“Financial liability of the parties to an employment contract” - Financial liability of the employer. If the amount of recovery does not exceed the average earnings for 1 month. Voluntary upon application or written undertaking. For the employee. Material liability of the employee Limited Full Individual Collective (team). By deduction from wages by order of the employer.
“Point oscillation” - 5. Linear oscillations. 7. Free vibrations with viscous resistance. 4. Examples of oscillations. Beating. 3. Examples of oscillations. The movement is damped and aperiodic. Shows how many times the oscillation amplitude exceeds the static deviation. Free vibrations caused by a driving force. 4) The period of damped oscillations is longer than that of undamped ones.
“Rectilinear motion” - Graphs for traffic control. Rectilinear uniform motion (RUM). Sx =X – X0= vx t - projection of movement onto the X axis. Rectilinear uniformly accelerated motion (RUM). Pond. X = X0 + sx - law of motion. POND charts. That is, the speed changes? - Law of motion. Example: X = X0 + Vx t - the law of motion for the PRD.
“Points of the celestial sphere” - The days of the solstices, like the days of the equinoxes, can change. In 1 radian 57°17?45". degree is the central angle corresponding to 1/360 of the circle. At the point of the summer solstice on June 22, the Sun has a maximum declination. The movement of the Sun along the ecliptic is caused by the annual movement of the Earth around the Sun.
“Distance from a point to a line” - In the unit cube A...D1, find the distance from point A to line CB1. Finding distances 2. In the unit cube A...D1, point E is the middle of edge C1D1. In the unit cube A...D1, find the distance from point A to straight line CD. In the unit cube A...D1, find the distance from point A to straight line CD1. In the unit cube A...D1, find the distance from point A to line BD.
“Four remarkable points of the triangle” - The height of the triangle. Median of a triangle. Segment AN is a perpendicular dropped from point A to straight line a, if. Median. The segment connecting a vertex to the middle of the opposite side is called. Bisector of a triangle. Task No. 2. Problem No. 1. A perpendicular dropped from the vertex of a triangle onto a straight line containing the opposite side is called.
The point moves rectilinearly according to the law S = t 4 +2t (S - in meters, t- in seconds). Find its average acceleration in the interval between moments t 1 = 5 s, t 2 = 7 s, as well as its true acceleration at the moment t 3 = 6 s.
Solution.
1. Find the speed of the point as the derivative of the path S with respect to time t, those.
2. Substituting instead of t its values t 1 = 5 s and t 2 = 7 s, we find the speeds:
V 1 = 4 5 3 + 2 = 502 m/s; V 2 = 4 7 3 + 2 = 1374 m/s.
3. Determine the speed increment ΔV for the time Δt = 7 - 5 =2 s:
ΔV = V 2 - V 1= 1374 - 502 = 872 m/s.
4. Thus, the average acceleration of the point will be equal to
5. To determine the true value of the acceleration of a point, we take the derivative of the speed with respect to time:
6. Substituting instead t value t 3 = 6 s, we get acceleration at this point in time
a av =12-6 3 =432 m/s 2 .
Curvilinear movement. During curvilinear motion, the speed of a point changes in magnitude and direction.
Let's imagine a point M, which during time Δt, moving along some curvilinear trajectory, moved to the position M 1(Fig. 6).
Velocity increment (change) vector ΔV will
For to find the vector ΔV, move the vector V 1 to the point M and construct a velocity triangle. Let's determine the vector of average acceleration:
Vector a Wed is parallel to the vector ΔV, since dividing the vector by a scalar quantity does not change the direction of the vector. The true acceleration vector is the limit to which the ratio of the velocity vector to the corresponding time interval Δt tends to zero, i.e.
This limit is called the vector derivative.
Thus, the true acceleration of a point during curvilinear motion is equal to the vector derivative with respect to speed.
From Fig. 6 it is clear that the acceleration vector during curvilinear motion is always directed towards the concavity of the trajectory.
For the convenience of calculations, the acceleration is decomposed into two components to the trajectory of motion: along a tangent, called tangential (tangential) acceleration A, and along the normal, called normal acceleration a n (Fig. 7).
In this case, the total acceleration will be equal to
Tangential acceleration coincides in direction with the speed of the point or is opposite to it. It characterizes the change in speed and is accordingly determined by the formula
Normal acceleration is perpendicular to the direction of the point’s velocity, and its numerical value is determined by the formula
where r - radius of curvature of the trajectory at the point under consideration.
Since the tangential and normal accelerations are mutually perpendicular, therefore the value of the total acceleration is determined by the formula
and its direction
If 
, then the tangential acceleration and velocity vectors are directed in one direction and the movement will be accelerated.
If 
, then the tangential acceleration vector is directed in the direction opposite to the velocity vector, and the movement will be slow.
The normal acceleration vector is always directed towards the center of curvature, which is why it is called centripetal.
Task. The point moves rectilinearly according to the law S(t) = 2 t? — 3 t Calculate the speed of the point: a) at time t; b) at time t=2s. Solution. a) b).
“Test “Functions and their properties”” - Testing. Find the smallest positive period of the function. The graph of which function is shown in the figure. Set of function values. Give the graph of an even function. Tasks for teams. Group task for teams. Properties of functions. Which of the figures shows the graph of an odd function? Find the intervals of increase of the function given graphically. Portrait. Specify all zeros of the function. Star relay. Star for the captain.
“Algebra “Derivatives”” - Mechanical meaning of derivatives. Structure of the topic study. Find the derivative of the function. Function graph. An example of finding the derivative. Algorithm for finding the derivative. Differentiation formulas. Tangent equation. Derivative function. Tangent to the graph of a function. Geometric meaning of derivative. Equation of a tangent to the graph of a function. Definition of derivative. Derivative. Origin of terms.
“Equations” - Appearance of the equality symbol. Geometry. Equations are all around us. Mathematics in Ancient India. Mathematics of the Islamic Middle Ages. Algebraic method. Analytical method. Methods for solving equations. The appearance of letter symbols. A little history. Unknown number. Mathematics in Ancient Egypt. Arithmetic of Diophantus. Graphic method. Solution. Where are equations used today? Physics. What is an equation?
“Problems on polynomials” - Pairwise distinct roots. Find all parameter values. Contradiction. Multiplying polynomials. Find the roots of the trinomial. Euclid's algorithm. Theory. Fundamental theorem of algebra. Historical reference. Remainder. The number A is called the root of the polynomial. Tasks. Division of polynomials. Roots of the first equation. Polynomials. Find the integers x and y. Four pairwise distinct natural numbers. Polynomial ax + b. Non-negative integer values.
“Gorner scheme” - Division according to Horner scheme. Horner Williams George. Calculation algorithm. Horner's scheme. Horner's scheme. Compact recording. Polynomial. Calculations using Horner's scheme. The resulting numbers. Factor the polynomial.
“Trigonometric functions of an angular argument” - Trigonometric functions of a numerical argument. Summarize and systematize educational material on the topic. Exercise. The cosine of angle A (cos A) is the abscissa (x) of a point. Values of trigonometric functions of basic angles. Values of trigonometric functions of the remaining angles of the table. Reduction formulas. Signs of trigonometric functions in the quarters of the unit circle. Independent work. Values of trigonometric functions of the angular argument.
There are a total of 52 presentations in the topic “Grade 10 Algebra”
Physical meaning of derivative. Tasks!
Physical meaning of derivative. The Unified State Exam in mathematics includes a group of problems for solving which requires knowledge and understanding of the physical meaning of the derivative. In particular, there are problems where the law of motion of a certain point (object) is given, expressed by an equation, and it is required to find its speed at a certain moment in time of movement, or the time after which the object will acquire a certain given speed. The tasks are very simple, they can be solved in one action. So:
Let the law of motion of a material point x (t) along the coordinate axis be given, where x is the coordinate of the moving point, t is time.
Velocity at a certain moment in time is the derivative of the coordinate with respect to time. This is the mechanical meaning of the derivative.
Likewise, acceleration is the derivative of speed with respect to time:
Thus, the physical meaning of the derivative is speed. This could be the speed of movement, the rate of change of a process (for example, the growth of bacteria), the speed of work (and so on, there are many applied problems).
In addition, you need to know the derivative table (you need to know it just like the multiplication table) and the rules of differentiation. Specifically, to solve the specified problems, knowledge of the first six derivatives is necessary (see table):
x (t) = t 2 – 7t – 20
where x t is the time in seconds measured from the beginning of the movement. Find its speed (in meters per second) at time t = 5 s.
The physical meaning of a derivative is speed (speed of movement, rate of change of a process, speed of work, etc.)
V (t) = x?(t) = 2t – 7 m/s.
The material point moves rectilinearly according to the law x (t) = 6t 2 – 48t + 17, where x- distance from the reference point in meters, t- time in seconds measured from the start of movement. Find its speed (in meters per second) at time t = 9 s.
The material point moves rectilinearly according to the law x (t) = 0.5t 3 – 3t 2 + 2t, where x- distance from the reference point in meters, t- time in seconds measured from the start of movement. Find its speed (in meters per second) at time t = 6 s.
A material point moves rectilinearly according to the law
x (t) = –t 4 + 6t 3 + 5t + 23
Where x- distance from the reference point in meters, t- time in seconds measured from the start of movement. Find its speed (in meters per second) at time t = 3 s.
x(t) = (1/6)t 2 + 5t + 28
where x is the distance from the reference point in meters, t is the time in seconds, measured from the beginning of the movement. At what point in time (in seconds) was its speed equal to 6 m/s?
Let's find the law of speed change:
In order to find at what point in time t the speed was 3 m/s, it is necessary to solve the equation:
The material point moves rectilinearly according to the law x (t) = t 2 – 13t + 23, where x- distance from the reference point in meters, t- time in seconds measured from the start of movement. At what point in time (in seconds) was its speed equal to 3 m/s?
A material point moves rectilinearly according to the law
x (t) = (1/3) t 3 – 3t 2 – 5t + 3
Where x- distance from the reference point in meters, t- time in seconds measured from the start of movement. At what point in time (in seconds) was its speed equal to 2 m/s?
I would like to note that you should not focus only on this type of tasks on the Unified State Exam. They may completely unexpectedly introduce problems that are the opposite of those presented. When the law of change of speed is given and the question will be about finding the law of motion.
Hint: in this case, you need to find the integral of the speed function (this is also a one-step problem). If you need to find the distance traveled at a certain point in time, you need to substitute time into the resulting equation and calculate the distance. However, we will also analyze such problems, don’t miss it! I wish you success!
matematikalegko.ru
Explain why the derivative of the formula for the motion of a point is taken
Velocity is the derivative of a coordinate with respect to time.
I can’t get a different answer at all, you somehow decide who knows how
everything is right here
x- distance from the reference point in meters, t- time in seconds measured from the beginning of movement). At what point in time (in seconds) was its speed equal to 3 m/s?
Let's find the law of speed change:
To find at what point in time the speed was 3 m/s, solve the equation:
A material point moves rectilinearly according to the law (where x- distance from the reference point in meters, t- time in seconds measured from the beginning of movement). At what point in time (in seconds) was its speed equal to 2 m/s?
Let's find the law of speed change: m/s. To find at what point in time the speed was equal to 2 m/s, solve the equation:
Material point M starts moving from a point A and moves in a straight line for 12 seconds. The graph shows how the distance from the point changed A to the point M with time. Time is plotted on the x-axis t in seconds, on the ordinate - distance s.
Determine how many times during the movement the speed of the point M turned to zero (do not take into account the beginning and end of the movement).
Instantaneous speed is equal to the derivative of displacement with respect to time. The derivative value is zero at the extremum points of the function s(t). There are 6 extreme points on the graph.
Derivative. Physical meaning of derivative. Task B8 (2015)
In this article we will introduce the concept derivative of a function, With physical meaning of the derivative and solve several problems from Tasks B9 from the Open Bank of Problems for preparing for the Unified State Exam in mathematics for use physical meaning of the derivative.
To understand what it is derivative, let's draw an analogy with instantaneous speed. Consider a material point that moves in a straight line with variable speed. Since the speed of a point changes all the time, we can talk about its speed only at a given moment in time. To find the speed of a point at a moment in time, consider a small period of time. During this period of time the point will travel a distance. Then the speed of the point will be approximately equal. The shorter the time period we take, the more accurate the speed value we will get. In the limit, at, we get the exact value of the instantaneous speed at the moment of time:
In a similar way we introduce the concept derivative.
Consider an arbitrary function and fix a point. The value of the function at this point is equal to. Let's take the argument increment. The value of the function at this point is equal to. We get the increment of the function
The derivative of a function is the limit of the ratio of the increment of the function to the increment of the argument when the increment of the argument tends to zero:
Physical meaning of derivative.
So, we see that, by analogy with instantaneous speed, the derivative of a function at a point. shows the rate of change of the function at this point.
If the dependence of distance on time is a function, then to find the speed of a body at a moment in time, you need to find the value of the derivative of the function at a point:
Example 1. Let's solve task B9 (No. 119975) from the Open Bank of tasks for preparing for the Unified State Exam in mathematics.
A material point moves rectilinearly according to the law, where - distance from the reference point in meters, - time in seconds measured from the start of movement. Find its speed (in meters per second) at the moment of time.
Solution.
1. Find the derivative of the function:
2. Find the value of the derivative at the point:
Example 2. Let's solve task B9 (No. 119978)
A material point moves rectilinearly according to the law, where is the distance from the reference point in meters, is the time in seconds, measured from the beginning of the movement. At what point in time (in seconds) was its speed equal to 3 m/s?
Solution.
If we know the speed of a point at a certain moment in time, then we know the value of the derivative at the point.
Let's find the derivative of the function
According to the condition, the speed of the point is 3 m/s, which means that the value of the derivative at the moment of time is 3.
Answer: 8
Example 3. Similar task. Task B9 (No. 119979)
A material point moves rectilinearly according to the law, where is the distance from the reference point in meters, is the time in seconds, measured from the beginning of the movement. At what point in time (in seconds) was its speed equal to 2 m/s?
Let's find the derivative of the function:
According to the condition, the speed of the point is 2 m/s, which means that the value of the derivative at the moment of time is 2.
, - does not fit the meaning of the problem: time cannot be negative.
The point of motion is rectilinear according to the law
Task 7. A material point moves rectilinearly according to the law (where x is the distance from the reference point in meters, t is the time in seconds measured from the beginning of the movement). Find its speed in (m/s) at time t=3 s.
The speed of movement is the derivative of the path with respect to time, that is, to find the law of change in speed, you need to calculate the derivative of the function x(t) with respect to t, we get:
At the moment of time t=3 s the speed of the material point is equal to
Physical meaning of derivative. The Unified State Exam in mathematics includes a group of problems for solving which requires knowledge and understanding of the physical meaning of the derivative. In particular, there are problems where the law of motion of a certain point (object) is given, expressed by an equation, and it is required to find its speed at a certain moment in time of movement, or the time after which the object will acquire a certain given speed.The tasks are very simple, they can be solved in one action. So:
Let the law of motion of a material point x (t) along the coordinate axis be given, where x is the coordinate of the moving point, t is time.
Velocity at a certain moment in time is the derivative of the coordinate with respect to time. This is the mechanical meaning of the derivative.
Likewise, acceleration is the derivative of speed with respect to time:
Thus, the physical meaning of the derivative is speed. This could be the speed of movement, the rate of change of a process (for example, the growth of bacteria), the speed of work (and so on, there are many applied problems).
In addition, you need to know the derivative table (you need to know it just like the multiplication table) and the rules of differentiation. Specifically, to solve the specified problems, knowledge of the first six derivatives is necessary (see table):
Let's consider the tasks:
x (t) = t 2 – 7t – 20
where x t is the time in seconds measured from the beginning of the movement. Find its speed (in meters per second) at time t = 5 s.
The physical meaning of a derivative is speed (speed of movement, rate of change of a process, speed of work, etc.)
Let's find the law of speed change: v (t) = x′(t) = 2t – 7 m/s.
At t = 5 we have:
Answer: 3
Decide for yourself:
The material point moves rectilinearly according to the law x (t) = 6t 2 – 48t + 17, where x- distance from the reference point in meters, t- time in seconds measured from the start of movement. Find its speed (in meters per second) at time t = 9 s.
The material point moves rectilinearly according to the law x (t) = 0.5t 3 – 3t 2 + 2t, where xt- time in seconds measured from the start of movement. Find its speed (in meters per second) at time t = 6 s.
A material point moves rectilinearly according to the law
x (t) = –t 4 + 6t 3 + 5t + 23
Where x- distance from the reference point in meters,t- time in seconds measured from the start of movement. Find its speed (in meters per second) at time t = 3 s.
A material point moves rectilinearly according to the law
x(t) = (1/6)t 2 + 5t + 28
where x is the distance from the reference point in meters, t is the time in seconds, measured from the beginning of the movement. At what point in time (in seconds) was its speed equal to 6 m/s?
Let's find the law of speed change:
In order to find at what point in timetthe speed was 3 m/s, it is necessary to solve the equation:
Answer: 3
Decide for yourself:
The material point moves rectilinearly according to the law x (t) = t 2 – 13t + 23, where x- distance from the reference point in meters, t- time in seconds measured from the start of movement. At what point in time (in seconds) was its speed equal to 3 m/s?
A material point moves rectilinearly according to the law
x (t) = (1/3) t 3 – 3t 2 – 5t + 3
Where x- distance from the reference point in meters, t- time in seconds measured from the start of movement. At what point in time (in seconds) was its speed equal to 2 m/s?
I would like to note that you should not focus only on this type of tasks on the Unified State Exam. They may completely unexpectedly introduce problems that are the opposite of those presented. When the law of change of speed is given and the question will be about finding the law of motion.
Hint: in this case, you need to find the integral of the speed function (this is also a one-step problem). If you need to find the distance traveled at a certain point in time, you need to substitute time into the resulting equation and calculate the distance. However, we will also analyze such problems, don’t miss it!I wish you success!
Sincerely, Alexander Krutitskikh.
P.S: I would be grateful if you tell me about the site on social networks.