through

Let the arrows line up for the second time after time z. The minute hand will travel a distance of 1 z, and the hour hand will travel one more circle. Let's write the equation:

Solving it, we get that .

So, through the arrows they will line up for the second time, another through - for the third, and even through - for the fourth time.

Therefore, if the start was at 8.00, then for the fourth time the arrows will line up through

4h = 60 * 4 min = 240 min.

Answer: 240 minutes.

Tasks (independently).

I.4.1. Clock with hands shows 4 hours 45 minutes. After how many minutes will the minute hand align with the hour hand for the seventh time?

I.4.2. A clock with hands shows exactly 2 o'clock. In how many minutes will the minute hand align with the hour hand for the tenth time?

I.4.3. The clock with hands shows 8 hours 20 minutes. After how many minutes will the minute hand align with the hour hand for the fourth time? fourth

II. Problems to find the length of a moving object.

A train moving at a uniform speed of 80 km/h passes a roadside post in 36 seconds. Find the length of the train in meters.

Decision. Since the speed of the train is indicated in hours, we will convert seconds into hours.

1) 36 sec =

2) find the length of the train in kilometers.

Answer: 800m.

Tasks (independently).

II.2. The train, moving uniformly at a speed of 60 km/h, passes a roadside post 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. Tasks for medium speed.

In a math exam, you may encounter the problem of finding the average speed. It must be remembered that the average speed is not equal to the arithmetic mean of speeds. The average speed is found by a special formula:

If there were two sections of the path, then .

The distance between the two villages is 18 km. The 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 for the entire journey?

Decision:

2 hours + 3 hours = 5 hours - spent on the whole movement,

A tourist walked at a speed of 4 km/h, then exactly the same time at a speed of 5 km/h. What is the average travel speed for the entire journey?

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 h he traveled 4t + 5t = 9t (km). The average speed of a tourist is = 4.5 (km/h).

Answer: 4.5 km/h.

We notice that the average speed of the tourist turned out to be equal to the arithmetic mean of these two speeds. It can be seen that if the time of movement on two sections of the path is the same, then the average speed of movement is equal to the arithmetic mean of the two given speeds. To do this, we solve the same problem in a general form.

The tourist walked at a speed of km / h, then exactly the same time at a speed of km / h. What is the average travel speed for the entire journey?

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 travel 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.

A car covered some distance uphill at a speed of km/h, and downhill at a speed of km/h.

What is the average speed of the car for the entire journey?

Let the length of the path segment be equal to s km. Then the car traveled 2 s km in both directions, spending the whole way .

The average movement speed 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. Equation required.

A cyclist was traveling uphill at a speed of 10 km/h, and downhill at some other constant speed. As he calculated, the average speed of movement was equal to 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 for 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 journey . What is the average speed for the entire journey 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 because the road is slightly downhill. Find the average speed of the cyclist all the way from home to work and back.

III.5. The 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 did he travel empty if the average speed was 70 km/h?.

III.6. The car drove the first 100 km at a speed of 50 km/h, the next 120 km at a speed of 90 km/h, and then 120 km at a speed of 100 km/h. Find the average speed of the car for the entire journey.

III.7. The car drove 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 for the entire journey.

III.8. The car drove 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 for the entire journey.

III. 9. The car drove 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 for the entire journey.

Problem 1. Two cars left point A for point B at the same time.
The first traveled all the way at a constant speed.
The second traveled the first half of the way at a speed
lower speed of the first by 14 km/h,
and the second half of the way at a speed of 105 km / h,
and therefore arrived at 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 as x.
Then, the time for which the first car traveled the entire distance,
equals 1/x.
At the second car speed the first half of the way, i.e. 1/2,
was 14 km/h less than the speed of the first car, x-14.
The time that the second car spent is 1/2: (x-14) = 1/2(x-14).
The second half of the way, i.e. 1/2, car passed
at a speed of 105 km/h.
The time he spent is 1/2: 105 = 1/2 * 105 = 1/210.
The time of the first and second are equal to each other.
We make an equation:
1/x = 1/2(x-14) + 1/210
We find a common denominator - 210x (x-14)
210(x-14) = 105x + x(x-14)
210x - 2940 \u003d 105x + x² - 14x
x² - 119x + 2940 = 0
Solving this quadratic equation through the discriminant, we find the roots:
x1 = 84
x2 \u003d 35. The second root does not fit the condition of the problem.
Answer: The speed of the first car is 84 km/h.

Problem 2. From point A of the circular track, 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.
After how many minutes the first motorist
will be ahead of the second 1 circle?
Decision: This task, despite the fact that it is given in the 11th grade,
can be solved at the elementary school level.
Let's ask only four questions and get four answers.
1. How many kilometers will the first motorist cover in 1 hour?
92 km.
2. How many kilometers will the second motorist cover 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 ahead of the second by 30 km?
30:15 = 2 hours = 120 minutes.
Answer: in 120 minutes.

Task 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 at one 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 task set before us,
you need to follow a certain plan.
And most importantly, we need to understand what we want out of it.
That is, what equation do we want to come to under the conditions that are given.
We will compare the time of each.
A car travels 90 km per hour more than a cyclist.
This means that the speed of the car is greater than the speed
cyclist at 90 km/h.
Assuming the speed of the cyclist is x km/h,
we get the speed of the car x + 90 km / h.
Travel time of a cyclist 60/s.
The travel time of the car is 60 / (x + 90).
5 hours 24 minutes is 5 24/60 hours = 5 2/5 = 27/5 hours
We make an equation:
60/x \u003d 60 / (x + 90) + 27/5 We 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 Does not fit the meaning of the task.
x2 = 10
Answer: The speed of the cyclist is 10 km/h.

Problem 4. A cyclist traveled 40 km from the city to the village.
On the way back he drove at the same speed
but after 2 hours of driving made a stop for 20 minutes.
After stopping, he increased his speed by 4 km/h
and therefore spent as much time on the way back from the village to the city as on the way from the city to the village.
Find the initial speed of the cyclist.
Solution: we solve this problem in relation to the time spent
first to the village and then back.
A cyclist traveled from the city to the village at the same speed x km/h.
In doing so, he spent 40/x hours.
He traveled 2 km back in 2 hours.
It remains for him to drive 40 - 2 km, which he passed
with a speed of x + 4 km/h.
The time it took him to get back
is made up of three terms.
2 hours; 20 minutes = 1/3 hour; (40 - 2x) / (x + 4) hours.
We make an equation:
40/x \u003d 2 + 1/3 + (40 - 2x) / (x + 4)
40/x \u003d 7/3 + (40 - 2x) / (x + 4) Common denominator 3x(x + 4)
40*3(x + 4) = 7x(x + 4) + 3x(40 - 2x)
120x + 480 \u003d 7x² + 28x + 120x - 6x²
x² + 28x - 480 = 0 Solving this equation through the discriminant or Vieta's theorem, we get:
x1 = 12
x2 = - 40 Not suitable for the condition of the problem.
Answer: The cyclist's initial speed 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 overtakes another, they speeds are subtracted.
If we take the speed of the third car as x km/h,
then it turns out that he caught up with the second car in 20/(x-40) hours.
Then he will overtake the first car in 25/(x - 50) hours.
We 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.