How to solve traffic problems? Methods for solving motion problems. Clarification when driving into the oncoming lane is allowed

In life, we often have to deal with quantities: distance, time, speed of movement. When solving such problems, we proceed from the fact that all bodies move at a constant speed and along a straight path. This is far from reality, but even with such a simplification of real conditions, quite digestible results can be obtained by finding the value of one of these quantities from the values ​​of the other two.

Task 1. From Leningrad to Tallinn 360 km, the bus covers this distance in6 h . Find the speed of the bus.

In this problem, the distance between cities is 360 km, the bus travel time is 6 hours. It is required to find the speed of the bus.

Solution. 360:60=60 (km per hour).

Answer. The speed of the bus is 60 km per hour.

Compose and solve inverse problems.

Task 2. From Leningrad to Tallinn 360 km. How long does it take the bus to cover this distance if it travels at a speed of 60 km per hour?

Solution. 360:60=6 (h)

Answer. Bus time? h.

Task 3. A bus moving at a speed of 60 km per hour covers the distance from Leningrad to Tallinn in 6 hours. Find the distance from Leningrad to Tallinn.

Solution. 60*?=360 (km).

Answer. The distance from Leningrad to Tallinn is 360 km.

If we denote the distance through, speed through, time of movement through, then the relationship between distance, speed and time of movement can be written by the formulas:

2. Tasks for oncoming traffic.

In life, we observe the oncoming movement. If we go out onto the streets of the city, we will see how pedestrians are moving towards each other along the sidewalk, along the pavement - trolleybuses, buses, trams, cars and trucks, cyclists, motorcyclists. On the rivers of the city boats go towards each other. On the railway, trains rush past each other, planes fly in the sky.

The tasks associated with oncoming traffic are varied. First of all, let's find out what quantities we have to deal with when there is an oncoming movement, and what is the relationship between them.

Let two pedestrians leave points A and B at the same time towards each other. One at a speed of 4 km per hour, the other at 5 km per hour.

4 km per hour 5 km per hour

In an hour, pedestrians will walk 4 + 5 = 9 (km) together. The distance between them will decrease by 9 km. In other words, they will approach each other in an hour of movement of 9 km. The distance that two pedestrians approach each other in an hour is called the speed of their convergence. 9 km per hour - approach speed pedestrians.

If the speed of convergence of pedestrians is known, then it is easy to find out how much the distance between them will decrease in 2 hours, 3 hours of movement towards each other. 9 * 2 \u003d 18 (km) - the distance between pedestrians will decrease by 18 km in 2 hours 9 * 3 = 27 (km) - the distance between pedestrians will decrease by 27 km in 3 hours.

Every hour the distance between pedestrians decreases. There will come a time when they meet.

Let the distance between A and B be 36 km. Find the distance between the pedestrians 1 hour after they left points A and B after 2 hours, 3 hours, 4 hours.

After 1 hour	After 2 hours	After 3 hours	After 4 hours
36 – 9= 27 (km)	36 – 9*2 = 18 (km)	36 – 9*3 = 9 (km)	38 – 9*4 = 0 (km)

4 hours after leaving points A and B, the pedestrians will meet.

Considering the oncoming movement of two pedestrians, we dealt with the following quantities:

one). The distance between the points from which the simultaneous movement begins;

2). approach speed;

3). The time from the start of the movement to the moment of meeting (movement time).

Knowing the value of two of these three quantities, you can find the value of the third quantity.

The table contains the conditions of the problems that can be compiled about the oncoming movement of two pedestrians.

	Approach speed	Time from the start of the movement to the moment of meeting per hour	Distance from A to B

We express the relationship between these quantities by the formula. Let us denote by – the distance between and; – the speed of approach; – the time from the moment of exit to the moment of meeting.

In problems for oncoming traffic, the approach speed is most often not given, but it can be easily found from the problem data.

A task. Two pedestrians left two points A and B at the same time towards each other. One at a speed of 4 km per hour, the other at 5 km per hour. They met after 3 hours. Find the distance between points A and B.

Graphical illustration of the task:

4 km per hour 5 km per hour

after 3 hours

To find the distance between points, you can multiply the speed of approach by the time of movement, the speed of approach is equal to the sum of the speeds of pedestrians. Solution formula: \u003d (4 + 5) * 3; \u003d 27.

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

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

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

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

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

Consider a table.

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

Examples of problem solving.

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

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

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

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

Let's consider other tasks.

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

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

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

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

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

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

You are already familiar with the concept of "average speed" and you know how the quantities speed, time and distance are related. We will solve more complex problems.

Two skiers left two villages towards each other at the same time and met after 3 hours. The first skier walked at an average speed of 12 km/h, the second - 14 km/h. Find the distance between villages. See illustration in figure 1.

Rice. 1. Illustration for problem 1

To find the distance between villages, we need to know how far each skier traveled. To find the distance that a skier has traveled, you need to know his average speed and the time he traveled.

We know that the skiers went out towards each other at the same time and were on the road for 3 hours. This means that each skier was on the road for three hours.

The average speed of one skier is 12 km/h, the travel time is 3 hours. If the speed is multiplied by the time, then we find out how far the first skier traveled:

The average speed of the second skier is 14 km / h, the travel time is the same as that of the first skier - three hours. To find out how far the second skier has traveled, multiply his average speed by his travel time:

Now we can find the distance between the villages.

Answer: the distance between the villages is 78 km.

In the first hour one skier walked 12 km, in the same hour the second skier walked 14 km towards the first skier. We can find the speed of convergence:

We know that in every hour the skiers moved closer to each other by 26 km. Then we can find how far they approached in 3 hours.

Multiplying the speed of approach by time, we found out how far the two skiers traveled, that is, we found out the distance between the villages.

Answer: the distance between the villages is 78 km.

From two villages, the distance between which is 78 km, two skiers left at the same time towards each other. The first skier walked at an average speed of 12 km/h, and the second - 14 km/h. How many hours later did they meet? (See figure 2).

Rice. 2. Illustration for problem 2

To find the time after which the skiers will meet, you need to know the distance that the skiers have traveled and the speed of both skiers.

We know that every hour the first skier approached the meeting point at 12 km, and the second skier approached the meeting point at 14 km. That is, together they approached for every hour by:

We found the speed of convergence of skiers.

We know the entire distance that the skiers have traveled and we know the rate of convergence. If the distance is divided by the speed, then we get the time after which the skiers met.

Answer: the skiers met after 3 hours.

From two villages, the distance between which is 78 km, two skiers left at the same time towards each other and met after 3 hours. The first skier walked at an average speed of 12 km/h. What was the average speed of the second skier? (See figure 3.)

Rice. 3. Illustration for problem 3

To find out the average speed of the second skier, you need to know how far the skier traveled to the meeting point and how long he traveled. To find out how far the second skier traveled to the meeting point, you need to know how far the first skier traveled and the total distance. The total distance covered by both skiers, we know - 78 km. To find the distance traveled by the first skier, you need to know his average speed and the time he traveled. The average speed of the first skier is 12 km / h, he was on the road for three hours. If the speed is multiplied by the time, we get the distance that the first skier traveled.

We know the total distance, 78 km, and the distance that the first skier traveled, 36 km. We can find how far the second skier traveled.

We now know how far the second skier traveled, and we know how long he traveled - 3 hours. If the distance traveled by the second skier is divided by the time he traveled, we get his average speed.

Answer: the average speed of the second skier is 14 km/h.

Today we learned to solve problems for oncoming traffic.

Bibliography

1. Maths. Textbook for 4 cells. early school At 2 o'clock / M.I. Moro, M.A. Bantova. - M.: Education, 2010.
2. Demidova T.E., Kozlova S.A., Tonkikh A.P. Maths. 4th grade. Textbook at 3 hours. 2nd ed., corrected. - M.: 2013.; Ch. 1 - 96 p., Ch. 2 - 96 p., Ch. 3 - 96 p.
3. Mathematics: textbook. for the 4th class. general education institutions with Russian. lang. learning. At 2 p.m. Part 2 / T.M. Chebotarevskaya, V.L. Drozd, A.A. joiner; per. with white lang. L.A. Bondareva. - 3rd ed., revised. - Minsk: Nar. asveta, 2008. - 135 p.: ill.

1. Uchit.rastu.ru ().
2. For6cl.uznateshe.ru ().
3. Volna.org().

Homework

1. Try to solve problem number 3 in a different way.
2. The distance between two cyclists is 240 m. They left at the same time towards each other and met after 30 seconds. What is the speed of the first cyclist if the speed of the second is 3 m/s?
3. Towards each other from two villages, the distance between which is 30 km, two pedestrians left at the same time. One was walking at a speed of 4 km/h and the other was walking at a speed of 5 km/h. How many kilometers will they converge in 1 hour of travel? And in three hours?

The Rules of the Road describe four cases that prohibit driving into the oncoming lane:

clause 9.2."On roads with two-way traffic having four lanes or more, it is forbidden to enter the side of the road intended for oncoming traffic. On such roads, left turns or U-turns can be made at intersections and in other places where it is not prohibited by the Rules, signs and (or ) markup.

clause 9.3."On two-way roads with three lanes marked with markings (with the exception of marking 1.9 (marking the boundaries of lanes on which reverse traffic is carried out)), of which the middle one is used for traffic in both directions, it is allowed to enter this lane only for overtaking , a detour, a left turn or a U-turn. It is prohibited to enter the leftmost lane intended for oncoming traffic. "

clause 11.4.“Overtaking with exit into the oncoming lane is prohibited:

• at regulated intersections, as well as at unregulated intersections when driving on a road that is not the main one;
• at pedestrian crossings;
• at railway crossings and closer than 100 meters in front of them;
• on bridges, overpasses, overpasses and under them, as well as in tunnels;
• at the end of a climb, on dangerous curves, and in other areas with limited visibility."

At the same time, limited visibility is determined by signs 1.11.1, 1.11.2 "Dangerous turn", 1.12.1, 1.12.2 - "Dangerous turns", 1.29 "tunnel", 1.31.1, 1.31.2 "Direction of turn on the rounding of the small road radius with limited visibility".

clause 15.3."... It is forbidden to go around with an exit to the oncoming traffic lane vehicles standing in front of the crossing ...".

According to the rules of the road, horizontal marking lines, and it is forbidden to cross!

As a result of violations of the requirements of road signs: that entailed traffic in the opposite direction on a road intended for one-way traffic, an administrative offense is also formed under Part 4 and Part 3 of the Code of Administrative Offenses of the Russian Federation.

Any violation by drivers of the requirements of road signs or markings, which led to the exit to the side of the carriageway intended for oncoming traffic, should be qualified under parts 3 and 4 of Article 12.15 and under part 3 of Article 12.16 of the Code of Administrative Offenses of the Russian Federation.

Oncoming traffic penalty

To date, the following oncoming punishment:

In addition to the absence of a marking transition to a solid line in court, an “emergency” can help you. Those. if you have performed this maneuver in order to prevent an emergency. For example, they avoided vehicles leaving a secondary road, from bricks flying at you from the body of a KAMAZ.

If everything is in order with the markup and there was no emergency, then even the best lawyer can not help here. In any case, you have only 10 days to appeal against the decision, so a lawyer's consultation will not hurt. An experienced lawyer will advise you on how to avoid deprivation of rights by simply paying a fine, because. going into the opposite lane in any case is punished.

Related video:

Mikhail Alekseev, automotive journalist
Publication date: 27.10.2011
Updated by Vitaly Dmitriev on 04/14/2016
Reprinting without an active link is prohibited!

Counsel's Comments

Yevgeny Lipatov, lawyer of the Spravami.ru Center for Assistance to Car Owners, answers the questions of the editors of our site.

Often there are cases when the markings on the road are erased or covered with snow. As a result, it turns out that the car crosses a double solid marking line. How to be in such situations?

In such cases, it is better for the driver to take photographs of the section of the road where the road markings are erased or covered with snow in order to prove the fact that the traffic violation was not the fault of the driver. Photographing is best done in the presence of witnesses who, if necessary, can confirm when and where it was taken.

If the sign "Go straight or right" hangs, and the driver makes a left turn, is this considered an oncoming lane? If so, what is the punishment for this action: a fine or imprisonment?

The driver's actions will be qualified as an exit to the "oncoming lane" in the event that he, having violated the requirements of the prescriptive sign, drove onto a one-way road. Otherwise, he will be liable under Art. 12.16 of the Code of Administrative Offenses of the Russian Federation, the sanction of which provides for a fine of 100 rubles. As a rule, on the side from which traffic on a one-way road is prohibited, a road sign "No Entry" ("brick") is installed. If, when making a maneuver, the driver crossed a solid marking line, then his actions will be qualified under Part 3 of Article 12.15 of the Code of Administrative Offenses of the Russian Federation (Entering tram tracks in the opposite direction, as well as leaving in violation of the Rules of the Road on the side of the road intended for oncoming traffic, combined with a U-turn, a left turn or a detour around an obstacle), which will entail the imposition of a fine from one thousand to one and a half thousand rubles.

However, there are times when driving into the oncoming lane will not be punished. Such cases are quite common. An oncoming lane will not be considered an obstacle detour in the oncoming lane if there are simply no other exits. You just need to do it carefully, without creating interference. For example, on a road with two lanes for traffic in one direction, the obstruction in the left lane is bypassed only on the right side. But if there is only one lane, then driving into the oncoming lane is allowed.

If you were caught "for an oncoming lane", most likely the inspector will draw up a protocol and withdraw your license. The review team then decides whether to impose a fine or take the matter to court. Usually they look at history on a single basis. If a person has not had serious problems with traffic rules over the past year, a fine is imposed directly in the analysis group. Although, some analysis groups, having seen Article 12.15.4 in the protocol, immediately issue a subpoena. Judges also look primarily at the "track record", and also listen to the violator. If the latter is remorseful or has extenuating circumstances (such as breaking the rules by a pregnant woman), the court may order a fine.

Therefore, if you are caught under article 12.15.4, it is best to agree with the traffic police officer and give him the rights. And then, when parsing, "repent" to the inspector. In extreme cases, you will be sent to court. But unlike traffic police officers, judges are more loyal to those who plead guilty and impose the minimum punishment - a fine. But if you are firmly convinced that you have not violated the rules, and the protocol was drawn up unfairly, start the "war" from the very first minutes. In the protocol in the column "Explanation of the offender" write: "I did not violate traffic rules, qualified legal assistance is required." Next, make a complaint (either with the help of a lawyer or on an online forum) and register it with the traffic police department. The answer to the complaint must be given within 10 days, but in reality it takes up to a month. And the court has only two months from the date of violation to deprive you of your rights. And then there are high chances that the case simply will not be considered in time.

Now let's take a closer look at the situations that can happen on the road. At first they look ambiguous, but one must always understand which violation should be followed by deprivation of rights, and for which - a fine.

Reviews and opinions in the automotive communities

Georgy Fedorov Sep 12, 2008 at 0:07 am
My brother once overtook a car that was driving halfway along the side of the road for 20 km. at one o'clock. The markup is intermittent. The sign was 3.20. Stopped. Wrote part 4. I went, photographed from all angles the place, markings, traces of the car on the side of the road, measured all the distances with a tape measure. I made a beautiful scheme with indications of the dimensions of the road, the width of the lane, the width of the car. He explained to the judge that two of these cars could easily fit in the same lane. I dictated an explanation to my brother. I made my own explanation. He filed a petition to call the inspector (I don’t know why). The inspector didn't show up. On the last day of the term, they considered it without an inspector and closed it for lack of staff.

Georgy Fedorov Sep 12, 2008 at 0:08 am
Explanation in the course of administrative proceedings:

On May 18, 2008, Nikolay Nikolaevich Zarucheinikov, senior police lieutenant, was drawn up by the inspector for the investigation of the OGIBDD of the city of Kolpino on administrative offense No. 78 AA in relation to Fedorov M.V.
From the explanation of Fedorov M.V. it can be seen that he did not enter the lane intended for the movement of vehicles in the opposite direction, and did not cross the marking line 1.5 separating oncoming traffic flows, but performed ahead of the vehicle moving along the side of the road.
The dimensions of vehicles and the roadway (which can be seen from the diagram provided by the defender to the court as evidence) allow you to perform this action without leaving the oncoming lane, even if the car that Fedorov M.V. did not move along the side of the road, but along the same lane.
In accordance with clause 1.2 of the Rules of the Road of the Russian Federation, overtaking is considered to be "the advance of one or more moving vehicles associated with the departure from the occupied lane."
In accordance with paragraph 9.1 of the SDA of the Russian Federation, “The number of lanes for roadless vehicles is determined by markings and (or) signs 5.15.1, 5.15.2, 5.15.7, 5.15.8, and if they are not there, then by the drivers themselves, taking into account the width the carriageway, the dimensions of vehicles and the necessary intervals between them.
On this road there is a marking 1.5, as can be seen from the photographs presented to the court as evidence and from the explanation of Fedorov M.The. Therefore, on this section of the road, the number of lanes for traffic is determined by the markings and is equal to two. One row in each direction.

I ask the court to pay attention to the fact that the place of the administrative offense in the protocol is house No. 55 on Zagorodnaya Street, and the place where the protocol was drawn up is house No. 47 on the same street where the traffic police car stood, in which inspector N.N. at the time of stopping Fedorov M.The. an intern of the traffic police squad, who was on duty together with inspector Zarucheinikov N.N. The numbering of houses along Zagorodnaya Street is one-way and sequential, and these two houses are separated by a distance of 250 meters and a turn of the road. In addition, the traffic police car was parked in a residential area among trees and shrubs. These facts indicate that Inspector Zarucheinikov N.N. could not see the alleged offense.
According to Article 28.1 of the Code of Administrative Offenses of the Russian Federation, the reasons for initiating a case on an administrative offense are:
1) direct discovery by officials authorized to draw up protocols on administrative offenses of sufficient data indicating the presence of an event of an administrative offense;
2) materials received from law enforcement agencies, as well as from other state bodies, local self-government bodies, from public associations, containing data indicating the presence of an event of an administrative offense;
3) messages and statements of individuals and legal entities, as well as messages in the media containing data indicating the presence of an event of an administrative offense.
Since Inspector Zarucheinikov N.N. is not a person who directly discovered sufficient data on the occurrence of an event of an administrative offense, he did not receive messages, statements and other materials from state and law enforcement agencies, as well as individuals and legal entities, did not receive messages from the media, and so on, he did not have the right to draw up a protocol on an administrative offense.

According to paragraph 2 of Article 28.2 of the Code of Administrative Offenses of the Russian Federation, witnesses, if any, must be indicated in the protocol on an administrative offense.
The Plenum of the Supreme Court of the Russian Federation in Resolution No. 5 dated March 24, 2005 in paragraph 4, he indicated that “A significant drawback of the protocol is the lack of data directly listed in part 2 of article 28.2 of the Code of Administrative Offenses of the Russian Federation, and other information depending on their significance for this particular case of an administrative offense”
The traffic police trainee, who was on duty together with inspector Zarucheinikov N.N., stopped the car of Fedorov M.V. allegedly for violating paragraph 1.3 of the traffic rules, after which a protocol was drawn up by inspector N.N. Zarucheinikov.
It follows that the traffic police trainee who stopped the car of Fedorov M.The. is a witness in this case and his data should have been recorded in the protocol on an administrative offense, which was not done.
Based on the foregoing, and guided by paragraph 2 of Article 28.2 of the Code of Administrative Offenses of the Russian Federation, as well as paragraph 3 of Article 26.2 of the Code of Administrative Offenses of the Russian Federation, which states that “it is not allowed to use evidence obtained in violation of the law”, I consider that it is unacceptable to use a protocol on an administrative offense as evidence in this case, as it was drawn up in violation of the law.

Also protection based on the words of Fedorov M.V. expresses its disagreement with the scheme drawn up by Inspector Zarucheynikov N.N., available in the case.

Since the explanations and diagram provided by Inspector Zarucheinikov N.N. to a certain extent conflict with the explanations and the scheme provided by Fedorov M.V., when considering the case, one should take into account the fact that the inspector is a person concerned, since when considering the case, facts of exceeding his official powers in terms of the illegality of drawing up the protocol may be established about an administrative offence.

In the event of a contradiction in the evidence presented to the court, and the impossibility of providing additional evidence, there are irremovable doubts about the guilt of the person brought to administrative responsibility.
Article 1.5 of the Code of Administrative Offenses of the Russian Federation establishes that irremovable doubts about the guilt of a person brought to administrative responsibility are interpreted in favor of this person. This is confirmed by the Supreme Court of the Russian Federation, which, in paragraph 13 of the Decree of the Plenum of the Supreme Court of the Russian Federation No. 5 dated March 24, 2005, indicated, “When considering cases of administrative offenses, as well as complaints against decisions or decisions in cases of administrative offenses, the judge must proceed from 1.5 of the Code of Administrative Offenses of the Russian Federation of the principle of administrative responsibility - the presumption of innocence of the person against whom the proceedings are carried out. The implementation of this principle lies in the fact that a person brought to administrative responsibility is not obliged to prove his innocence, guilt in committing an administrative offense is established by judges, bodies, officials authorized to consider cases of administrative offenses. Irremovable doubts about the guilt of a person brought to administrative responsibility must be interpreted in favor of this person.

Based on the foregoing, I declare that Fedorov M.The. did not commit an action qualified by the Rules as “overtaking” and, therefore, did not violate the requirements of sign 3.20 prohibiting overtaking vehicles, and did not drive into the oncoming traffic lane, for which liability is provided for in article 12.15, paragraph 4 of the Code of the Russian Federation on Administrative Offenses.
I also consider what was started in relation to Fedorov M.V. case of an administrative offense groundless due to the absence of an event of an administrative offense.

If the court disagrees with any information contained in the evidence and explanations presented, I ask the court to request this information from the relevant authorities and question the witnesses.
I ask the court to pay attention to the fact that for a three-year driving experience Fedorov M.The. committed only one minor offense, which is confirmed by a certificate of offenses available in the case, and is an indirect evidence of the decency and law-abidingness of the accused.

Based on the foregoing and based on Article 1.5; subparagraph 1 of paragraph 1 of Article 24.5; paragraph 5 of Article 25.5 of the RF Code of Administrative Offenses, I ask the court to stop what was started in relation to Fedorov M.The. proceedings in the case of an administrative offense.

I request that this explanation be attached to the case file.

Vyacheslav Sukhov Sep 27, 2008 at 11:11 am
Friends I want to share my experience in resolving sedebnlgl proceedings in my direction. August 02, 2008 at 11:15 am following the Luga River from SB with my wife in order to have a rest. In the area of ​​Krasnoye Selo, at the exit from the viaduct (Lenin St. 1.), overtook the truck together where the marking 1.11 was applied. After driving 390m from the viaduct, a traffic police officer stops me and claims that I violated the overtaking truck in the area where the overtaking is prohibited sign (which actually stood at the entrance to the viaduct.

After long and persistent arguments, we got into their garbage truck, and a divorce began, article 12.15 of the CRF on the AP. ch4. deprivation of money in general, but let's not say exactly how much.

In connection with my complete innocence was sent nah .... in the literal sense.
when signing the protocol, I will tell him that they will not deprive me of my rights ...

Preparation began for 2-days made speeches for the court.

Your Honor!
On August 02, 2008 at 11:15 a.m., the traffic police inspector in relation to me, Vyacheslav Sukhov, drew up a protocol on an administrative offense AD No. 46 84 31
In the Protocol, the Inspector indicated that I (literally) while overtaking, crossed the road markings 1.1, made an exit into the oncoming traffic lane.

On the fact of the administrative offense case initiated against me, I consider it necessary to explain the following:
While overtaking a car, GAZ actually crossed the road markings at the exit from the viaduct (the place where the road markings are crossed by my car photo No. 4, BUT in the place where the road markings applied to the roadway correspond to marking 1.11 - this marking separates traffic flows of opposite or passing directions on sections of roads where changing lanes are allowed only from one lane; indicates places intended for a U-turn, entry and exit from parking lots and the like, where traffic is only allowed in one direction.

And in connection with the extreme need for this maneuver, because. A.M. The gas moved in violation of the row (To put it simply, it chatted along the road)

I also feel the need to clarify the following:
That the distance from the end of the exit from the viaduct to the location of the traffic police officer who drew up the protocol on an administrative offense was about 350-370 meters Photo No. 2 This fact casts doubt on the fact that the traffic police officer could clearly see the intersection of the road markings Photo No. traffic police officer with an optical 3-fold approximation.
At the entrance to the viaduct there is a sign 3.20 "No overtaking" Photo #1
“Signs 3.20 and 3.22 are installed with one of the plates 8.5.4-8.5.7 (“Validity time”) on roads with three or less traffic lanes in both directions in cases of increased risk of collision with oncoming and passing vehicles, depending on the intensity traffic, width and condition of the carriageway.
Sign 3.20 is installed on sections of roads with unsecured visibility of an oncoming vehicle (Table 3), the area of ​​the sign in this case is determined by the length of the dangerous section.

Thus, there are two options for installing the sign "Overtaking is prohibited" - with the duration of the sign and with the area of ​​the sign. If there is no sign "Validity time", then the sign prohibiting overtaking is installed due to limited visibility and therefore the effect of the sign according to GOST should be limited to this zone.
To limit the coverage area, the sign "Overtaking is prohibited" (clause 5.4.31 GOST R 52289-2004) is placed with the sign "Sign coverage area", or the sign "End of the overtaking prohibition zone" is installed.

If, after the “Overtaking is prohibited” sign, intermittent markings begin, in fact, the markings begin where the limited visibility zone ended, that is, I drove into the oncoming traffic lane where exit is not prohibited. Marking and permanently installed road signs cannot contradict each other. The statements of the traffic police inspector that the sign has priority over the markings - a fairy tale - because. this is not written in the SDA, GOST or any other Regulatory Legal Act.
Only a temporary sign has an advantage over the markup. A temporary sign means a sign on a portable stand

Due to the absence of an event of an administrative offense in my actions, I consider it necessary to terminate the case of an administrative offense against me.

Attached to the speech were photographs of the markings, a photo from the place where the traffic cop stood, and the entrance to the viaduct in order to prove that the overtaking sign is not temporary.

The judge ruled to summon the traffic cop to testify...

After that, there were 4 trials for 3 of them, the traffic cop did not appear, the judge who decided to call the traffic cop went on vacation and I was sent to another station, where the judge had to explain everything again.

A traffic cop came to the 4th court and for the first time they showed me the diagram that he drew and to my surprise at the meeting, 3-cars were drawn right in front of me.

When overtaking visibility was 1.5-2 km. and not a single oncoming car.

Well, then it struck me that your honor, how could he see in what place I was overtaking if it was 390 m before him and there were 3 cars right in front of me.

Judge ... really HOW???

And then the traffic cop started up ... I saw everything, I saw everything, I saw everything.

In general, the judge kicked him out and said that if I again appear in the 105 precinct, they will definitely deprive me of ...

By the way, for 5 visits I have seen enough of such a number of people who in court could not connect 2 words, well, as a result, 0.4-0.6-1.5-2 years of deprivation.
At the 5th court, having seen enough of how our justice is going on with the rights, I said goodbye ahead of schedule.

Memo "Learning to solve problems for movement"

In motion problems, three interrelated quantities are considered:

S - distance (traveled path),

t - travel time and

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

Distance is the product of speed and time of travel.

S = V t

Speed ​​is the quotient of distance divided by time traveled.

V=S:t

Time is the quotient of distance divided by speed.

t = S : V

Tasks for oncoming traffic

If two bodies simultaneously move towards each other, then the distance between them constantly changes by the same number, equal to the sum of the distances that the bodies travel per unit time.

The speed of approach is the sum of the speeds of bodies moving towards each other. V approx. = 1V + 2V

Example 1 Two cyclists left two villages towards each other at the same time and met after 3 hours. The first cyclist was traveling at a speed of 12 km/h, and the second - 14 km/h. How far are the villages?

Scheme for the task:

Solution:

S = V t

V approx. = 1V + 2V

1) 12 3 \u003d 36 (km) - the first cyclist rode before the meeting

2) 14 3 \u003d 42 (km) - the second cyclist rode before the meeting

3) 36 + 42 = 78 (km)

1) 12 + 14 = 26 (km/h) – closing speed

2) 26 3 = 78 (km)

Answer: the distance between the settlements is 78 km.

Example 2 Two cars left two cities towards each other. The speed of the first is 80 km/h, the speed of the second is 60 km/h. In how many hours will the cars meet if the distance between cities is 280 km?

Scheme for the task:

Solution:

V approx. = 1V + 2V

t = S : V

1) 80 + 60 = 140 (km/h) – closing speed

2) 280: 140 = 2 (h)

Answer: cars will meet in 2 hours.

Example 3 From two cities, the distance between which is 340 km, two cars left at the same time towards each other. The speed of the first is 80 km / h. What was the speed of the second car if they met after 2 hours?

Scheme for the task:

Solution:

V=S:t

2V = V approx. - 1V

1) 340: 2 = 170 (km/h) – closing speed

2) 170 - 80 = 90 (km/h)

Answer: 90 km/h. second car speed

Tasks for movement in opposite directions

If two bodies simultaneously move in opposite directions, then the distance between them gradually increases.

The speed of removal is the distance that bodies travel in 1 hour when moving in opposite directions. V remote = 1V + 2V

Example 1 Two skiers left point A at the same time in opposite directions. The first skier walked at a speed of 12 km/h, and the second - 14 km/h. How far apart will they be in 3 hours?

Scheme for the task:

Solution:

S = V t

1 way

1) 12 3 \u003d 36 (km) - the distance that the first skier traveled in 3 hours

2) 14 3 \u003d 42 (km) - the distance that the second skier traveled in 3 hours

3)36 + 42 = 78 (km)

2 way

V remote = 1V + 2V

S = V t

1) 12 + 14 \u003d 26 (km / h) - removal speed

2)26 3 = 78 (km)

Answer: in 3 hours they will be at a distance of 78 km from each other.

Example 2 Two cars left the city in opposite directions. The speed of the first is 80 km/h, the speed of the second is 60 km/h. In how many hours will the distance between the cars be 280 km?

Scheme for the task:

V remote = 1V + 2V

t = S : V

1) 80 + 60 \u003d 140 (km / h) - removal speed

2) 280: 140 = 2 (h)

Answer: after 2 hours the distance between the cars will be 280 km

Example 3 Two cars left the city at the same time in opposite directions. The speed of the first is 80 km / h. How fast was the second car traveling if after 2 hours the distance between them was 340 km?

Scheme for the task:

Solution:

V=S:t

2V= V remote - 1V

1) 340: 2 = 170 (km/h) – vehicle removal speed

2) 170 - 80 = 90 (km/h)

Answer: The speed of the second car is 90 km/h.

Tasks for movement in one direction

Example 1. A car traveled 192 km in 2 hours. For the next 3 hours, he moved at a speed of 6 km/h less. How many kilometers did the car drive in total?

Scheme for the task:

1) 192: 2 = 96 (km/h) - first speed

2) 96 - 6 = 90 (km / h) - second speed

3) 90 3 \u003d 270 (km) - second distance

4) 192 + 270 = 462 (km)

Answer: 462 km.

Example 2. From two points, the distance between which is 24 km, an athlete left and a cyclist left at the same time. The speed of an athlete is 6 km/h and the speed of a cyclist is 18 km/h.

1). In how many hours will the cyclist overtake the athlete?

2). At what distance from point B will the cyclist overtake the athlete?

3). By how many kilometers is the cyclist's distance longer than the athlete's?

18 km/h 6 km/h?

V approx. = 2V-1V , where 2Vֺ > 1V

t = S : V

one). 18 - 6 \u003d 12 (km / h) - the speed of approach of the cyclist and athlete

2). 24: 12 \u003d 2 (h) - the time after which the cyclist will catch up with the athlete.

3). 6 ●2 = 12 (km) – the distance at which the cyclist will overtake the athlete.

Answer: after 2 hours; 12 km.

Example 3. How long will it take for a motorcycle to overtake a truck if the distance between them is 45 km and the speed of the motorcycle is 15 km/h greater than the speed of the truck?

Answer: after 3 hours.

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