Equally accelerated motion is a motion in which the acceleration vector does not change in magnitude and direction. Examples of such movement: a bicycle that is rolling down a hill; a stone thrown at an angle to the horizon. Uniform movement is a special case uniformly accelerated motion with acceleration equal to zero.
Let's consider the case of free fall (the body is thrown at an angle to the horizon) in more detail. Such movement can be represented as the sum of movements about the vertical and horizontal axes.
At any point of the trajectory, the body is affected by the acceleration of gravity g →, which does not change in magnitude and is always directed in one direction.
Along the X axis, the motion is uniform and rectilinear, and along the Y axis, uniformly accelerated and rectilinear. We will consider the projections of the vectors of speed and acceleration on the axis.
Formula for uniformly accelerated speed:
Here v 0 - the initial speed of the body, a \u003d c o n s t - acceleration.
Let us show on the graph that with uniformly accelerated motion the dependence v (t) has the form of a straight line.
Acceleration can be determined from the slope of the speed graph. In the picture above, the acceleration module equal to ratio sides of triangle ABC.
a \u003d v - v 0 t \u003d B C A C
The greater the angle β, the greater the slope (steepness) of the graph with respect to the time axis. Accordingly, the greater the acceleration of the body.
For the first graph: v 0 \u003d - 2 ms; a \u003d 0.5 m s 2.
For the second graph: v 0 \u003d 3 ms; a \u003d - 1 3 m s 2.
Using this graph, you can also calculate the movement of the body in time t. How to do it?
Let us select a small time interval ∆ t on the graph. We will assume that it is so small that the motion in the time ∆ t can be considered uniform motion with the speed, equal speed bodies in the middle of the interval ∆ t. Then, the displacement ∆ s during the time ∆ t will be equal to ∆ s \u003d v ∆ t.
We divide all time t into infinitely small intervals ∆ t. The displacement s during time t is equal to the area of \u200b\u200bthe trapezoid O D E F.
s \u003d O D + E F 2 O F \u003d v 0 + v 2 t \u003d 2 v 0 + (v - v 0) 2 t.
We know that v - v 0 \u003d a t, so the final formula for moving the body will take the form:
s \u003d v 0 t + a t 2 2
In order to find the coordinate of the body at a given time, you need to add displacement to the initial coordinate of the body. The change in coordinates depending on time expresses the law of uniformly accelerated motion.
Law of uniformly accelerated motion
Law of uniformly accelerated motiony \u003d y 0 + v 0 t + a t 2 2.
Another common problem in kinematics that arises in the analysis of uniformly accelerated motion is finding the coordinate at given values \u200b\u200bof the initial and final velocities and accelerations.
Eliminating t from the above equations and solving them, we get:
s \u003d v 2 - v 0 2 2 a.
From the known initial speed, acceleration and displacement, you can find the final speed of the body:
v \u003d v 0 2 + 2 a s.
For v 0 \u003d 0 s \u003d v 2 2 a and v \u003d 2 a s
Important!
The quantities v, v 0, a, y 0, s included in the expressions are algebraic quantities. Depending on the nature of the movement and the direction of the coordinate axes in the conditions of a specific task, they can take both positive and negative values.
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A straight line in space can always be defined as the intersection line of two non-parallel planes. If the equation of one plane, the equation of the second plane, then the equation of the straight line is given in the form
here 
noncollinear 
... These equations are called general equations
straight in space.
Canonical equations of the line
Any nonzero vector lying on a given line or parallel to it is called the direction vector of this line.
If the point is known 
line and its direction vector 
, then the canonical equations of the straight line have the form:
.
(9)
Parametric equations of a straight line
Let the canonical equations of the line
.
Hence, we obtain the parametric equations of the straight line:
(10)
These equations are useful for finding the point of intersection of a line and a plane.
Equation of a straight line passing through two points 
and 
looks like:
.
Angle between straight lines
Angle between straight lines
and 
is equal to the angle between their direction vectors. Therefore, it can be calculated using the formula (4):
The condition for parallelism of lines:
.
Condition of perpendicularity of planes:
Distance of a point from a straight line
P 
the point is given 
and straight
.
From the canonical equations of the straight line, the point 
belonging to a straight line and its direction vector 
... Then the distance of the point 
from a straight line is equal to the height of a parallelogram built on vectors 
and 
... Consequently,
.
Line intersection condition
Two non-parallel lines
,
intersect if and only if
.
The relative position of a straight line and a plane.
Let the straight line 
and plane. Angle 
between them can be found by the formula
.
Problem 73. Write the canonical equations of the line
(11)
Decision... In order to write down the canonical equations of the straight line (9), it is necessary to know any point belonging to the straight line and the direction vector of the straight line.
Find the vector 
parallel to the given line. Since it must be perpendicular to the normal vectors of the given planes, i.e.
,
then
.
From the general equations of the line we have that 
,
... Then
.
Since the point 
any point of a straight line, then its coordinates must satisfy the equations of a straight line and one of them can be specified, for example, 
, the other two coordinates are found from system (11):
Hence, 
.
Thus, the canonical equations of the desired line have the form:
or 
.
Problem 74.
and 
.
Decision. The coordinates of the point are known from the canonical equations of the first straight line 
belonging to the straight line and the coordinates of the direction vector 
... The coordinates of the point are also known from the canonical equations of the second line 
and the coordinates of the direction vector 
.
The distance between parallel lines is equal to the distance of a point 
from the second straight line. This distance is calculated by the formula
.
Find the coordinates of the vector 
.
We calculate the cross product 
:
.
Problem 75. Find point 
symmetrical point 
relatively straight
.
Decision... We write the equation of the plane perpendicular to the given straight line and passing through the point 
... As its normal vector 
you can take the directing vector of a straight line. Then 
... Consequently,
Find a point 
the point of intersection of this straight line and the plane P. To do this, we write down the parametric equations of the straight line, using Eqs. (10), we obtain
Consequently, 
.
Let be 
point symmetrical to point 
relative to this straight line. Then the point 
midpoint 
... To find the coordinates of a point 
we use the formulas for the coordinates of the midpoint of the segment:
,
,
.
So, 
.
Problem 76. Write the equation of the plane passing through the straight line 
and
a) through the point 
;
b) perpendicular to the plane.
Decision. Let us write down the general equations of this line. To do this, consider two equalities:
This means that the desired plane belongs to a bundle of planes with generators and its equation can be written in the form (8):
a) Find 
and 
from the condition that the plane passes through the point 
therefore, its coordinates must satisfy the plane equation. Substitute the coordinates of the point 
into the equation of the beam of planes:
Found value 
substitute in equation (12). we get the equation of the desired plane:
b) Find 
and 
from the condition that the desired plane is perpendicular to the plane. The normal vector of a given plane 
, the normal vector of the desired plane (see the equation of the bundle of planes (12).
Two vectors are perpendicular if and only if their dot product is zero. Consequently,
Substitute the found value 
into the equation of the beam of planes (12). We get the equation of the desired plane:
Tasks for independent decision
Problem 77. Bring the equations of straight lines to canonical form:
1)
2)
Problem 78. Write the parametric equations of the line 
, if a:
1)
,
;
2)
,
.
Assignment 79... Write the equation of the plane passing through the point 
perpendicular to straight 
Problem 80. Write the equations of the straight line passing the point 
perpendicular to the plane.
Problem 81. Find the angle between straight lines:
1)
and 
;
2)
and 
Problem 82. Prove the parallelism of lines:
and 
.
Problem 83. Prove the perpendicularity of the lines:
and 
Problem 84. Calculate point distance 
from straight line:
1)
;
2)
.
Problem 85. Calculate the distance between parallel lines:
and 
.
Assignment 86... In the equations of the straight line 
define parameter 
so that this line intersects with the line and find the point of their intersection.
Assignment 87... Show that straight 
parallel to plane 
and straight 
lies in this plane.
Assignment 88... Find point 
symmetrical point 
relative to the plane 
, if a:
1)
,
;
2)
,
;.
Problem 89. Write the equation of the perpendicular dropped from the point 
on a straight line 
.
Assignment 90... Find point 
symmetrical point 
relatively straight 
.
In July 2020, NASA will launch an expedition to Mars. Spacecraft will deliver to Mars an electronic carrier with the names of all registered members of the expedition.
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One of these code variants should be copied and pasted into the code of your web page, preferably between tags
and or right after the tag ... According to the first option, MathJax loads faster and slows down the page less. But the second option automatically tracks and loads the latest versions of MathJax. If you insert the first code, then it will need to be updated periodically. If you insert the second code, the pages will load more slowly, but you will not need to constantly monitor MathJax updates.The easiest way to connect MathJax is in Blogger or WordPress: in your site's dashboard, add a widget for inserting third-party JavaScript code, copy the first or second version of the download code above into it, and place the widget closer to the beginning of the template (by the way, this is not necessary at all because the MathJax script is loaded asynchronously). That's all. Now, learn the MathML, LaTeX, and ASCIIMathML markup syntax, and you're ready to embed math formulas into your website's web pages.
Another New Years Eve ... frosty weather and snowflakes on the window pane ... All this prompted me to write again about ... fractals, and what Wolfram Alpha knows about it. There is an interesting article about this, which contains examples of two-dimensional fractal structures. Here we will consider more complex examples three-dimensional fractals.
A fractal can be visualized (described) as a geometric figure or a body (meaning that both are a set, in this case, a set of points), the details of which have the same shape as the original figure itself. That is, it is a self-similar structure, considering the details of which with magnification, we will see the same shape as without magnification. Whereas in the case of the usual geometric shape (not a fractal), when we zoom in we will see details that have more simple formthan the original shape itself. For example, at a high enough magnification, part of the ellipse looks like a line segment. This does not happen with fractals: with any increase in them, we will again see the same complex shape, which with each increase will repeat again and again.
Benoit Mandelbrot, the founder of the science of fractals, wrote in his article Fractals and Art for Science: “Fractals are geometric shapes that are as complex in their details as in their general form... That is, if a part of the fractal is enlarged to the size of a whole, it will look like a whole, or exactly, or perhaps with a slight deformation. "
OoooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooTherefore, we will proceed to the first section, I hope, by the end of the article, I will remain cheerful.
The relative position of two straight lines
The case when the audience sings along with the chorus. Two straight lines can:
1) match;
2) be parallel:;
3) or intersect at a single point:.
Help for Dummies : please remember the mathematical sign of the intersection, it will be very common. The notation indicates that the line intersects with the line at a point.
How to determine the relative position of two straight lines?
Let's start with the first case:
Two straight lines coincide if and only if their corresponding coefficients are proportional, that is, there is such a number "lambda" that the equalities
Consider straight lines and compose three equations from the corresponding coefficients:. It follows from each equation that, therefore, these lines coincide.
Indeed, if all the coefficients of the equation 
multiply by -1 (change signs), and all coefficients of the equation 
reduced by 2, you get the same equation:.
The second case, when the lines are parallel:
Two straight lines are parallel if and only if their coefficients for the variables are proportional: 
but.
As an example, consider two lines. We check the proportionality of the corresponding coefficients for the variables: 
However, it is quite clear that.
And the third case, when the lines intersect:
Two straight lines intersect if and only if their coefficients for variables are NOT proportional, that is, there is NO such "lambda" value to make the equalities 
So, for straight lines we will compose the system 
From the first equation it follows that, and from the second equation:, therefore, the system is inconsistent (no solutions). Thus, the coefficients of the variables are not proportional.
Conclusion: lines intersect
In practical problems, you can use the solution scheme just considered. By the way, it is very similar to the algorithm for checking vectors for collinearity, which we considered in the lesson The concept of linear (non) dependence of vectors. Vector basis... But there is a more civilized packaging:
Example 1
Find out the relative position of the straight lines: 
Decision based on the study of direction vectors of straight lines:
a) From the equations we find the direction vectors of the straight lines: 
.
, so the vectors are not collinear and the lines intersect.
Just in case, I'll put a stone with pointers at the crossroads:
The rest jump over the stone and follow on, straight to Kashchei the Immortal \u003d)
b) Find the direction vectors of straight lines: 
Lines have the same direction vector, which means that they are either parallel or coincide. There is no need to count the determinant here.
It is obvious that the coefficients for the unknowns are proportional, while.
Let us find out whether the equality is true: 
Thus,
c) Find the direction vectors of straight lines: 
Let us calculate the determinant composed of the coordinates of these vectors: 
hence the direction vectors are collinear. Lines are either parallel or coincide.
The proportionality coefficient "lambda" is not difficult to see directly from the ratio of collinear direction vectors. However, it can also be found through the coefficients of the equations themselves: 
.
Now let's find out if the equality is true. Both free terms are zero, so:
The resulting value satisfies this equation (any number generally satisfies it).
Thus, the lines coincide.
Answer:
Very soon you will learn (or even have already learned) to solve the considered problem orally in just a few seconds. In this regard, I see no reason to propose something for an independent solution, it is better to lay another important brick in the geometric foundation:
How to build a straight line parallel to a given one?
For ignorance of this simplest task, the Nightingale the Robber severely punishes.
Example 2
The straight line is given by the equation. Equate a parallel line that goes through a point.
Decision: Let's denote the unknown direct letter. What does the condition say about her? The straight line goes through the point. And if the straight lines are parallel, then it is obvious that the directing vector of the straight line "tse" is also suitable for constructing the straight line "de".
We take out the direction vector from the equation:
Answer:
The geometry of the example looks simple: 
Analytical verification consists of the following steps:
1) We check that the lines have the same direction vector (if the equation of the line is not simplified properly, then the vectors will be collinear).
2) Check if the point satisfies the obtained equation.
Analytical review is in most cases easy to do orally. Look at the two equations, and many of you will quickly figure out the parallelism of straight lines without any drawing.
Examples for self-solution today will be creative. Because you still have to compete with Baba Yaga, and she, you know, is a lover of all kinds of riddles.
Example 3
Make an equation of a straight line passing through a point parallel to a straight line if
There is a rational and not very rational solution. The shortest way is at the end of the lesson.
We've done a bit of work with parallel lines and we'll come back to them later. The case of coinciding straight lines is of little interest, so consider a problem that is well known to you from the school curriculum:
How to find the intersection point of two lines?
If straight 
intersect at a point, then its coordinates are the solution systems of linear equations 
How to find the point of intersection of lines? Solve the system.
So much for you geometric meaning of the system of two linear equations with two unknowns Are two intersecting (most often) straight lines on a plane.
Example 4
Find the point of intersection of lines
Decision: There are two ways of solving - graphical and analytical.
The graphical way is to simply draw the data lines and find out the intersection point directly from the drawing: 
Here's our point:. To check it, you should substitute its coordinates in each equation of the line, they should fit both there and there. In other words, the coordinates of a point are the solution of the system. Basically, we looked at a graphical way to solve systems of linear equations with two equations, two unknowns.
The graphical method, of course, is not bad, but there are noticeable disadvantages. No, the point is not that seventh graders decide so, the point is that it will take time to get a correct and EXACT drawing. In addition, some straight lines are not so easy to construct, and the intersection point itself may be located somewhere in the thirty kingdom outside the notebook sheet.
Therefore, it is more expedient to look for the intersection point using the analytical method. Let's solve the system: 
To solve the system, the method of term-by-term addition of equations was used. Visit the lesson to develop relevant skills. How to solve a system of equations?
Answer:
The check is trivial - the coordinates of the intersection point must satisfy every equation in the system.
Example 5
Find the intersection point of the lines if they intersect.
This is an example for a do-it-yourself solution. It is convenient to divide the task into several stages. Analysis of the condition suggests what is needed:
1) Make the equation of the straight line.
2) Make the equation of the straight line.
3) Find out the relative position of the straight lines.
4) If the lines intersect, then find the intersection point.
The development of an algorithm of actions is typical for many geometric problems, and I will focus on this repeatedly.
Full solution and answer at the end of the tutorial:
A pair of shoes has not yet been worn down, as we got to the second section of the lesson:
Perpendicular straight lines. Distance from point to line.
Angle between straight lines
Let's start with a typical and very important task. In the first part, we learned how to build a straight line parallel to this one, and now the hut on chicken legs will turn 90 degrees:
How to build a line perpendicular to a given one?
Example 6
The straight line is given by the equation. Equate a perpendicular line through a point.
Decision: By condition it is known that. It would be nice to find the direction vector of the straight line. Since the lines are perpendicular, the trick is simple:
From the equation we "remove" the normal vector:, which will be the direction vector of the straight line.
Let us compose the equation of a straight line by a point and a direction vector:
Answer: 
Let's expand the geometric sketch:
Hmmm ... Orange sky, orange sea, orange camel.
Analytical verification solutions:
1) Take out the direction vectors from the equations 
and with the help dot product of vectors we come to the conclusion that the straight lines are indeed perpendicular:.
By the way, you can use normal vectors, it's even easier.
2) Check if the point satisfies the obtained equation 
.
The check is, again, easy to do orally.
Example 7
Find the point of intersection of perpendicular lines if the equation is known 
and point.
This is an example for a do-it-yourself solution. There are several actions in the task, so it is convenient to draw up the solution point by point.
Our exciting journey continues:
Distance from point to line
Before us is a straight strip of the river and our task is to reach it by the shortest route. There are no obstacles, and the most optimal route will be movement along the perpendicular. That is, the distance from a point to a straight line is the length of a perpendicular line.
Distance in geometry is traditionally denoted by the Greek letter "ro", for example: - the distance from the point "em" to the straight line "de".
Distance from point to line 
expressed by the formula
Example 8
Find the distance from point to line 
Decision: all you need is to carefully plug the numbers into the formula and perform the calculations:
Answer: 
Let's execute the drawing: 
The distance from the point to the line found is exactly the length of the red line. If you draw up a drawing on checkered paper on a scale of 1 unit. \u003d 1 cm (2 cells), then the distance can be measured with an ordinary ruler.
Consider another task for the same blueprint:
The task is to find the coordinates of a point that is symmetrical to a point with respect to a straight line 
... I propose to perform the actions yourself, but I will outline the solution algorithm with intermediate results:
1) Find a line that is perpendicular to the line.
2) Find the point of intersection of the lines: 
.
Both actions are detailed in this lesson.
3) The point is the midpoint of the line segment. We know the coordinates of the middle and one of the ends. By formulas for the coordinates of the midpoint of the segment we find.
It will not be superfluous to check that the distance is also 2.2 units.
Difficulties here can arise in calculations, but in the tower a micro calculator helps out great, allowing you to count common fractions... Repeatedly advised, will advise and again.
How to find the distance between two parallel lines?
Example 9
Find the distance between two parallel lines
This is another example for an independent solution. Let me give you a little hint: there are infinitely many ways to solve it. Debriefing at the end of the lesson, but better try to guess for yourself, I think you managed to disperse your ingenuity quite well.
Angle between two straight lines
Every angle is a jamb: 
In geometry, the angle between two straight lines is taken as the SMALLEST angle, from which it automatically follows that it cannot be obtuse. In the figure, the angle indicated by the red arc is not considered the angle between intersecting straight lines. And his "green" neighbor is considered as such, or oppositely oriented "Crimson" corner.
If the straight lines are perpendicular, then any of the 4 angles can be taken as the angle between them.
How do angles differ? Orientation. First, the direction of the corner "scrolling" is of fundamental importance. Second, a negatively oriented angle is written with a minus sign, for example, if.
Why did I tell this? It seems that you can do with the usual concept of an angle. The fact is that in the formulas by which we will find the angles, you can easily get a negative result, and this should not take you by surprise. An angle with a minus sign is no worse, and has a very specific geometric meaning. In the drawing, for a negative angle, be sure to indicate its orientation with an arrow (clockwise).
How to find the angle between two straight lines? There are two working formulas:
Example 10
Find the angle between straight lines
Decision and Method one
Consider two straight lines given by equations in general form: 
If straight not perpendicularthen oriented the angle between them can be calculated using the formula: 
Let's pay close attention to the denominator - this is exactly scalar product direction vectors of straight lines:
If, then the denominator of the formula vanishes, and the vectors will be orthogonal and straight lines are perpendicular. That is why a reservation was made about the non-perpendicularity of the straight lines in the formulation.
Based on the foregoing, it is convenient to arrange a solution in two steps:
1) Calculate the scalar product of the direction vectors of straight lines:
, so the straight lines are not perpendicular.
2) The angle between the straight lines is found by the formula:
Through inverse function the corner itself is easy to find. In this case, we use the oddness of the arctangent (see. Graphs and properties of elementary functions):
Answer: 
In the answer, we indicate the exact value, as well as the approximate value (preferably both in degrees and in radians), calculated using the calculator.
Well, minus, so minus, that's okay. Here's a geometric illustration: 
It is not surprising that the angle turned out to be of negative orientation, because in the problem statement the first number is a straight line and the "twisting" of the angle began with it.
If you really want to get a positive angle, you need to swap the straight lines, that is, take the coefficients from the second equation 
, and the coefficients are taken from the first equation. In short, you must start with a straight line 
.