In the previous tasks for movement in one direction, the movement of bodies began simultaneously from the same point. Consider solving problems for movement in one direction, when the movement of bodies begins at the same time, but from different points.
Let a cyclist and a pedestrian depart from points A and B, the distance between which is 21 km, and go in the same direction: a pedestrian at a speed of 5 km per hour, a cyclist at 12 km per hour
12 km per hour 5 km per hour
A B
The distance between a cyclist and a pedestrian at the start of their movement is 21 km. For an hour of their joint movement in one direction, the distance between them will decrease by 12-5=7 (km). 7 km per hour - the speed of convergence of a cyclist and a pedestrian:
A B
Knowing the speed of approach of the cyclist and the pedestrian, it is easy to find out how many kilometers the distance between them will decrease after 2 hours, 3 hours of their movement in the same direction.
7*2=14 (km) - the distance between the cyclist and the pedestrian will decrease by 14 km after 2 hours;
7*3=21 (km) - the distance between the cyclist and the pedestrian will decrease by 21 km after 3 hours.
Every hour the distance between the cyclist and the pedestrian decreases. After 3 hours, the distance between them becomes equal to 21-21=0, i.e. the cyclist overtakes the pedestrian:
A B
In tasks “to catch up” we deal with quantities:
1) the distance between the points from which the simultaneous movement begins;
2) approach speed
3) the time from the moment the movement begins to the moment when one of the moving bodies overtakes the other.
Knowing the value of two of these three quantities, you can find the value of the third quantity.
The table contains the conditions and solutions to problems that can be compiled to “catch up” with a pedestrian cyclist:
|
Approach speed of cyclist and pedestrian in km per hour |
Time from the start of the movement to the moment when the cyclist catches up with the pedestrian, in hours |
Distance from A to B in km | ||
We express the relationship between these quantities by the formula. Denote by the distance between the points and, - the speed of approach, the time from the moment of exit to the moment when one body catches up with another.
In catch-up problems, the convergence rate is most often not given, but it can be easily found from the problem data.
Task. A cyclist and a pedestrian left simultaneously in the same direction from two collective farms, the distance between which is 24 km. A cyclist was traveling at a speed of 11 km per hour, and a pedestrian was walking at a speed of 5 km per hour. In how many hours after his exit will the cyclist overtake the pedestrian?
To find how long after his exit the cyclist will catch up with the pedestrian, you need to divide the distance that was between them at the beginning of the movement by the speed of approach; the speed of approach is equal to the difference between the speeds of the cyclist and the pedestrian.
Solution formula: =24: (11-5);=4.
Answer. In 4 hours the cyclist will overtake the pedestrian. Conditions and solutions of inverse problems are written in the table:
|
The speed of the cyclist in km per hour |
Pedestrian speed in km per hour |
Distance between collective farms in km |
Time per hour | ||
Each of these tasks can be solved in other ways, but they will be irrational compared to these solutions.
Tasks on movement in one direction belong to one of the three main types of tasks on movement.
Now we will talk about tasks in which objects have different speeds.
When moving in one direction, objects can both approach and move away.
Here we consider problems for movement in one direction, in which both objects leave the same point. Next time we will talk about the movement in pursuit, when objects move in the same direction from different points.
If two objects left the same point at the same time, then, since they have different speeds, the objects move away from each other.
To find the speed of removal, it is necessary to subtract the smaller from the greater speed:
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If one object left one point, and after some time another object left it in the same direction, then they can both approach and move away from each other.
If the speed of the object moving in front is less than the object moving after it, then the second catches up with the first and they approach each other.
To find the speed of approach, subtract the smaller speed from the larger one:
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If the speed of the object that goes ahead is greater than the speed of the object that moves behind, then the second one will not be able to catch up with the first one and they move away from each other.
We find the removal rate in the same way - subtract the smaller one from the larger one:
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Speed, time and distance are related:
Task 1.
Two cyclists left the same village in the same direction at the same time. The speed of one of them is 15 km/h, the speed of the other is 12 km/h. How far will they be in 4 hours?
Decision:
The condition of the problem is most conveniently written in the form of a table:
1) 15-12=3 (km/h) removal speed of cyclists
2) 3∙4=12 (km) this distance will be between cyclists after 4 hours.
Answer: 12 km.
A bus leaves from point A to point B. After 2 hours, a car left behind him. At what distance from point A will the car overtake the bus if the speed of the car is 80 km/h and the speed of the bus is 40 km/h?
1) 80-40=40 (km/h) vehicle and bus approach speed
2) 40∙2=80 (km) at this distance from point A there is a bus when the car leaves A
3) 80:40=2 (h) the time after which the car will overtake the bus
4) 80∙2=160 (km) the distance that the car will cover from point A
Answer: at a distance of 160 km.
Task 3
A pedestrian left the village and a cyclist left the station at the same time. After 2 hours, the cyclist was ahead of the pedestrian by 12 km. Find the speed of the pedestrian if the speed of the cyclist is 10 km/h.
Decision:
1) 12:2=6 (km/h) removal speed of cyclist and pedestrian
2) 10-6=4 (km/h) walking speed.
Answer: 4 km/h.
Page 1
Beginning in 5th grade, students often come across these problems. Also in primary school students are given the concept of "general speed". As a result, they form not entirely correct ideas about the speed of approach and the speed of removal (there is no such terminology in elementary school). Most often, when solving a problem, students find the sum. It is best to start solving these problems with the introduction of the concepts: “rapprochement rate”, “removal rate”. For clarity, you can use the movement of the hands, explaining that bodies can move in one direction and in different directions. In both cases, there may be an approach speed and a removal speed, but in different cases they are found in different ways. After that, students write down the following table:
Table 1.
Methods for finding the speed of approach and speed of removal
|
Movement in one direction |
Movement in different directions |
|
|
Removal speed |
| |
|
Approach speed | ||
When analyzing the problem, the following questions are given.
Using the movement of the hands, we find out how the bodies move relative to each other (in one direction, in different ones).
We find out what action is the speed (addition, subtraction)
We determine what speed it is (approach, removal). Write down the solution to the problem.
Example #1. From the cities A and B, the distance between which is 600 km, at the same time, a truck and a car left towards each other. The speed of the passenger car is 100 km/h, and the speed of the truck is 50 km/h. In how many hours will they meet?
Students use their hands to show how cars move and draw the following conclusions:
cars move in different directions;
the speed will be found by addition;
since they are moving towards each other, then this is the speed of convergence.
100+50=150 (km/h) – closing speed.
600:150=4 (h) - the time of movement before the meeting.
Answer: after 4 hours
Example #2. The man and the boy left the state farm for the garden at the same time and go the same way. The man's speed is 5 km/h and the boy's speed is 3 km/h. How far apart will they be after 3 hours?
With the help of hand movements, we find out:
the boy and the man are moving in the same direction;
speed is the difference;
the man walks faster, i.e., moves away from the boy (removal speed).
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2. SPEED OF THE BODY. RECTILINEAR UNIFORM MOVEMENT.
Speed is a quantitative characteristic of the movement of the body.
average speed is a physical quantity equal to the ratio of the point displacement vector to the time interval Δt during which this displacement occurred. The direction of the average velocity vector coincides with the direction of the displacement vector . The average speed is determined by the formula:
Instant Speed, that is, the speed at a given moment of time is a physical quantity equal to the limit to which the average speed tends with an infinite decrease in the time interval Δt:
In other words, the instantaneous speed at a given moment of time is the ratio of a very small movement to a very small period of time during which this movement occurred.
The instantaneous velocity vector is directed tangentially to the trajectory of the body (Fig. 1.6).
Rice. 1.6. Instantaneous velocity vector.
In the SI system, speed is measured in meters per second, that is, the unit of speed is considered to be the speed of such uniform rectilinear motion, in which in one second the body travels a distance of one meter. The unit of speed is denoted m/s. Often speed is measured in other units. For example, when measuring the speed of a car, train, etc. The commonly used unit of measure is kilometers per hour:
1 km/h = 1000 m / 3600 s = 1 m / 3.6 s
1 m/s = 3600 km / 1000 h = 3.6 km/h
Addition of speeds (perhaps not necessarily the same question will be in 5).
The velocities of the body in different reference systems are connected by the classical law of addition of speeds.
body speed relative to fixed frame of reference is equal to the sum of the velocities of the body in moving frame of reference and the most mobile frame of reference relative to the fixed one.
For example, a passenger train is moving along a railroad at a speed of 60 km/h. A person is walking along the carriage of this train at a speed of 5 km/h. If we consider the railway to be stationary and take it as a reference frame, then the speed of a person relative to the reference system (that is, relative to the railway) will be equal to the addition of the speeds of the train and the person, that is
60 + 5 = 65 if the person is walking in the same direction as the train
60 - 5 = 55 if the person and the train are moving in different directions
However, this is only true if the person and the train are moving along the same line. If a person moves at an angle, then this angle will have to be taken into account, remembering that speed is vector quantity.
An example is highlighted in red + The law of displacement addition (I think this does not need to be taught, but for general development you can read it)
Now let's look at the example described above in more detail - with details and pictures.
So, in our case, the railway is fixed frame of reference. The train that is moving along this road is moving frame of reference. The car on which the person is walking is part of the train.
The speed of a person relative to the car (relative to the moving frame of reference) is 5 km/h. Let's call it C.
The speed of the train (and hence the wagon) relative to a fixed frame of reference (that is, relative to the railway) is 60 km/h. Let's denote it with the letter B. In other words, the speed of the train is the speed of the moving reference frame relative to the fixed frame of reference.
The speed of a person relative to the railway (relative to a fixed frame of reference) is still unknown to us. Let's denote it with a letter.
We will associate the XOY coordinate system with the fixed reference system (Fig. 1.7), and the X P O P Y P coordinate system with the moving reference system. Now let's try to find the speed of a person relative to the fixed reference system, that is, relative to the railway.
For a short period of time Δt, the following events occur:
Then for this period of time the movement of a person relative to the railway:
This is displacement addition law. In our example, the movement of a person relative to the railway is equal to the sum of the movements of a person relative to the wagon and the wagon relative to the railway.
Rice. 1.7. The law of addition of displacements.
The law of addition of displacements can be written as follows:
= ∆ H ∆t + ∆ B ∆t
The speed of a person relative to the railroad is:
The speed of a person relative to the car:
Δ H \u003d H / Δt
The speed of the car relative to the railway:
Therefore, the speed of a person relative to the railway will be equal to:
This is the lawspeed addition:
Uniform movement- this is movement at a constant speed, that is, when the speed does not change (v \u003d const) and there is no acceleration or deceleration (a \u003d 0).
Rectilinear motion- this is movement in a straight line, that is, the trajectory of rectilinear movement is a straight line.
Uniform rectilinear motion is a movement in which the body makes the same movements for any equal intervals of time. For example, if we divide some time interval into segments of one second, then with uniform motion the body will move the same distance for each of these segments of time.
The speed of uniform rectilinear motion does not depend on time and at each point of the trajectory is directed in the same way as the movement of the body. That is, the displacement vector coincides in direction with the velocity vector. In this case, the average speed for any period of time is equal to the instantaneous speed:
Speed of uniform rectilinear motion is a physical vector quantity equal to the ratio of the displacement of the body for any period of time to the value of this interval t:
Thus, the speed of uniform rectilinear motion shows what movement a material point makes per unit of time.
moving with uniform rectilinear motion is determined by the formula:
Distance traveled in rectilinear motion is equal to the displacement modulus. If the positive direction of the OX axis coincides with the direction of movement, then the projection of the velocity on the OX axis is equal to the velocity and is positive:
v x = v, i.e. v > 0
The projection of displacement onto the OX axis is equal to:
s \u003d vt \u003d x - x 0
where x 0 is the initial coordinate of the body, x is the final coordinate of the body (or the coordinate of the body at any time)
Motion equation, that is, the dependence of the body coordinate on time x = x(t), takes the form:
If the positive direction of the OX axis is opposite to the direction of motion of the body, then the projection of the body velocity on the OX axis is negative, the velocity is less than zero (v< 0), и тогда уравнение движения принимает вид.
§ 1 Simultaneous motion formula
We encounter formulas for simultaneous motion when solving problems for simultaneous motion. The ability to solve one or another task for movement depends on several factors. First of all, it is necessary to distinguish between the main types of tasks.
Tasks for simultaneous movement are conditionally divided into 4 types: tasks for oncoming movement, tasks for movement in opposite directions, tasks for movement in pursuit, and tasks for movement with a lag.
The main components of these types of tasks are:
distance traveled - S, speed - ʋ, time - t.
The relationship between them is expressed by the formulas:
S = ʋ t, ʋ = S: t, t = S: ʋ.
In addition to the above main components, when solving problems for movement, we may encounter such components as: the speed of the first object - ʋ1, the speed of the second object - ʋ2, the speed of approach - ʋsbl., the speed of removal - ʋsp., the meeting time - tin., the initial distance - S0 etc.
§ 2 Tasks for oncoming traffic
When solving problems of this type, the following components are used: the speed of the first object - ʋ1; speed of the second object - ʋ2; approach speed - ʋsbl.; time before meeting - tvstr.; the path (distance) traveled by the first object - S1; the path (distance) traveled by the second object - S2; the entire path traveled by both objects - S.
The dependence between the components of tasks for oncoming traffic is expressed by the following formulas:
1. The initial distance between objects can be calculated using the following formulas: S = ʋsbl. · tvstr. or S = S1 + S2;
2. Approach speed is found by the formulas: ʋsbl. = S: tint. or ʋsl. = ʋ1 + ʋ2;
3.meeting time is calculated as follows:
Two boats are sailing towards each other. The speeds of motor ships are 35 km/h and 28 km/h. After what time will they meet if the distance between them is 315 km?
ʋ1 = 35 km/h, ʋ2 = 28 km/h, S = 315 km, tint. = ? h.
To find the meeting time, you need to know the initial distance and speed of approach, since tin. = S: ʋsbl. Since the distance is known by the condition of the problem, we will find the speed of approach. ʋsbl. = ʋ1 + ʋ2 = 35 + 28 = 63 km/h. Now we can find the desired meeting time. tint. = S: ʋsbl = 315: 63 = 5 hours. We got that the ships will meet in 5 hours.
§ 3 Tasks for moving after
When solving problems of this type, the following components are used: the speed of the first object - ʋ1; speed of the second object - ʋ2; approach speed - ʋsbl.; time before meeting - tvstr.; the path (distance) traveled by the first object - S1; the path (distance) traveled by the second object - S2; initial distance between objects - S.
The scheme for tasks of this type is as follows:
The dependence between the components of the tasks for the movement in pursuit is expressed by the following formulas:
1. The initial distance between objects can be calculated using the following formulas:
S = ʋsbl. tbuilt-in or S = S1 - S2;
2. Approach speed is found by the formulas: ʋsbl. = S: tint. or ʋsl. = ʋ1 - ʋ2;
3.The meeting time is calculated as follows:
tint. = S: ʋbl., tint. = S1: ʋ1 or tint. = S2: ʋ2.
Consider the application of these formulas on the example of the following problem.
The tiger chased the deer and caught up with it after 7 minutes. What is the initial distance between them if the tiger's speed is 700 m/min and the deer's speed is 620 m/min?
ʋ1 = 700 m/min, ʋ2 = 620 m/min, S = ? m, tvstr. = 7 min.
To find the initial distance between a tiger and a deer, it is necessary to know the meeting time and the speed of approach, since S = tin. · ʋsbl. Since the meeting time is known by the condition of the problem, we find the speed of approach. ʋsbl. = ʋ1 - ʋ2 = 700 - 620 = 80 m/min. Now we can find the desired initial distance. S =tin. · ʋsbl = 7 · 80 = 560 m. We found that the initial distance between the tiger and the deer was 560 meters.
§ 4 Tasks for movement in opposite directions
When solving problems of this type, the following components are used: the speed of the first object - ʋ1; speed of the second object - ʋ2; removal rate - ʋud.; travel time - t.; the path (distance) traveled by the first object - S1; the path (distance) traveled by the second object - S2; initial distance between objects - S0; the distance that will be between objects after a certain time - S.
The scheme for tasks of this type is as follows:
The dependence between the components of tasks for movement in opposite directions is expressed by the following formulas:
1. The final distance between objects can be calculated using the following formulas:
S = S0 + ʋsp t or S = S1 + S2 + S0; and the initial distance - according to the formula: S0 \u003d S - ʋsp. t.
2. The removal rate is found by the formulas:
ʋud. = (S1 + S2) : t orʋsp. = ʋ1 + ʋ2;
3.The travel time is calculated as follows:
t = (S1 + S2) : ʋsp, t = S1: ʋ1 or t = S2: ʋ2.
Consider the application of these formulas on the example of the following problem.
Two cars left the car parks at the same time in opposite directions. The speed of one is 70 km/h, the other is 50 km/h. What will be the distance between them after 4 hours if the distance between the fleets is 45 km?
ʋ1 = 70 km/h, ʋ2 = 50 km/h, S0 = 45 km, S = ? km, t = 4 h.
To find the distance between cars at the end of the journey, you need to know the travel time, the initial distance and the speed of removal, since S = ʋsp. · t+ S0 Since the time and the initial distance are known by the condition of the problem, let's find the speed of removal. ʋud. = ʋ1 + ʋ2 = 70 + 50 = 120 km/h. Now we can find the desired distance. S = ʋud. t+ S0 = 120 4 + 45 = 525 km. We got that after 4 hours there will be a distance of 525 km between the cars
§ 5 Tasks for moving with a lag
When solving problems of this type, the following components are used: the speed of the first object - ʋ1; speed of the second object - ʋ2; removal rate - ʋud.; travel time - t.; initial distance between objects - S0; the distance that will become between objects after a certain amount of time - S.
The scheme for tasks of this type is as follows:
The dependence between the components of tasks for movement with a lag is expressed by the following formulas:
1. The initial distance between objects can be calculated using the following formula: S0 = S - ʋsp t; and the distance that will become between objects after a certain time is according to the formula: S = S0 + ʋsp. t;
2. The removal rate is found by the formulas: ʋsp. = (S - S0) : t or ʋsp. = ʋ1 - ʋ2;
3. Time is calculated as follows: t = (S - S0) : ʋsp.
Consider the application of these formulas on the example of the following problem:
Two cars left two cities in the same direction. The speed of the first is 80 km/h, the speed of the second is 60 km/h. In how many hours will there be 700 km between the cars if the distance between the cities is 560 km?
ʋ1 = 80 km/h, ʋ2 = 60 km/h, S = 700 km, S0 = 560 km, t = ? h.
To find the time, you need to know the initial distance between objects, the distance at the end of the path and the speed of removal, since t = (S - S0) : ʋsp. Since both distances are known by the condition of the problem, we will find the removal rate. ʋud. = ʋ1 - ʋ2 = 80 - 60 = 20 km/h. Now we can find the desired time. t \u003d (S - S0) : ʋsp \u003d (700 - 560) : 20 \u003d 7h. We got that in 7 hours there will be 700 km between the cars.
§ 6 Brief summary of the topic of the lesson
With simultaneous oncoming and chasing movement, the distance between two moving objects decreases (until the meeting). For a unit of time, it decreases by ʋsbl., and for the entire time of movement before the meeting it will decrease by the initial distance S. Hence, in both cases, the initial distance is equal to the speed of approach multiplied by the time of movement to the meeting: S = ʋsbl. · tvstr.. The only difference is that with oncoming traffic ʋsbl. = ʋ1 + ʋ2, and when moving after ʋsbl. = ʋ1 - ʋ2.
When moving in opposite directions and with a lag, the distance between objects increases, so the meeting will not occur. For a unit of time, it increases by ʋsp., and for the entire time of movement it will increase by the value of the product ʋsp. · t. Hence, in both cases, the distance between objects at the end of the path is equal to the sum of the initial distance and the product of ʋsp. t. S = S0 + ʋsp. t. The only difference is that with the opposite movement ʋsp. = ʋ1 + ʋ2, and when moving with a lag, ʋsp. = ʋ1 - ʋ2.
List of used literature:
- Peterson L.G. Mathematics. 4th grade. Part 2. / L.G. Peterson. – M.: Yuventa, 2014. – 96 p.: ill.
- Mathematics. 4th grade. Guidelines to the textbook of mathematics "Learning to learn" for grade 4 / L.G. Peterson. – M.: Yuventa, 2014. – 280 p.: ill.
- Zak S.M. All tasks for the mathematics textbook for grade 4 L.G. Peterson and a set of independent and control works. GEF. – M.: UNVES, 2014.
- CD-ROM. Mathematics. 4th grade. Lesson scenarios for the textbook for part 2 Peterson L.G. – M.: Yuventa, 2013.
Used images: