Graphic tasks. Graphical problems in physics and graphical problem solving

All constructions in the process of graphical reckoning are performed using a spacer tool:

navigation protractor,

parallel ruler,

measuring compass,

drawing compass with pencil.

The lines are drawn with a simple pencil and removed with a soft eraser.

Take the coordinates of a given point from the map. This task can be most accurately performed using a measuring compass. To measure latitude, one leg of the compass is placed at a given point, and the other is brought to the nearest parallel so that the arc described by the compass touches it.

Without changing the angle of the legs of the compass, bring it to the vertical frame of the map and place one leg on the parallel to which the distance was measured.
The other leg is placed on the inner half of the vertical frame towards the given point and the latitude reading is taken with an accuracy of 0.1 of the smallest division of the frame. The longitude of a given point is determined in the same way, only the distance is measured to the nearest meridian, and the longitude reading is taken along the upper or lower frame of the map.

Place a point at the given coordinates. The work is usually carried out using a parallel ruler and a measuring compass. The ruler is applied to the nearest parallel and one half of it is moved to the specified latitude. Then, using a compass solution, take the distance from the nearest meridian to a given longitude along the upper or lower frame of the map. One leg of the compass is placed at the cut of the ruler on the same meridian, and with the other leg a weak injection is made also at the cut of the ruler in the direction of the given longitude. The injection site will be the given point

Measure the distance between two points on a map or plot a known distance from a given point. If the distance between the points is small and can be measured with one compass solution, then the legs of the compass are placed at one and the other point, without changing its solution, and placed on the side frame of the map at approximately the same latitude in which the measured distance lies.

When measuring a large distance, it is divided into parts. Each part of the distance is measured in miles in the latitude of the area. You can also use a compass to take a “round” number of miles (10,20, etc.) from the side frame of the map and count how many times to place this number along the entire line being measured.
In this case, miles are taken from the side frame of the map approximately opposite the middle of the measured line. The remainder of the distance is measured in the usual way. If you need to set aside a small distance from a given point, then remove it with a compass from the side frame of the map and set it off on the laid line.
The distance is taken from the frame approximately at the latitude of a given point, taking into account its direction. If the distance being set aside is large, then they take it from the map frame approximately opposite the middle of the given distance 10, 20 miles, etc. and postpone the required number of times. The remainder of the distance is measured from the last point.

Measure the direction of the true course or bearing line drawn on the map. A parallel ruler is applied to the line on the map and a protractor is placed on the edge of the ruler.
The protractor is moved along the ruler until its central stroke coincides with any meridian. The division on the protractor through which the same meridian passes corresponds to the direction of course or bearing.
Since two readings are marked on the protractor, when measuring the direction of the laid line, one should take into account the quarter of the horizon in which the given direction lies.

Draw a line of true course or bearing from a given point. To perform this task, use a protractor and a parallel ruler. The protractor is placed on the map so that its central stroke coincides with any meridian.

Then the protractor is turned in one direction or the other until the stroke of the arc corresponding to the reading of the given course or bearing coincides with the same meridian. A parallel ruler is applied to the lower edge of the protractor ruler, and, having removed the protractor, they move it apart, bringing it to a given point.

A line is drawn along the cut of the ruler in the desired direction. Move a point from one map to another. The direction and distance to a given point from any lighthouse or other landmark marked on both maps is taken from the map.
On another map, by plotting the desired direction from this landmark and plotting the distance along it, the given point is obtained. This task is a combination

If a linear programming problem has only two variables, then it can be solved graphically.

Consider a linear programming problem with two variables and :
(1.1) ;
(1.2)
Here, there are arbitrary numbers. The task can be either to find the maximum (max) or to find the minimum (min). The system of restrictions may contain both signs and signs.

Construction of the domain of feasible solutions

The graphical method for solving problem (1) is as follows.
First, we draw the coordinate axes and select the scale. Each of the inequalities of the system of constraints (1.2) defines a half-plane bounded by the corresponding straight line.

So, the first inequality
(1.2.1)
defines a half-plane bounded by a straight line. On one side of this straight line, and on the other side. On the very straight line. To find out on which side inequality (1.2.1) holds, we choose an arbitrary point that does not lie on the line. Next, we substitute the coordinates of this point into (1.2.1). If the inequality holds, then the half-plane contains the selected point. If the inequality does not hold, then the half-plane is located on the other side (does not contain the selected point). Shade the half-plane for which inequality (1.2.1) holds.

We do the same for the remaining inequalities of system (1.2). This way we get shaded half-planes. The points of the region of feasible solutions satisfy all inequalities (1.2). Therefore, graphically, the region of feasible solutions (ADA) is the intersection of all constructed half-planes. Shading the ODR. It is a convex polygon whose faces belong to the constructed straight lines. Also, an ODF can be an unlimited convex figure, a segment, a ray or a straight line.

The case may also arise that the half-planes do not contain common points. Then the domain of feasible solutions is the empty set. This problem has no solutions.

The method can be simplified. You don’t have to shade each half-plane, but first construct all the straight lines
(2)
Next, select an arbitrary point that does not belong to any of these lines. Substitute the coordinates of this point into the system of inequalities (1.2). If all inequalities are satisfied, then the region of feasible solutions is limited by the constructed straight lines and includes the selected point. We shade the region of feasible solutions along the boundaries of the lines so that it includes the selected point.

If at least one inequality is not satisfied, then choose another point. And so on until one point is found whose coordinates satisfy system (1.2).

Finding the extremum of the objective function

So, we have a shaded region of feasible solutions (ADA). It is limited by a broken line consisting of segments and rays belonging to the constructed straight lines (2). The ODS is always a convex set. It can be either a bounded set or not bounded along some directions.

Now we can look for the extremum of the objective function
(1.1) .

To do this, choose any number and build a straight line
(3) .
For the convenience of further presentation, we assume that this straight line passes through the ODR. On this line the objective function is constant and equal to . such a straight line is called a function level line. This straight line divides the plane into two half-planes. On one half-plane
.
On another half-plane
.
That is, on one side of straight line (3) the objective function increases. And the further we move the point from the straight line (3), the greater the value will be. On the other side of straight line (3), the objective function decreases. And the further we move the point from straight line (3) to the other side, the smaller the value will be. If we draw a straight line parallel to line (3), then the new straight line will also be a level line of the objective function, but with a different value.

Thus, in order to find the maximum value of the objective function, it is necessary to draw a straight line parallel to straight line (3), as far as possible from it in the direction of increasing values, and passing through at least one point of the ODD. To find the minimum value of the objective function, it is necessary to draw a straight line parallel to straight line (3) and as far as possible from it in the direction of decreasing values, and passing through at least one point of the ODD.

If the ODR is unlimited, then a case may arise when such a direct line cannot be drawn. That is, no matter how we remove the straight line from the level line (3) in the direction of increasing (decreasing), the straight line will always pass through the ODR. In this case it can be arbitrarily large (small). Therefore, there is no maximum (minimum) value. The problem has no solutions.

Let us consider the case when the extreme line parallel to an arbitrary line of the form (3) passes through one vertex of the ODR polygon. From the graph we determine the coordinates of this vertex. Then the maximum (minimum) value of the objective function is determined by the formula:
.
The solution to the problem is
.

There may also be a case when the straight line is parallel to one of the faces of the ODR. Then the straight line passes through two vertices of the ODR polygon. We determine the coordinates of these vertices. To determine the maximum (minimum) value of the objective function, you can use the coordinates of any of these vertices:
.
The problem has infinitely many solutions. The solution is any point located on the segment between the points and , including the points and themselves.

An example of solving a linear programming problem using the graphical method

The task

The company produces dresses of two models A and B. Three types of fabric are used. To make one dress of model A, 2 m of fabric of the first type, 1 m of fabric of the second type, 2 m of fabric of the third type are required. To make one dress of model B, 3 m of fabric of the first type, 1 m of fabric of the second type, 2 m of fabric of the third type are required. The stocks of fabric of the first type are 21 m, of the second type - 10 m, of the third type - 16 m. The release of one product of type A brings in an income of 400 den. units, one product type B - 300 den. units

Draw up a production plan that provides the company with the greatest income. Solve the problem graphically.

Solution

Let the variables and denote the number of dresses produced, models A and B, respectively. Then the amount of fabric of the first type consumed will be:
(m)
The amount of fabric of the second type consumed will be:
(m)
The amount of fabric of the third type consumed will be:
(m)
Since the number of dresses produced cannot be negative, then
And .
The income from the dresses produced will be:
(den. units)

Then the economic-mathematical model of the problem has the form:

We solve it graphically.
We draw the coordinate axes and .

We are building a straight line.
At .
At .
Draw a straight line through the points (0; 7) and (10.5; 0).

We are building a straight line.
At .
At .
Draw a straight line through the points (0; 10) and (10; 0).

We are building a straight line.
At .
At .
Draw a straight line through the points (0; 8) and (8; 0).

We shade the area so that the point (2; 2) falls into the shaded part. We get the quadrilateral OABC.

(A1.1) .
At .
At .
Draw a straight line through the points (0; 4) and (3; 0).

We further note that since the coefficients of and of the objective function are positive (400 and 300), it increases as and increases. We draw a straight line parallel to straight line (A1.1), as far as possible from it in the direction of increasing , and passing through at least one point of the quadrilateral OABC. Such a line passes through point C. From the construction we determine its coordinates.
.

The solution of the problem: ;

Answer

.
That is, to obtain the greatest income, it is necessary to make 8 dresses of model A. The income will be 3200 den. units

Example 2

The task

Solve a linear programming problem graphically.

Solution

We solve it graphically.
We draw the coordinate axes and .

We are building a straight line.
At .
At .
Draw a straight line through the points (0; 6) and (6; 0).

We are building a straight line.
From here.
At .
At .
Draw a straight line through points (3; 0) and (7; 2).

We are building a straight line.
We build a straight line (abscissa axis).

The region of admissible solutions (ADA) is limited by the constructed straight lines. To find out which side, we notice that the point belongs to the ODR, since it satisfies the system of inequalities:

We shade the area along the boundaries of the constructed lines so that point (4; 1) falls into the shaded part. We get triangle ABC.

We build an arbitrary line of the level of the objective function, for example,
.
At .
At .
Draw a straight level line through points (0; 6) and (4; 0).
Since the objective function increases with increasing and , we draw a straight line parallel to the level line and as far away from it as possible in the direction of increasing , and passing through at least one point of triangle ABC. Such a line passes through point C. From the construction we determine its coordinates.
.

The solution of the problem: ;

Answer

Example of no solution

The task

Solve a linear programming problem graphically. Find the maximum and minimum value of the objective function.

Solution

We solve the problem graphically.
We draw the coordinate axes and .

We are building a straight line.
At .
At .
Draw a straight line through the points (0; 8) and (2.667; 0).

We are building a straight line.
At .
At .
Draw a straight line through the points (0; 3) and (6; 0).

We are building a straight line.
At .
At .
Draw a straight line through points (3; 0) and (6; 3).

The straight lines are the coordinate axes.

The region of admissible solutions (ADA) is limited by the constructed straight lines and coordinate axes. To find out which side, we notice that the point belongs to the ODR, since it satisfies the system of inequalities:

We shade the area so that the point (3; 3) falls into the shaded part. We obtain an unbounded area bounded by the broken line ABCDE.

We build an arbitrary line of the level of the objective function, for example,
(A3.1) .
At .
At .
Draw a straight line through the points (0; 7) and (7; 0).
Since the coefficients of and are positive, it increases with increasing and .

To find the maximum, you need to draw a parallel line, which is as far away as possible in the direction of increasing , and passing through at least one point of the region ABCDE. However, since the area is unlimited on the side of large values ​​of and , such a straight line cannot be drawn. No matter what line we draw, there will always be points in the region that are more distant in the direction of increasing and . Therefore there is no maximum. you can make it as big as you like.

We are looking for the minimum. We draw a straight line parallel to straight line (A3.1) and as far away from it as possible in the direction of decreasing , and passing through at least one point of the region ABCDE. Such a line passes through point C. From the construction we determine its coordinates.
.
Minimum value of the objective function:

Answer

There is no maximum value.
Minimum value
.

Problems of this type include those in which all or part of the data is specified in the form of graphical dependencies between them. In solving such problems, the following stages can be distinguished:

Stage 2 - find out from the given graph what quantities the relationship is between; find out which physical quantity is independent, i.e. an argument; what quantity is dependent, i.e., a function; determine by the type of graph what kind of dependence it is; find out what is required - define a function or an argument; if possible, write down the equation that describes the given graph;

Stage 3 - mark the given value on the abscissa (or ordinate) axis and restore the perpendicular to the intersection with the graph. Lower the perpendicular from the intersection point to the ordinate (or abscissa) axis and determine the value of the desired quantity;

Stage 4 - evaluate the result obtained;

Stage 5 - write down the answer.

Reading the coordinate graph means that from the graph you should determine: the initial coordinate and speed of movement; write down the coordinate equation; determine the time and place of meeting of the bodies; determine at what point in time the body has a given coordinate; determine the coordinate that the body has at a specified moment in time.

Problems of the fourth type - experimental . These are problems in which to find an unknown quantity it is necessary to measure part of the data experimentally. The following operating procedure is suggested:

Stage 2 - determine what phenomenon, law underlies the experience;

Stage 3 - think over the experimental design; determine a list of instruments and auxiliary items or equipment for conducting the experiment; think over the sequence of the experiment; if necessary, develop a table for recording the results of the experiment;

Stage 4 - perform the experiment and write the results in the table;

Stage 5 - make the necessary calculations, if required according to the conditions of the problem;

Stage 6 - think about the results obtained and write down the answer.

Particular algorithms for solving problems in kinematics and dynamics have the following form.

Algorithm for solving problems in kinematics:

Stage 2 - write down the numerical values ​​of the given quantities; express all quantities in SI units;

Stage 3 - make a schematic drawing (trajectory of movement, vectors of velocity, acceleration, displacement, etc.);

Stage 4 - choose a coordinate system (you should choose a system so that the equations are simple);

Stage 5 - compile basic equations for a given movement that reflect the mathematical relationship between the physical quantities shown in the diagram; the number of equations must be equal to the number of unknown quantities;

Stage 6 - solve the compiled system of equations in general form, in letter notation, i.e. get the calculation formula;

Stage 7 - select a system of units of measurement (“SI”), substitute the names of units in the calculation formula instead of letters, perform actions with the names and check whether the result results in a unit of measurement of the desired quantity;

Stage 8 - express all given quantities in the selected system of units; substitute into the calculation formulas and calculate the values ​​of the required quantities;

Stage 9 - analyze the solution and formulate an answer.

Comparing the sequence of solving problems in dynamics and kinematics makes it possible to see that some points are common to both algorithms, this helps to remember them better and apply them more successfully when solving problems.

Algorithm for solving dynamics problems:

Stage 2 - write down the condition of the problem, expressing all quantities in SI units;

Stage 3 - make a drawing indicating all the forces acting on the body, acceleration vectors and coordinate systems;

Stage 4 - write down the equation of Newton’s second law in vector form;

Stage 5 - write down the basic equation of dynamics (the equation of Newton’s second law) in projections on the coordinate axes, taking into account the direction of the coordinate axes and vectors;

Stage 6 - find all the quantities included in these equations; substitute into equations;

Stage 7 - solve the problem in general form, i.e. solve an equation or system of equations for an unknown quantity;

Stage 8 - check the dimension;

Stage 9 - obtain a numerical result and correlate it with real values.

Algorithm for solving problems on thermal phenomena:

Stage 1 - carefully read the problem statement, find out how many bodies are involved in heat exchange and what physical processes occur (for example, heating or cooling, melting or crystallization, vaporization or condensation);

Stage 2 - briefly write down the conditions of the problem, supplementing with the necessary tabular values; express all quantities in the SI system;

Stage 3 - write down the heat balance equation taking into account the sign of the amount of heat (if the body receives energy, then put the “+” sign, if the body gives it away, put the “-” sign);

Stage 4 - write down the necessary formulas for calculating the amount of heat;

Stage 5 - write down the resulting equation in general form relative to the required quantities;

Stage 6 - check the dimension of the resulting value;

Stage 7 - calculate the values ​​of the required quantities.

CALCULATION AND GRAPHIC WORKS

Job No. 1

INTRODUCTION BASIC CONCEPTS OF MECHANICS

Key points:

Mechanical movement is a change in the position of a body relative to other bodies or a change in the position of body parts over time.

A material point is a body whose dimensions can be neglected in this problem.

Physical quantities can be vector and scalar.

A vector is a quantity characterized by a numerical value and direction (force, speed, acceleration, etc.).

A scalar is a quantity characterized only by a numerical value (mass, volume, time, etc.).

Trajectory is a line along which a body moves.

The distance traveled is the length of the trajectory of a moving body, designation - l, SI unit: 1 m, scalar (has a magnitude, but no direction), does not uniquely determine the final position of the body.

Displacement is a vector connecting the initial and subsequent positions of the body, designation - S, unit of measurement in SI: 1 m, vector (has a module and direction), uniquely determines the final position of the body.

Speed ​​is a vector physical quantity equal to the ratio of the movement of a body to the period of time during which this movement occurred.

Mechanical motion can be translational, rotational and oscillatory.

Progressive movement is a movement in which any straight line rigidly connected to the body moves while remaining parallel to itself. Examples of translational motion are the movement of a piston in an engine cylinder, the movement of ferris wheel cabs, etc. During translational motion, all points of a rigid body describe the same trajectories and at each moment of time have the same velocities and accelerations.

Rotational the motion of an absolutely rigid body is a motion in which all points of the body move in planes perpendicular to a fixed straight line, called axis of rotation, and describe circles whose centers lie on this axis (rotors of turbines, generators and engines).

Oscillatory motion is a movement that repeats itself periodically in space over time.

Reference system is a combination of a body of reference, a coordinate system and a method of measuring time.

Reference body- any body chosen arbitrarily and conventionally considered motionless, in relation to which the location and movement of other bodies is studied.

Coordinate system consists of directions identified in space - coordinate axes intersecting at one point, called the origin and the selected unit segment (scale). A coordinate system is needed to quantitatively describe movement.

In the Cartesian coordinate system, the position of point A at a given time relative to this system is determined by three coordinates x, y and z, or radius vector.

Trajectory of movement of a material point is the line described by this point in space. Depending on the shape of the trajectory, the movement can be straightforward And curvilinear.

The motion is called uniform if the speed of the material point does not change over time.

Actions with vectors:

Speed– a vector quantity showing the direction and speed of movement of a body in space.

Every mechanical movement has absolute and relative nature.

The absolute meaning of mechanical motion is that if two bodies approach or move away from each other, then they will approach or move away in any frame of reference.

The relativity of mechanical motion is that:

1) it makes no sense to talk about motion without indicating the body of reference;

2) in different reference systems the same movement can look different.

Law of addition of speeds: The speed of a body relative to a fixed frame of reference is equal to the vector sum of the speed of the same body relative to a moving frame of reference and the speed of the moving system relative to a stationary one.

Control questions

1. Definition of mechanical motion (examples).

2. Types of mechanical movement (examples).

3. The concept of a material point (examples).

4. Conditions under which the body can be considered a material point.

5. Forward movement (examples).

6. What does the frame of reference include?

7. What is uniform motion (examples)?

8. What is called speed?

9. Law of addition of velocities.

Complete the tasks:

1. The snail crawled straight for 1 m, then made a turn, describing a quarter circle with a radius of 1 m, and crawled further perpendicular to the original direction of movement for another 1 m. Make a drawing, calculate the distance traveled and the displacement module, do not forget to show the snail’s movement vector in the drawing.

2. A moving car made a U-turn, describing half a circle. Make a drawing showing the path and movement of the car in a third of the turning time. How many times is the distance traveled during the specified period of time greater than the modulus of the vector of the corresponding displacement?

3. Can a water skier move faster than a boat? Can a boat move faster than a skier?

Experts prove the advantage of technical education over the humanities, they prove that Russia is in dire need of highly qualified engineers and technical specialists, and this trend will continue not only in 2014, but also over the coming years. According to personnel selection specialists, if the country expects economic growth in the coming years (and there are prerequisites for this), then it is very likely that the Russian educational base will not be able to cope with many sectors (high technology, industry). “At the moment, there is an acute shortage of specialists in the labor market in the field of engineering and technical specialties, in the field of IT: programmers, software developers. Engineers of almost all specializations remain in demand. At the same time, the market is oversaturated with lawyers, economists, journalists, psychologists,” - says General Director of the Recruitment Agency for Unique Specialists Ekaterina Krupina. Analysts, making long-term forecasts until 2020, are confident that the demand for technical specialties will grow rapidly every year. Relevance of the problem. Therefore, the quality of preparation for the Unified State Exam in physics is important. Mastering methods for solving physical problems is crucial. A variety of physical tasks are graphical tasks. 1) Solving and analyzing graphical problems allows you to understand and remember the basic laws and formulas of physics. 2) In KIMs for the Unified State Examination in physics, tasks with graphic content are included.

Download work with presentation.

OBJECTIVE OF THE PROJECT WORK:

Studying the types of graphic problems, varieties, features and solution methods .

OBJECTIVES OF THE WORK:

1. Studying literature about graphic tasks; 2. Study of Unified State Exam materials (prevalence and level of complexity of graphic tasks); 3. Study of general and specific graphic problems from different branches of physics, degree of complexity. 4. Study of solution methods; 5. Conducting a sociological survey among school students and teachers.

Physics problem

In methodological and educational literature, educational physical tasks are understood as appropriately selected exercises, the main purpose of which is to study physical phenomena, form concepts, develop students’ physical thinking and instill in them the ability to apply their knowledge in practice.

Teaching students to solve physical problems is one of the most difficult pedagogical problems. I think this problem is very relevant. My project aims to solve two problems:

1. Help in teaching schoolchildren the ability to solve graphic problems;

2. Involve students in this type of work.

Solving and analyzing a problem allows you to understand and remember the basic laws and formulas of physics, create an idea of ​​their characteristic features and limits of application. Problems develop skills in using the general laws of the material world to solve specific issues of practical and educational significance. The ability to solve problems is the best criterion for assessing the depth of study of program material and its assimilation.

In studies to identify the degree to which students have mastered individual operations included in the ability to solve problems, it has been found that 30-50% of students in various classes indicate that they lack such skills.

Inability to solve problems is one of the main reasons for decreased success in studying physics. Studies have shown that the inability to solve problems independently is the main reason for irregular homework completion. Only a small part of students master the ability to solve problems, which they consider as one of the most important conditions for improving the quality of knowledge in physics.

This state of learning practice can be explained by the lack of clear requirements for the formation of this skill, the lack of internal motivations and cognitive interest among students.

Solving problems in the process of teaching physics has multifaceted functions:

• Mastering theoretical knowledge.
• Mastering the concepts of physical phenomena and quantities.
• Mental development, creative thinking and special abilities of students.
• Introduces students to the achievements of science and technology.
• Develops hard work, perseverance, will, character, and determination.
• It is a means of monitoring the knowledge, skills and abilities of students.

Graphic task.

Graphical tasks are those tasks in the process of solving which graphs, diagrams, tables, drawings and diagrams are used.

For example:

1. Construct a graph of the path of uniform motion if v = 2 m/s or uniformly accelerated motion if v 0 = 5 m/s and a = 3 m/s 2 .

2. What phenomena are characterized by each part of the graph...

3. Which body moves faster

4. In which area did the body move faster?

5. Determine the distance traveled from the speed graph.

6. In what part of the motion was the body at rest. The speed increased and decreased.

Solving graphic problems helps to understand the functional relationship between physical quantities, develop skills in working with graphs, and develop the ability to work with scales.

Based on the role of graphs in solving problems, they can be divided into two types: - problems, the answer to the question of which can be found as a result of constructing a graph; - tasks for which the answer can be found by analyzing the graph.

Graphic tasks can be combined with experimental ones.

For example:

Using a beaker filled with water, determine the weight of a wooden block...

Preparation for solving graphic problems.

To solve graphic problems, the student must know various types of functional dependencies, which means the intersection of graphs with axes and graphs with each other. You need to understand how the dependencies differ, for example, x = x 0 + vt and x = v 0 t + at 2 /2 or x = x m sinω 0 t and x = - x m sinω 0 t; x =x m sin(ω 0 t+ α) and x =x m cos (ω 0 t+ α), etc.

The preparation plan should contain the following sections:

· a) Repeat graphs of functions (linear, quadratic, power) · b) Find out what role graphs play in physics, what information they carry. · c) Systematize physical problems according to the significance of the graphs in them. · d) Study methods and techniques for analyzing physical graphs · e) Develop an algorithm for solving graphic problems in various branches of physics · f) Find out the general pattern in solving graphic problems. To master problem solving methods, it is necessary to solve a large number of different types of problems, observing the principle - “From simple to complex.” Starting with simple ones, master solution methods, compare, generalize different problems both on the basis of graphs and on tables, diagrams, diagrams. You should pay attention to the designation of quantities along the coordinate axes (units of physical quantities, the presence of submultiple or multiple prefixes), the scale, the type of functional dependence (linear, quadratic, logarithmic, trigonometric, etc.), the angles of inclination of the graphs, the points of intersection of the graphs with coordinate axes or graphs among themselves. It is necessary to approach problems with inherent “errors” especially carefully, as well as problems with photographs of measuring instrument scales. In this case, it is necessary to correctly determine the division value of the measuring instruments and accurately read the values ​​of the measured quantities. In problems involving geometric optics, it is especially important to carefully and accurately construct rays and determine their intersections with axes and with each other.

How to solve graphics problems

Mastering the general algorithm for solving physical problems

1. Carrying out an analysis of the problem conditions with the identification of system tasks, phenomena and processes described in the problem, with the determination of the conditions for their occurrence

2. Coding the problem conditions and the solution process at various levels:

a) a brief statement of the problem conditions;

b) making drawings and electrical diagrams;

c) execution of drawings, graphs, vector diagrams;

d) writing an equation (system of equations) or constructing a logical conclusion

3. Identification of the appropriate method and methods for solving a specific problem

4. Application of a general algorithm to solve problems of various types

Solving the problem begins with reading the conditions. You need to make sure that all terms and concepts in the condition are clear to students. Unclear terms are clarified after initial reading. At the same time, it is necessary to highlight what phenomenon, process or property of bodies is being described in the problem. Then the problem is read again, but with the data and required quantities highlighted. And only after this a brief recording of the conditions of the problem is carried out.

Planning

The action of orientation allows for a secondary analysis of the perceived conditions of the task, as a result of which physical theories, laws, equations that explain a specific task are identified. Then methods for solving problems of one class are identified and the optimal method for solving this problem is found. The result of student activity is a solution plan, which includes a chain of logical actions. The correctness of the actions to draw up a plan for solving the problem is monitored.

Solution process

First, it is necessary to clarify the content of already known actions. The action of orientation at this stage involves once again highlighting the method for solving the problem and clarifying the type of problem to be solved by the method of setting the conditions. The next step is planning. A method for solving the problem is planned, the apparatus (logical, mathematical, experimental) with the help of which it is possible to carry out its further solution.

Solution Analysis

The last stage of the problem solving process is to check the result obtained. It is carried out again by the same actions, but the content of the actions changes. The action of orientation is finding out the essence of what needs to be checked. For example, the results of the solution can be the values ​​of coefficients, physical constant characteristics of mechanisms and machines, phenomena and processes.

The result obtained from solving the problem must be plausible and consistent with common sense.

Prevalence of graphical tasks in computer simulation machines in Unified State Examination tasks

The study of Unified State Exam materials for a number of years (2004 - 2013) showed that graphical problems in various sections of physics are common in Unified State Exam assignments in various sections of physics. In tasks A: in mechanics - 2-3 in molecular physics - 1 in thermodynamics - 3 in electrodynamics - 3-4 in optics - 1-2 in quantum physics - 1 in atomic and nuclear physics - 1 In tasks B: in mechanics - 1 in molecular physics - 1 in thermodynamics - 1 in electrodynamics - 1 in optics - 1 in quantum physics - 1 in atomic and nuclear physics - 1 In tasks C: in mechanics - in molecular physics - in thermodynamics - 1 in electrodynamics - 1 in optics - 1 in quantum physics - in atomic and nuclear physics - 1

Our research

A. Analysis of errors when solving graphic problems

Analysis of solving graphic problems showed that the following common errors occur:

Errors in reading charts;

Errors in operations with vector quantities;

Errors when analyzing isoprocess graphs;

Errors in the graphical dependence of electrical quantities;

Errors when constructing using the laws of geometric optics;

Errors in graphic tasks on quantum laws and the photoelectric effect;

Errors in the application of the laws of atomic physics.

B. Sociological survey

In order to find out how school students are aware of graphic tasks, we conducted a sociological survey.

We asked the students and teachers of our school the following questions: profiles:

1. 1. What is a graphics task?

a) problems with pictures;

b) tasks containing diagrams, diagrams;

c) I don’t know.

1. 2. What are graphical tasks for?

b) to develop the ability to build graphs;

c) I don’t know.

3. Can you solve graphic problems?

a) yes; b) no; c) not sure ;

4. Do you want to learn how to solve graphic problems?

A) yes ; b) no; c) I find it difficult to answer.

50 people were interviewed. As a result of the survey, the following data were obtained:

CONCLUSIONS:

1. As a result of working on the “Graphical Tasks” project, we studied the features of graphic tasks.
2. We studied the features of the methodology for solving graphic problems.
3. We analyzed typical errors.
4. Conducted a sociological survey.

Activity reflection:

1. It was interesting for us to work on the problem of graphics tasks.
2. We learned how to conduct research activities, compare and contrast research results.
3. We found that mastery of methods for solving graphical problems is necessary for understanding physical phenomena.
4. We found out that mastery of methods for solving graphic problems is necessary for successfully passing the Unified State Exam.

Graphic puzzles

1. Connect the four points with three lines without lifting your hands and return to the starting point.

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1. Connect nine dots with four lines without lifting your hand.

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1. Show how to cut a rectangle with lines of 4 and 9 units into two equal parts so that when added they form a square.

1. The cube, painted on all sides, was sawn as shown in Fig.

a) How many cubes will you get?

Not painted at all?

b) How many cubes have colored

Will there be one edge?

c) How many cubes will have

Are two edges painted?

d) How many cubes are colored?

Will there be three sides?

e) How many cubes are colored?

Will there be four sides?

Situational, design

And technological challenges

Task. Balls of three sizes, under the influence of their own weight, roll down an inclined tray in a continuous stream. How to continuously sort balls into groups depending on size?

Solution. It is necessary to develop the design of a calibrating device.

The balls, having left the tray, roll further along a wedge-shaped gauge. In the place where the width of the slot coincides with the diameter of the ball, it falls into the corresponding receiver.

Task. The heroes of one science fiction story take on a flight, instead of thousands of necessary spare parts, a synthesizer-machine that can do everything. When landing on another planet, the ship is damaged. You need 10 identical parts for repair. Here it turns out that the synthesizer does everything in one copy. How to find a way out of this situation?

Solution. You need to order the synthesizer to produce itself. The second synthesizer gives them another one, etc.

Answers to graphic puzzles.

1. . .

2. . . .

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