"Pecuniary liability of the parties to the employment contract" - Material liability of the employer. If the amount of recovery does not exceed the average earnings for 1 month. Voluntary upon application or written commitment. For an employee. Liability of an employee Limited Full Individual Collective (team). By deduction from wages by order of the employer.

"Point fluctuation" - 5. Linear fluctuations. 7. Free vibrations with viscous resistance. 4. Examples of vibrations. beat. 3. Examples of oscillations. The motion is damped and aperiodic. Shows how many times the amplitude of oscillations exceeds the static deviation. Free vibrations caused by a driving force. 4) The period of damped oscillations is greater than that of undamped ones.

"Rectilinear motion" - Graphs for PRD. Rectilinear uniform motion (PRD). Sx \u003d X - X0 \u003d vx t - projection of movement on the X axis. Rectilinear uniformly accelerated motion (POND). Pond. X = X0 + sx is the law of motion. POND charts. Does that mean the speed changes? - The law of motion. Example: X = X0 + Vx t - the law of motion for the PRD.

"Points of the celestial sphere" - The days of the solstice, like the days of the equinox, can change. At 1 radian, 57 ° 17? 45 ", a degree is a central angle corresponding to 1/360 of a circle. At the point of the summer solstice on June 22, the Sun has its 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 midpoint of the edge C1D1. In the unit cube A…D1 find the distance from point A to line CD. In the unit cube A…D1 find the distance from point A to 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. The median of a triangle. Segment AN is a perpendicular dropped from point A to line a, if. Median. A line segment that connects a vertex to the midpoint of the opposite side is called. Bisector of a triangle. Task number 2. Problem No. 1. The perpendicular dropped from the vertex of the triangle to the line containing the opposite side is called.

The physical meaning of the derivative. The USE in mathematics includes a group of tasks for the solution of which knowledge and understanding of the physical meaning of the derivative is necessary. In particular, there are tasks 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 acquires a certain given speed.The tasks are very simple, they are solved in one step. 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 the time.

Velocity at a given point in time is the derivative of the coordinate with respect to time. This is the mechanical meaning of the derivative.

Similarly, acceleration is the derivative of speed with respect to time:

Thus, the physical meaning of the derivative is speed. This can be the speed of movement, the speed of a change in a process (for example, the growth of bacteria), the speed of work (and so on, there are many applied tasks).

In addition, you need to know the table of derivatives (you need to know it as well as the multiplication table) and the rules of differentiation. Specifically, to solve the specified problems, it is necessary to know the first six derivatives (see table):

Consider the tasks:

x (t) \u003d t 2 - 7t - 20

where x t is the time in seconds measured from the start of the movement. Find its speed (in meters per second) at time t = 5 s.

The physical meaning of the derivative is speed (speed of movement, speed of process change, speed of work, etc.)

Let's find the law of speed change: v (t) = x′(t) = 2t – 7 m/s.

For t = 5 we have:

Answer: 3

Decide on your own:

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 the 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 the movement. Find its speed (in meters per second) at time t = 6 s.

The material point moves in a straight line 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 the movement. Find its speed (in meters per second) at time t = 3 s.

The material point moves in a straight line 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 start of the movement. At what point in time (in seconds) was her speed equal to 6 m/s?

Let's find the law of speed change:

To find out at what point in timetthe speed was equal to 3 m / s, it is necessary to solve the equation:

Answer: 3

Decide for yourself:

A material point moves in a straight line according to the law x (t) \u003d t 2 - 13t + 23, where x- distance from the reference point in meters, t- time in seconds, measured from the start of the movement. At what point in time (in seconds) was her speed equal to 3 m/s?

The material point moves in a straight line according to the law

x (t) \u003d (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 the movement. At what point in time (in seconds) was her speed equal to 2 m/s?

I note that focusing only on this type of tasks on the exam is not worth it. They can quite unexpectedly introduce tasks inverse to those presented. When the law of change of speed is given, the question of finding the law of motion will be raised.

Hint: in this case, you need to find the integral of the speed function (these are also tasks in one action). If you need to find the distance traveled for a certain point in time, then you need to substitute the time in the resulting equation and calculate the distance. However, we will also analyze such tasks, do not miss it!I wish you success!

Sincerely, Alexander Krutitskikh.

P.S: I would be grateful if you tell about the site in social networks.

The point moves in a straight line according to the law S \u003d t 4 +2t (S - in meters t- in seconds). Find its average acceleration between the 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 a derivative of the path S with respect to time t, those.

2. Substituting instead of t its values ​​t 1 \u003d 5 s and t 2 \u003d 7 s, we find the speeds:

V 1 \u003d 4 5 3 + 2 \u003d 502 m / s; V 2 \u003d 4 7 3 + 2 \u003d 1374 m / s.

3. Determine the speed increment ΔV over time Δt = 7 - 5 = 2 s:

ΔV \u003d 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 the point, we take the derivative of the speed with respect to time:

6. Substituting instead t value t 3 \u003d 6 s, we get the acceleration at this point in time

a cf \u003d 12-6 3 \u003d 432 m / s 2.

curvilinear movement. In curvilinear motion, the speed of a point changes in magnitude and direction.

Imagine a point M, which during the time Δt, moving along some curvilinear trajectory, moved to the position M 1(Fig. 6).

Increment (change) vector of velocity ΔV will be

For finding the vector ΔV we move the vector V 1 to the point M and construct a triangle of speeds. Let's define the average acceleration vector:

Vector a wed is parallel to the vector ΔV, since dividing the vector by a scalar value 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.

Such a limit is called the vector derivative.

In this way, the true acceleration of a point during curvilinear motion is equal to the vector derivative with respect to velocity.

From fig. 6 shows 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: tangentially, called tangential (tangential) acceleration a, and along the normal, called the normal acceleration a n (Fig. 7).

In this case, the total acceleration will be

The tangential acceleration coincides in direction with the speed of the point or opposite to it. It characterizes the change in the velocity value and, accordingly, is 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 considered point.

Since the tangent and normal accelerations are mutually perpendicular, therefore, the magnitude of the total acceleration is determined by the formula

and its direction

If a , then the tangential acceleration and velocity vectors are directed in the same direction and the movement will be accelerated.

If a , then the tangential acceleration vector is directed in the direction opposite to the velocity vector, and the movement will be slow.

The vector of normal acceleration is always directed towards the center of curvature, so it is called centripetal.