Many people do not like motion problems, because they often misunderstand how to solve them. But, as you know, nothing is impossible, and therefore you can learn how to solve problems for movement, there would be a desire.

How to solve motion problems: theory

All tasks related to movement are solved according to one formula, which you must know by heart. Here it is: S=Vt. S is distance, V is speed, and t is time.

This formula is the key to solving all these problems, and everything else is written in the text of the problem, the main thing is to read and understand the problem carefully.

Second important point, is the reduction of all data in the problem of quantities to single units measurements. That is, if the time is given in hours, then the distance should be measured in kilometers, if in seconds, then the distance in meters, respectively.

Problem solving

So, let's look at three main examples for solving motion problems.

Two objects left one after another.

Suppose you are given the following task: the first car left the city at a speed of 60 km/h, after half an hour the second car left at a speed of 90 km/h. After how many kilometers will the second car overtake the first one? To solve such a problem, we have a formula: t = S / (v1 - v2). Since we know the time, but not the distance, we will transform it S = t (v1 - v2) .We substitute the numbers: S=0.5(90-60), S=15 km. That is, both cars will meet in 15 km.

Two objects left in the opposite direction

If you are given a task in which two objects left towards each other, and you need to find out when they will meet, then you need to apply the following formula: t \u003d S / (v1 + v2). For example, from point A and B, between which 43 km , a car was driving at a speed of 80 km/h, and a bus was driving from point B to A at a speed of 60 km/h. After how long will they meet? Solution: 43/(80+60)=0.30 hours.

Two objects left at the same time in the same direction

The problem is given: a pedestrian moving from point A to point B left at a speed of 5 km/h, and a cyclist left at a speed of 15 km/h. How many times faster will a cyclist get from point A to point B if it is known that the distance between these points is 10 km. First you need to find the time it takes for a pedestrian to cover this distance. We redo the formula S=Vt, we get t=S/V. We substitute the numbers 10/5=2. that is, the pedestrian will spend 2 hours on the road.

Now we calculate the time for the cyclist. t \u003d S / V or 10/15 \u003d 0.7 hours. The third step is very simple, we must find the time difference between a pedestrian and a person on a bicycle. 2/0.7=2.8. The answer is that the cyclist will get to point B 2.8 times faster than the pedestrian.

Thus, by applying these simple formulas, you will always know how motion problems are solved. You just need to read the problem very carefully, take into account all the data, bring them into one measurement system, and then choose the right formula for solving.

But be careful, it is not necessary that your task will have only one action, sometimes, before applying our formulas, you will have to perform a number of intermediate actions in order to find the necessary data. Do not forget about them, and then you will certainly succeed.

Task 1.

From the village and the city towards each other, two buses left at the same time. One bus traveled 100 km before the meeting at a speed of 25 km/h. How many kilometers before the meeting the second bus traveled if its speed is 50 km/h.

Decision:
• 1) 100: 25 = 4 (one bus drove for hours)
• 2) 50 * 4 = 200
• Expression: 50 * (100: 25) = 200
• Answer: the second bus traveled 200 km before the meeting.

Task 2.

The distance between the two marinas is 90 km. From each of them, two ships left at the same time towards each other. How many hours will it take them to meet if the speed of the first is 20 km/h and the second is 25 km/h?

Decision:
• 1) 25 + 20 \u003d 45 (the sum of the speeds of the ships)
• 2) 90: 45 = 2
• Expression: 90: (20 + 25) = 2
• Answer: The boats will meet in 2 hours.

Task 3.

From two stations, the distance between which is 564 km., Two trains left at the same time towards each other. The speed of one of them is 63 km/h. What is the speed of the second one if the trains meet after 4 hours?

Decision:
• 1) 63 * 4 = 252 (1 train passed)
• 2) 564 - 252 \u003d 312 (train 2 passed)
• 3) 312: 4 = 78
• Expression: (63 * 4 - 252) : 4 = 78
• Answer: the speed of the second train is 78 km/h.

Task 4.

After how many seconds will two swallows meet, flying towards each other, if the speed of each of them is 23 meters per second, and the distance between them is 920 m.

Decision:
• 1) 23 * 2 = 46 (the sum of the speeds of the swallows)
• 2) 920: 46 = 20
• Expression: 920: (23 * 2) = 20
• Answer: The swallows will meet in 20 seconds.



Task 5

From two villages, a cyclist and a motorcyclist left at the same time towards each other. The speed of a motorcyclist is 54 km/h, of a cyclist 16 km/h. How many kilometers did the motorcyclist travel before the meeting if the cyclist traveled 48 km?

Decision:
• 1) 48: 16 = 3 (the cyclist spent hours)
• 2) 54 * 3 = 162
• Expression: 54 * (48: 16) = 162
• Answer: A motorcyclist traveled 162 km.

Task 6

Two boats, the distance between which is 90 km, began to move towards each other. The speed of one of the boats is 10 km/h, the other is 8 km/h. How many hours will it take the boats to meet?

Decision:
• 1) 10 + 8 = 18 (speed of two boats together)
• 2) 90: 18 = 5
• Expression: 90: (10 + 8) = 5
• Answer: the boats will meet in 5 hours.

Task 7

Along the path, which is 200 meters long, two boys ran towards each other. One of them ran at a speed of 5 m/s. What is the speed of the second boy if they meet after 20 seconds?

Decision:
• 1) 20 * 5 = 100 (the first boy ran meters)
• 2) 200 - 100 = 100 (the second boy ran meters)
• 3) 100: 20 = 5
• Expression: (200 - 5 * 20) : 20 = 5
• Answer: the speed of the second boy is 5 km/s.

Task 8

Two trains left towards each other. The speed of one of them is 35 km/h, the other 29 km/h. What was the distance between the trains at first if they met after 5 hours?

Decision:
• 1) 35 + 29 = 64 (speed of two trains together)
• 2) 64 * 5 = 320
• Expression: (35 + 29) * 5 = 320
• Answer: The distance between trains was 320 km.

Task 9

Two horsemen rode out of two villages towards each other. The speed of one of them is 13 km / h, they met after 4 hours. How fast did the second rider move if the distance between the villages is 100 km.

Decision:
• 1) 13 * 4 \u003d 52 (the first rider rode)
• 2) 100 - 52 = 48 (the second rider rode)
• 3) 48: 4 = 12
• Expression: (100 - 13 * 4) : 4 = 12
• Answer: The speed of the second rider is 12 km/h.



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 with constant speed and in a straight line. This is far from reality, but even with such a simplification real conditions one can obtain quite digestible results 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.

Decision. 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?

Decision. 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.

Decision. 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 ​​see oncoming traffic. 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.

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.

AT motion tasks commonly used formulas expressing the law uniform motion, i.e.

s = v t.

When compiling equations in such problems, it is convenient to use a geometric illustration of the motion process.

When moving around a circle, it is convenient to use the concept angular velocity, i.e. the angle by which a moving object rotates around its center per unit of time. It happens that in order to complicate the task, its condition is formulated in different units measurements. In such cases, in order to formulate equations, it is necessary to express all given values ​​in terms of the same unit of measurement.

The following considerations serve as a source for compiling equations in motion problems:

1) Objects that started moving towards each other at the same time move the same time until the moment of meeting. The time after which they meet is found by the formula

t = s/(v 1 + v 2) (*).

2) If one body catches up with another, then the time after which the first catches up with the second is calculated by the formula

t \u003d s / (v 1 - v 2) (**).

3) If the objects have traveled the same distance, then it is convenient to take the value of this distance as the common unknown of the problem.

4) If, with the simultaneous movement of two objects along a circle from one point, one of them catches up with the other for the first time, then the difference between the distances they have traveled by this moment is equal to the circumference

5) For time new meeting when moving in opposite directions, we get the formula (*), if in one direction, then the formula (**).

6) When moving along the river, the speed of an object is equal to the sum of the speeds in standing water and flow speed. When moving against the current, the speed of movement is the difference between these speeds.

Analytical solution of motion problems

Task 1.

Two pedestrians simultaneously went out towards each other and met after 3 hours and 20 minutes. How long did it take for each pedestrian to cover the entire distance, if it is known that the first one arrived at the point from which the second left, 5 hours later than the second arrived at the point from which the first left?

Decision.

There is no information about the distance traveled in this problem. This is her main feature. In such cases, it will be convenient to take the entire distance as a unit, then the speed of the first pedestrian will be equal to
v 1 = 1/x, and the second - v 2 = 1/y, where x hours is the travel time of the first, and y is the travel time of the second pedestrian.

The conditions of the problem allow us to compose a system of equations:

(3⅓ 1/x + 3⅓ 1/y = 1,
(x - y = 5.

Solving this system, we get that y = 5, x = 10.

Answer: 10 o'clock and 5 o'clock.

Task 2.

A cyclist left point A for point B. After 3 hours, a motorcyclist left point B towards him, at a speed 3 times greater than the speed of a cyclist. The meeting of the cyclist and the motorcyclist takes place in the middle, between points A and B. If the motorcyclist leaves 2 hours later than the cyclist, their meeting would take place 15 kilometers closer to point A. Find the distance AB.

Decision.

Let's make an illustration for the problem (Fig. 1).

Let AB = s km, v km/h is the speed of the cyclist, 3v km/h is the speed of the motorcyclist.

t 1 \u003d 0.5 s / v hours - the time before the meeting of the cyclist,

t 2 \u003d 0.5 s / 3v hours - the time until the meeting of the motorcyclist.

By condition t 1 - t 2 \u003d 3, then 0.5 s / v - 0.5s / 3v \u003d 3, from where s \u003d 9v.

If the motorcyclist left 2 hours later than the cyclist, they would meet at point F.

AF = 0.5s - 15, BF = 0.5s + 15.

Let's make the equation: (0.5s - 15) / v - (0.5s + 15) / 3v = 2, whence s - 60 = 6v.

We get a system of equations:

(s=9v,
(s = 60 + 6v.

(v=20,
(s = 180.

Answer: v = 20 km/h, s = 180 km.

Graphical method for solving motion problems

There is also graphic method task solutions. Let us consider the application of this method for solving motion problems. Graphic image functions that describe the condition of the problem is often a very convenient technique that allows you to visualize the situation of the problem. It also allows you to compose new equations or replace the algebraic solution of the problem with a purely geometric one.

Task 3.

The pedestrian went from point A to point B. Following him, a cyclist left point A, but with a delay of 2 hours. After another 30 minutes, a motorcyclist left in the direction of point B. A pedestrian, a cyclist and a motorcyclist moved to point B without stopping and evenly. Some time after the motorcyclist left, it turned out that by this moment all three had overcome the same part way from A to B. How many minutes before the pedestrian did the cyclist arrive at point B if the motorcyclist arrived at point B 1 hour before the pedestrian?

Decision.

For algebraic solution it requires the introduction of many variables and the compilation of a cumbersome system. Graphically, the situation described in the problem is shown in Figure 2.

Using the similarity of triangles AOL and KOM, as well as triangles AOP and KON, you can make a proportion:

x = 4/5 h = 48 minutes.

Answer: 48 minutes.

Task 4.

Two messengers left the two cities towards each other at the same time. After the meeting, one of them was on the road for another 16 hours, and the second - 9 hours. Determine how long each messenger traveled.

Decision.

Let the time of movement until the meeting of each messenger be t. According to the condition of the problem, we build a graph (Fig. 3).

Similar to problem 3, it is necessary to use the similarity of triangles.

So, 12 + 16 = 28 (hours) - the first one was on the way, 12 + 9 = 21 (hours) - the second one was on the way.

Answer: 21 hours and 28 hours.

So we have analyzed the main methods for solving problems on movement. In the exam, they are very common, so be sure to practice solving these problems.

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One of the basic topics in mathematics of elementary grades is "Movement and tasks for movement." You can start studying it after you have mastered the basic mathematical operations(addition, difference, product and quotient), mental counting. It is not necessary for children of this age to show formulas that connect the path, speed and time. As a rule, children begin to understand this intuitively. Of course, this topic prepares the student for the future study of physics, but this is still very far away. However, it is worth discussing with the child, for example, the reality of the speeds that are present in the tasks being solved, asking the student what moves the fastest, what or who is the slowest. You can pick up a lot of questions that will coincide with the plot of the problem.

Task 1. At the same time, two trains set off towards each other from two cities. One of them covers 13 km in 1/4 hour, and the second travels 16 km in 1/3 hour. After 2 hours, these trains met. How many kilometers between these cities?

Task 2. A cyclist and a pedestrian are moving towards each other. On the this moment the distance between them is 52 km. A cyclist has a speed of 9 km / h, a pedestrian has a speed of 5 km / h less, a. What will be the distance between them after 6 hours?

Task 3. Two cyclists left villages A and B at the same time. The distance between the villages is 117 km. The cyclists set off towards each other. The first cyclist has a speed of 17 km/h, the second cyclist has a speed of 24 km/h. What was the distance between the cyclists after 2 hours.

Task 4. A train left from a certain city. The second train left the same city at opposite side 2 hours later. When 3 hours had passed from that moment, the distance between the trains became 402 km. The speed of the first train is 6 km/h less than the speed of the second. What are the speeds of the trains?

Task 5. At the same time, two planes flew towards each other. In 10 minutes they moved away for 270 km. The first aircraft has a speed of 15 km/min. What is the speed of the second aircraft if the distance between the airfields is 540 km? At what time will the second plane arrive at the opposite airfield if it took off at 10:15?

Task 6. At 9 o'clock in the morning, a train left city A with a speed of 67 km/h. On the same day at 12 o'clock another train departed from city B towards him, its speed was 50 km/h. After 7 hours after the second train left, there were 365 km between them. Find out how many kilometers between cities A and B.

Task 7. A car left point A for point B at a speed of 65 km/h. After 2 hours, a motorcycle drove out from point B towards him, its speed is 80 km / h. At a distance of 240 km from point B, he met a car. Find the distance from point A to point B.

Task 8. Two cyclists are riding towards each other on the highway. Between them now 2700 meters, cyclists will meet in 6 minutes. The speed of one is 50 m/min more than the speed of the other. Determine their speed.

Task 9. Two cars left at the same time towards each other. How long will it take for the distance between them to be equal to 150 km if the first one has traveled 180 km up to this moment.

Task 10. From one city to another 250 km, two motorcyclists set off from these cities towards each other at the same time. When 2 hours had passed, it turned out that the distance between the motorcyclists was 30 km. The first motorcyclist has a speed of 10 km/h more than the speed of the second. Find the speed of each motorcyclist.

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