Graphic puzzles. Graphic tasks solved on sea charts
Enrolled bypassing exams. Even in our time, this riddle is considered one of the better ways testing attention and logic of thinking.
Well, let's get started!
- How many tourists live in this camp?
- When did they come here: today or a few days ago?
- Why did they come here?
- Is it far from the camp to the nearest village?
- Where does the wind blow: from the north or south?
- What time of day is it?
- Where did Shura go?
- Who was on duty yesterday (say by name)?
- What day of what month is today?
Answers:
- Four. If you look closely, you can see: cutlery for 4 people, and there are 4 names on the duty list.
- Not today, judging by the web between the tree and the tent, the guys arrived a few days ago.
- On the boat. There are oars near the tree.
- No. There is a chicken in the picture, which means that the village is somewhere nearby.
- From South. There is a flag on the tent by which you can determine where the wind is blowing from. There is a tree in the picture: on one side the branches are shorter, on the other longer. As a rule, at
- trees on the south side of the branch are longer.
- Morning. In the previous question, we determined where the north-south is, now you can understand where the east-west is, and look at the shadows that objects cast.
- He is catching butterflies. A net is visible from behind the tent.
- Kolya. Today, Kolya is looking for something in a backpack with the letter “K”, Shura is catching butterflies, and Vasya is taking pictures of nature (because a tripod from the camera is visible from the backpack with the letter “B”).
- So, today Petya is on duty, and yesterday, according to the list, Kolya was on duty.
- 8 August. Judging by the list, since Petya is on duty today, the number is 8. And since there is a watermelon in the clearing, it means August.
According to statistics, only 7% correctly answer all questions.
The riddle is really very difficult, in order to correctly answer all the questions you need to understand some aspects, and of course you need to connect logic and attentiveness. The riddle is complicated by a not very high-quality image. I wish you success.
Looking at the picture, answer the following questions:
- How long have the guys been involved in tourism?
- Are they familiar with home economics?
- Is the river navigable?
- In which direction does it flow?
- What is the depth and width of the river at the next rift?
- How long will the laundry take to dry?
- How much more sunflower will grow?
- Is there a tourist camp far from the city?
- What transport did the guys get here?
- Do they like dumplings in these places?
- Is the newspaper up to date? (Newspaper dated August 22)
- Which city is the plane flying to?
Answers:
- Obviously, recently: experienced tourists will not pitch a tent in a hollow.
- In all likelihood, not very much: they don’t clean the fish from the head, it’s inconvenient to sew on a button with too long a thread, it’s necessary to chop a branch with an ax on a block of wood.
- navigable. This is evidenced by the navigation mast standing on the shore.
- From left to right. Why? See the answer to the next question.
- The navigation sign on the river bank is set in a strictly defined way. If you look from the side of the river, then signs are hung downstream to the right, showing the width of the river at the nearest rift, and to the left - signs showing the depth. The depth of the river is 125 cm (rectangle 1 m, large circle 20 cm and small circle 5 cm), the width of the river is 30 m (large circle 20 m and 2 small circles 5 m each). Such signs are installed 500 m before the roll.
- Not for long. There is a wind: the floats of the fishing rods were carried against the current.
- The sunflower is obviously broken and stuck in the ground, since its "hat" is not facing the sun, and a broken plant will no longer grow.
- No further than 100 km, at a greater distance, the body antenna would be of a more complex design.
- The guys have, in all likelihood, bicycles: there is a bicycle wrench on the ground.
- No. They love dumplings here. The hut, the pyramidal poplar and the high altitude of the sun above the horizon (63° - in the shadow of the sunflower) show that this is a Ukrainian landscape.
- Judging by the height of the sun above the horizon, it takes place in June. For Kyiv, for example, 63° is the highest angular height of the sun. This happens only at noon on June 22. The newspaper is dated August - therefore, it is at least last year.
- None. The aircraft produces agricultural work.
Here is a problem in the 60s of the last century offered to solve the students of the second grade.
Looking at the picture, answer the following questions:
- Is a steamboat going up or down the river?
- What season is shown here?
- Is the river deep in this place?
- Is the harbor far?
- Is it on the right or left bank of the river?
- What time of day did the artist show in the drawing?
Answers:
- The wooden triangles on which the buoys are fixed are always directed against the current. The ship is sailing up the river.
- The figure shows a flock of birds; they fly in the form of an angle, one of its sides is shorter than the other: these are cranes. Flocking flight of cranes occurs in spring and autumn. From the crowns of trees on the edge of the forest, you can determine where the south is: they always grow thicker on the side that faces south. Cranes fly south. So, the picture shows autumn.
- The river in this place is shallow: a sailor, standing on the bow of the steamer, measures the depth of the fairway with a sixth.
- Obviously, the ship is approaching the pier: a group of passengers, taking their things, prepared to get off the ship.
- Answering the 1st question, we determined in which direction the river flows. In order to indicate where the right and where the left bank of the river is, one must stand facing downstream. We know that the ship is mooring at the wharf. It can be seen that the passengers are getting ready to go to the side from where you are looking at the picture. So the nearest pier is on the right bank of the river.
- On the beacons - lanterns; put them before evening and take off early in the morning. It can be seen that the shepherds are driving the flock to the village. From here we come to the conclusion that the figure shows the end of the day.
Looking at the picture, answer the following questions:
- What time of year is this apartment shown?
- What month?
- Is the boy you see going to school now, or is he on vacation?
- Does the apartment have running water?
- Who lives in this apartment besides the father and son you see in the picture?
- What is the profession of the father?
Answers:
- The apartment is shown in winter: a boy in felt boots; the stove is heated - this is indicated by an open air vent.
- Month of December: the last sheet of the calendar is open.
- The first 7 numbers are crossed out on the calendar: they have already passed. The winter vacation start later. So the boy goes to school.
- If the apartment had running water, then you would not have to use the washstand, which is shown in the figure.
- The dolls indicate that there is a girl in the family, probably of preschool age.
- A tube and a hammer for listening to patients indicate that the father is a doctor by profession.
Soviet riddles for logic: 8 questions for attentiveness
Another Soviet riddle, this one will be more difficult than the previous one. Only 4% of people can answer all 8 questions correctly.
Looking at the picture, answer the following questions:
- What time of day is shown in the picture?
- Does the drawing depict early spring or late autumn?
- Is this river navigable?
- In which direction does the river flow: south, north, west or east?
- Is the river deep near the bank where the boat is parked?
- Is there a bridge across the river nearby?
- Is the railroad far from here?
- Do cranes fly north or south?
Answers:
- Having examined the picture, you see that sowing is going on in the field (a tractor with a seeder and wagons with grain). As you know, sowing is done in autumn or early spring. Autumn sowing takes place when there are still leaves on the trees. In the picture, the trees and bushes are completely bare. It should be concluded that the artist depicted early spring.
- In spring, cranes fly from south to north.
- Buoys, that is, signs marking the fairway, are placed only on navigable rivers.
The buoy is fixed on a wooden float, which is always directed at an angle against the flow of the river. - Having determined by the flight of cranes where the north is, and paying attention to the position of the triangle with the buoy, it is not difficult to decide that in this place the river flows from north to south.
- The direction of the shadow from the tree shows that the sun is in the southeast. In the spring, on this side of the sky, the sun is at 8 - 10 o'clock in the morning.
- A railway conductor with a lantern is sent to the boat; he obviously lives somewhere near the station.
- The footbridges and stairs descending to the river, as well as a boat with passengers, show that a constant transportation across the river has been established in this place. He is needed here because there is no bridge nearby.
- On the shore you see a boy with a fishing rod. Only when fishing in a deep place can you move the float so far away from the hook.
If you liked this riddle, then try another one
Soviet logic puzzle about the railway (near the road)
Looking at the picture, answer the following questions:
- How long before the new moon?
- Will the night come soon?
- What time of year does the drawing belong to?
- In which direction does the river flow?
- Is she navigable?
- How fast is the train moving?
- How long has the previous train passed here?
- How long will the car move along the railroad?
- What should the driver prepare for now?
- Is there a bridge nearby?
- Is there an airfield in the area?
- Is it easy for drivers of oncoming trains to slow down the train in this section?
- Does the wind blow?
Answers:
- A little. The month is old (you can see its reflection in the water).
- Not soon. The old month is visible at dawn.
- Autumn. By the position of the sun, it is easy to figure out that the cranes are flying south.
- Rivers flowing in the Northern Hemisphere have a steep right bank. So the river flows from us to the horizon.
- navigable. Beacons are visible.
- The train is standing. The lower eye of the traffic light is lit - red.
- Recently. He is now at the nearest blocking area.
- Road sign indicates that there is a railroad crossing ahead.
- To braking. The road sign shows that there is a steep descent ahead.
- Probably there is. There is a sign obliging the driver to close the blower.
- In the sky, the trace of the plane that made the loop. Aerobatics are allowed to be done only not far from airfields.
- A sign near the railway track indicates that the oncoming train will have to climb up the slope. It will be easy to slow him down.
- Duet. The smoke of the locomotive spreads, but the train, as we know, is motionless.
These are the Soviet riddles for logic in pictures (riddles of the USSR for children). Did everyone get it right? - I don't think so! But it was still time well spent!
Write comments, perhaps there will be questions or new riddles from you.
Graphic puzzles
- Connect the four points with three lines without taking your hands off and return to the starting point.
. .
- Connect nine dots with four lines without taking your hands off.
. . .
. . .
. . .
- Show how to cut a rectangle with rows 4 and 9 units into two equal parts so that when they are added, they get a square.
- A cube, colored on all sides, was sawn as shown in fig.
a) How many cubes
Not dyed at all?
b) How many cubes of colored
Will there be one edge?
c) How many cubes will have
Are two faces painted?
d) How many cubes are colored
Will there be three edges?
e) How many cubes are colored
Will there be four edges?
Situational, design
And technological challenges
Task. Balls of three sizes under the influence of their own weight roll down the 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 the calibrating device.
The balls, leaving the tray, roll further along the wedge-shaped caliber. 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 fantastic 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 to repair. It turns out that the synthesizer does everything in one instance. How to find a way out of this situation?
Solution. It is necessary to order the synthesizer to produce itself. The second synth gives them another one, and so on.
Answers to graphic puzzles.
1. . .
2. . . .
. . .
. . .
If there are only two variables in a linear programming problem, then it can be solved graphic method.
Consider a linear programming problem with two variables and :
(1.1)
;
(1.2)
Here , are arbitrary numbers. The task can be both to find the maximum (max) and to find the minimum (min). In the system of restrictions, both signs and signs can be present.
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 constraint system (1.2) defines a half-plane bounded by the corresponding line.
So the first inequality
(1.2.1)
defines a half-plane bounded by a line. On one side of this line, and on the other side. On the straightest line. To find out from which side the inequality (1.2.1) is satisfied, we choose an arbitrary point that does not lie on the line. Next, we substitute the coordinates of this point in (1.2.1). If the inequality holds, then the half-plane contains the chosen point. If the inequality is not satisfied, then the half-plane is located on the other side (does not contain the selected point). We shade the half-plane for which inequality (1.2.1) is satisfied.
We do the same for the remaining inequalities of system (1.2). So we get the shaded half-planes. Area points feasible solutions satisfy all inequalities (1.2). Therefore, graphically, the area of feasible solutions (ODD) is the intersection of all constructed half-planes. We shade ODR. It is a convex polygon whose faces belong to the constructed lines. Also, ODR 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 admissible solutions is the empty set. This problem has no solutions.
You can simplify the method. You can not shade each half-plane, but first build all the lines
(2)
Next, choose 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 area of feasible solutions is limited by the constructed lines and includes the chosen point. We shade the area of admissible 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, the coordinates of which satisfy the system (1.2).
Finding the extremum of the objective function
So, we have a shaded area of feasible solutions (ODD). It is bounded by a broken line consisting of segments and rays belonging to the constructed lines (2). ODR is always a convex set. It can be either a bounded set or an unbounded set 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 ODS. On this straight line, the objective function is constant and equal to . such a straight line is called the level line of the function. This line divides the plane into two half-planes. On one half plane
.
On the other half plane
.
That is, on one side of the straight line (3), the objective function increases. And the further we move the point away from the line (3), the greater the value will be. On the other side of the straight line (3), the objective function decreases. And the further we move the point away from the line (3) to the other side, the smaller the value will be. If we draw a line parallel to line (3), then the new line will also be the objective function level line, 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 the straight line (3), as far as possible from it in the direction of increasing values of , and passing through at least one point of the ODT. To find the minimum value of the objective function, it is necessary to draw a straight line parallel to the straight line (3) and as far as possible from it in the direction of decreasing values of , and passing through at least one point of the ODT.
If the ODE is unbounded, then a case may arise when such a straight 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.
Consider the case when the extreme line parallel to an arbitrary line of the form (3) passes through one vertex of the ODD 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 ODS. Then the line passes through two vertices of the ODD 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 themselves and .
An example of solving a linear programming problem by a graphical method
The task
The company produces dresses of two models A and B. Three types of fabric are used. For the manufacture of one model A dress, 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. For the manufacture of 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, the second type - 10 m, the third type - 16 m. The release of one product of type A brings an income of 400 den. unit, one product of 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 produced dresses of models A and B, respectively. Then the amount of tissue used of the first type will be:
(m)
The amount of fabric used of the second type will be:
(m)
The amount of fabric used of the third type will be:
(m)
Since the number of dresses produced cannot be negative, then
And .
The income from the produced dresses will be:
(den. units)
Then the economic-mathematical model of the problem has the form:
We solve it graphically.
Draw the coordinate axes and .
We build a straight line.
At .
At .
We draw a straight line through the points (0; 7) and (10.5; 0).
We build a straight line.
At .
At .
We draw a straight line through the points (0; 10) and (10; 0).
We build a straight line.
At .
At .
We 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.
(P1.1) .
At .
At .
We draw a straight line through the points (0; 4) and (3; 0).
Further, we note that since the coefficients for and of the objective function are positive (400 and 300), then it increases with increasing and . We draw a straight line parallel to the straight line (A1.1), as far as possible from it in the direction of increase, and passing through at least one point of the quadrilateral OABC. Such a straight line passes through the 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 in this case will be 3200 den. units
Example 2
The task
Solve a linear programming problem using a graphical method.
Solution
We solve it graphically.
Draw the coordinate axes and .
We build a straight line.
At .
At .
We draw a straight line through the points (0; 6) and (6; 0).
We build a straight line.
From here.
At .
At .
We draw a straight line through the points (3; 0) and (7; 2).
We build a straight line.
We build a straight line (abscissa axis).
The domain of admissible solutions (DDR) is limited by the constructed straight lines. To find out from which side, we notice that the point belongs to the ODT, since it satisfies the system of inequalities:
We shade the area along the boundaries of the constructed lines so that the point (4; 1) falls into the shaded part. We get triangle ABC.
We construct an arbitrary level line of the objective function, for example,
.
At .
At .
We draw a straight level line through the 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 as possible from it in the direction of increasing , and passing through at least one point of the triangle ABC. Such a straight line passes through the point C. From the construction, we determine its coordinates.
.
The solution of the problem: ;
Answer
Example of no solution
The task
Solve graphically the problem of linear programming. Find the maximum and minimum value of the objective function.
Solution
We solve the problem graphically.
Draw the coordinate axes and .
We build a straight line.
At .
At .
We draw a straight line through the points (0; 8) and (2.667; 0).
We build a straight line.
At .
At .
We draw a straight line through the points (0; 3) and (6; 0).
We build a straight line.
At .
At .
We draw a straight line through the points (3; 0) and (6; 3).
The lines and are the coordinate axes.
The domain of admissible solutions (SDR) is limited by the constructed straight lines and coordinate axes. To find out from which side, we notice that the point belongs to the ODT, since it satisfies the system of inequalities:
We shade the area so that the point (3; 3) falls into the shaded part. We get an unlimited area bounded by the broken line ABCDE.
We construct an arbitrary level line of the objective function, for example,
(P3.1) .
At .
At .
We draw a straight line through the points (0; 7) and (7; 0).
Since the coefficients at and are positive, then increases with increasing and .
To find the maximum, you need to draw a parallel line, as far as possible in the direction of increase, and passing through at least one point of the region ABCDE. However, since the region is unbounded on the side of large values of and , such a straight line cannot be drawn. Whatever straight line we draw, there will always be points in the region that are more distant in the direction of increase and . Therefore, there is no maximum. you can make it as big as you want.
We are looking for the minimum. We draw a straight line parallel to the straight line (A3.1) and as far as possible from it in the direction of decreasing , and passing through at least one point of the region ABCDE. Such a straight line passes through the point C. From the construction, we determine its coordinates.
.
The minimum value of the objective function:
Answer
There is no maximum value.
Minimum value
.
Tasks of this type include those in which all or part of the data is given in the form of graphical dependencies between them. In solving such problems, the following stages can be distinguished:
Stage 2 - to find out from the above graph, between which quantities the relationship is presented; find out which physical quantity is independent, i.e., an argument; what value is dependent, i.e., a function; determine by the type of graph what kind of dependence it is; find out what is required - to 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 point of intersection to the y-axis (or abscissa) and determine the value of the desired value;
Stage 4 - evaluate the result;
Stage 5 - write down the answer.
To read the graph of the coordinates means that from the graph one should determine: the initial coordinate and the speed of movement; write down the coordinate equation; determine the time and place of the meeting of the bodies; determine at what point in time the body has a given coordinate; determine the coordinate that the body has at the specified time.
Tasks of the fourth type - experimental . These are tasks in which, in order to find an unknown quantity, it is required to measure a part of the data empirically. The following workflow is suggested:
Stage 2 - to determine what phenomenon, the law underlies the experience;
Stage 3 - think over the scheme of experience; determine the list of devices 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 a table;
Stage 5 - make the necessary calculations, if required according to the condition of the problem;
Stage 6 - think about the results 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 out the numerical values of the given values; express all quantities in SI units;
Stage 3 - make a schematic drawing (trajectory of motion, vectors of speed, acceleration, displacement, etc.);
Stage 4 - choose a coordinate system (in this case, you should choose such a system so that the equations are simple);
Stage 5 - to compose for a given movement the basic equations 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 view, in letter notation, i.e. get the calculation formula;
Stage 7 - select a system of units of measurement ("SI"), substitute the names of the units in the calculation formula instead of letters, perform actions with the names and check whether the result is a unit of measurement of the desired value;
Stage 8 - Express all the given values in the chosen system of units; substitute in the calculation formulas and calculate the values of the required quantities;
Stage 9 - analyze the solution and formulate an answer.
Comparison of 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 in solving problems.
Algorithm for solving problems in dynamics:
Stage 2 - write down the condition of the problem, expressing all quantities in units of "SI";
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 the equations;
Stage 7 - solve the problem in a general way, i.e. solve an equation or system of equations for an unknown quantity;
Stage 8 - check the dimension;
Stage 9 - get a numerical result and correlate it with the real values of the quantities.
Algorithm for solving problems for thermal phenomena:
Stage 1 - carefully read the condition of the problem, find out how many bodies are involved in heat transfer and what physical processes occur (for example, heating or cooling, melting or crystallization, vaporization or condensation);
Stage 2 - briefly write down the condition 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 - the “-” sign);
Stage 4 - write down the necessary formulas for calculating the amount of heat;
Stage 5 - write down the resulting equation in general terms with respect to the desired values;
Stage 6 - check the dimension of the obtained value;
Stage 7 - calculate the values of the desired quantities.
CALCULATION AND GRAPHIC WORKS
Job #1
INTRODUCTION BASIC CONCEPTS OF MECHANICS
Basic provisions:
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 are vector and scalar.
A vector is a quantity characterized by numerical value and direction (force, speed, acceleration, etc.).
A scalar is a quantity characterized only by a numerical value (mass, volume, time, etc.).
Trajectory - the line along which the body moves.
The distance traveled - the length of the trajectory of a moving body, the designation - l, SI unit: 1 m, scalar (has a modulus but no direction), does not unambiguously determine the final position of the body.
Displacement - a vector connecting the initial and subsequent positions of the body, designation - S, unit of measure in SI: 1 m, vector (has a module and direction), uniquely determines the final position of the body.
Velocity is a vector physical quantity equal to the ratio of the movement of the body to the time interval during which this movement occurred.
Mechanical motion is translational, rotational and oscillatory.
Translational motion is a motion in which any straight line, rigidly connected with 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. In translational motion, all points solid body describe the same trajectories and at each moment of time have the same speeds and accelerations.
rotational motion of an absolutely rigid body is such 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).
vibrational motion is a motion that periodically repeats itself in space over time.
Reference system is called the totality of the body of reference, the coordinate system and the method of measuring time.
Reference body- any body, chosen arbitrarily and conditionally considered to be motionless, relative to which the location and movement of other bodies is studied.
Coordinate system consists of directions allocated in space - coordinate axes intersecting at one point, called the origin and the selected single segment(scale). The coordinate system is needed for a quantitative description of the movement.
In the Cartesian coordinate system, the position of point A at a given moment of time with respect to this system is determined by three x, y and z coordinates, or radius vector .
Trajectory of movementmaterial point the line described by this point in space is called. Depending on the shape of the trajectory, the movement can be straightforward And curvilinear.
The motion is called uniform if the speed of a material point does not change over time.
Actions with vectors:
Speed- a vector quantity showing the direction and speed of movement of the body in space.
Every mechanical movement has absolute and relative character.
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 is meaningless to talk about motion without specifying the reference body;
2) in different reference systems, the same movement may look different.
The 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 a moving frame relative to a fixed one.
1. Definition of mechanical movement (examples).
2. Types of mechanical movement (examples).
3. The concept of a material point (examples).
4. Conditions under which a body can be considered a material point.
5. Translational movement (examples).
6. What does the reference system include?
7. What is uniform motion (examples)?
8. What is called speed?
9. The law of addition of speeds.
Complete the tasks:
1. The snail crawled straight for 1 m, then made a turn, describing a quarter of a circle with a radius of 1 m, and crawled further perpendicular to the original direction of movement for another 1 m.
2. A moving car made a U-turn, describing half a circle. Make a drawing on which to indicate the path and movement of the car in a third of the turnaround time. How many times is the path traveled in the specified time interval 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 superiority of technical education over the humanities, 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 in subsequent years. According to recruiters, 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 handle many industries (high technology, industry). "At the moment, there is an acute shortage of specialists 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," speaks CEO recruitment agency unique specialists Ekaterina Krupin. Analysts, making long-term forecasts until 2020, are sure: the demand for technical specialties will grow rapidly every year. The urgency of the problem. Therefore, the quality of preparation for the exam in physics is relevant. Mastering the methods for solving physical problems is decisive. A variety of physical tasks are graphic tasks. 1) The solution and analysis of graphic problems allow you to understand and remember the basic laws and formulas in physics. 2) In CIMs for conducting the exam in physics included tasks with graphic content.Download work with presentation.
PURPOSE OF THE PROJECT WORK:
The study of types of graphic tasks, varieties, features and methods of solution .WORK OBJECTIVES:
1. Study of literature on graphic tasks; 2. Study USE materials(prevalence and level of complexity of graphic tasks); 3. The study of general and special graphic problems from different branches of physics, the degree of complexity. 4. Study of solution methods; 5. Conducting a sociological survey among students and teachers of the school.Physical challenge
In methodical and educational literature educational physical tasks are understood as expediently selected exercises, the main purpose of which is to study physical phenomena, form concepts, develop the physical thinking of students 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 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.
The solution and analysis of the problem allow us to understand and remember the basic laws and formulas of physics, create an idea of their characteristic features and limits of application. Tasks develop skills in using the general laws of the material world to solve specific issues of practical and cognitive importance. The ability to solve problems is the best criterion for assessing the depth of study of the program material and its assimilation.
In studies to identify the degree of assimilation by students of individual operations that are part of the ability to solve problems, it was found that 30-50% of students in various classes indicate that they do not have such a skill.
The inability to solve problems is one of the main reasons for the decline in success in the study of physics. Studies have shown that the inability to independently solve problems is the main reason for irregular homework. Only a small part of students masters the ability to solve problems, considers it as one of the most important conditions for improving the quality of knowledge in physics.
This state in the practice of teaching can be explained by the lack of clear requirements for the formation of this skill, the lack of internal incentives and cognitive interest among students.
Solving problems in the process of teaching physics has many-sided 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.
- It brings up diligence, perseverance, will, character, purposefulness.
- It is a means of monitoring the knowledge, skills and abilities of students.
Graphic task.
Graphics tasks- these are tasks in the process of solving which graphs, diagrams, tables, drawings and diagrams are used.
For example:
1. Build a path graph uniform motion, if v \u003d 2 m / s or uniformly accelerated at v 0 \u003d 5 m / s and a \u003d 3 m / s 2.
2. What phenomena characterizes each part of the graph ...
3. Which body is moving faster
4. In what area did the body move faster
5. Determine the distance traveled from the speed graph.
6. On what part of the movement the body rested. The speed increased and decreased.
Solving graphical problems helps to understand the functional relationship between physical quantities, instilling skills in working with graphs, and developing the ability to work with scales.
According to the role of graphs in solving problems, they can be divided into two types: - tasks, the answer to the question of which can be found as a result of constructing a graph; - tasks, the answer to the question of which 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 piece of wood...
Preparation for solving graphic problems.
To solve graphic problems, the student must know different kinds functional dependencies, which means the intersection of graphs with axes, graphs among themselves. You need to understand how the dependencies differ, for example, x \u003d x 0 + vt and x \u003d v 0 t + at 2 / 2 or x \u003d x m sinω 0 t and x \u003d - 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 the 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 importance of graphs in them. d) Study methods and techniques for analyzing physical graphs e) Develop an algorithm for solving graphic problems in various areas of physics f) Find out general pattern in solving graphic problems. To master the methods of solving problems, 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 the methods of solving, compare, generalize different problems both on the basis of graphs and on the basis of tables, diagrams, diagrams. Attention should be paid to the designation of quantities along the coordinate axes (units physical quantities, the presence of longitudinal or multiple prefixes), scale, type of functional dependence (linear, quadratic, logarithmic, trigonometric, etc.), on the angles of inclination of the graphs, the points of intersection of graphs with coordinate axes or graphs with each other. Particular attention should be paid to tasks with embedded "mistakes", as well as to tasks with photographs of measuring instrument scales. In this case, it is necessary to correctly determine the division value of measuring instruments and accurately read the values of the measured quantities. In geometrical optics problems, it is especially important to carefully and accurately build rays and determine their intersections with the axes and with each other.
How to solve graphics problems
Mastering the general algorithm for solving physical problems
1. Implementation of the analysis of the conditions of the problem with the allocation of the tasks of the system, phenomena and processes described in the problem, with the definition of the conditions for their flow
2. Implementation of coding the condition of the problem and the solution process at various levels:
a) a brief statement of the condition of the problem;
b) execution of drawings, electrical circuits;
c) execution of drawings, graphs, vector diagrams;
d) writing an equation (system of equations) or building a logical conclusion
3. Selection of the appropriate method and methods for solving a specific problem
4. Application of a general algorithm for solving problems of various types
The solution of the problem begins with reading the condition. You need to make sure that all the terms and concepts in the condition are clear to the students. Incomprehensible terms are clarified after the initial reading. At the same time, it is necessary to single out which phenomenon, process or property of bodies is described in the problem. Then the task is read again, but with the selection of data and the desired values. And only after that, a brief record of the condition of the problem is carried out.
Planning
The action of orientation makes it possible to carry out a secondary analysis of the perceived condition of the task, as a result of which physical theories, laws, equations, explaining a specific problem. Then, methods for solving problems of one class are singled out and the optimal method for solving this problem is found. The result of the students' 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 controlled.
Solution Process
First, it is necessary to clarify the content of already known actions. Orientation action this stage involves once again highlighting the method of solving the problem and clarifying the type of problem being solved by the method of setting the condition. The next step is planning. A method for solving the problem is planned, that apparatus (logical, mathematical, experimental) with the help of which it is possible to carry out its further solution.
Solution Analysis
The last step in the process of solving the problem is to check the result. It is carried out again by the same actions, but the content of the actions changes. The act of orientation is to ascertain the essence of what needs to be tested. For example, the results of the solution can be the values of the coefficients, physical constant characteristics of mechanisms and machines, phenomena and processes.
The result obtained in the course of solving the problem must be plausible and consistent with common sense.
The prevalence of graphics tasks in CMMs in USE assignments
The study of the USE materials for a number of years (2004 - 2013) showed that in the USE tasks in various sections of physics, graphic tasks in various sections of physics are common. 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 - 1Our research
A. Analysis of errors in solving graphic problems
An analysis of the solution of graphic problems showed that the following common errors occur:
Errors in reading charts;
Errors in operations with vector quantities;
Errors in the analysis of graphs of isoprocesses;
Errors in the graphical dependence of electrical quantities;
Errors in construction using the laws of geometric optics;
Errors in graphic assignments for quantum laws and the photoelectric effect;
Errors in the application of the laws of atomic physics.
B. Opinion poll
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. What is a graphic task?
a) tasks with pictures;
b) tasks containing schemes, diagrams;
c) I don't know.
- 2. What are graphics tasks for?
b) to develop the ability to build graphics;
c) I don't know.
3. Can you solve graphics problems?
a) yes; b) no; c) not sure ;
4. Do you want to learn how to solve graphic problems?
A) yes ; b) no; c) find it difficult to answer.
50 people were interviewed. As a result of the survey, the following data were obtained:
CONCLUSIONS:
- As a result of work on the project "Graphic Tasks", we studied the features of graphic tasks.
- We studied the features of the methodology for solving graphic problems.
- An analysis of typical errors was carried out.
- Conducted a sociological survey.
Reflection of activity:
- It was interesting for us to work on the problem of graphic tasks.
- We have learned to research activities, compare and compare research results.
- We have found that mastering the methods of solving graphic problems is necessary for understanding physical phenomena.
- We found out that mastering the methods of solving graphic problems is necessary for the successful passing of the exam.
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