The material point moves rectilinearly according to the online law. The physical meaning of the derivative

− Teacher Dumbadze V.A.
from school 162 of the Kirovsky district of St. Petersburg.

Our VKontakte group
Mobile applications:

(where x t- time in seconds, measured from the beginning of the movement). Find its speed (in m/s) at the time t= 9 s.

At t= 9 c we have:

Why don't we factor in the number 17 from the original equation?

find the derivative of the original function.

there is no number 17 in the derivative

Why find the derivative?

Velocity is the derivative of a coordinate with respect to time.

The problem asks you to find the speed

x- distance from the reference point in meters, t- time in seconds, measured from the beginning of the movement). Find its speed in (m/s) at time t= 6 s.

Let's find the law of speed change:

(6)=3/2*36-6*6+2=54-38=16 not 20

remember the procedure

Since when is addition better than subtraction?

Multiplication takes precedence over addition and subtraction. Remember childish school example: 2 + 2 2. Let me remind you that here it turns out not 8, as some people think, but 6.

You didn't understand the guest's answer.

1,5*36 — 6*6 + 2 = 54 — 36 + 2 = 18 + 2 = 20.

So that's right, count for yourself.

2) multiplication / division (depends on the order in the equation, that the first one is - then it is solved first of all);

3) addition / subtraction (similarly depends on the order in the example).

Multiplication = division, addition = subtraction =>

Not 54 - (36+2), but 54-36+2 = 54+2-36 = 20

First, for you - Sergey Batkovich. Secondly, did you yourself understand what you wanted to say and to whom? I did not understand you.

The 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 c.

Let's find the law of speed change: m/s. When we have:

Lesson on the topic: "Rules of differentiation", 11th grade

Sections: Mathematics

Lesson type: generalization and systematization of knowledge.

Lesson Objectives:

educational:
- generalize, systematize the material of the topic by finding the derivative;
- fix the rules of differentiation;
- open polytechnic for students, applied value themes;
developing:
- control the assimilation of knowledge and skills;
- develop and improve the ability to apply knowledge in a changed situation;
- develop a culture of speech and the ability to draw conclusions and generalize;
educational:
- develop the cognitive process;
- instill in students accuracy in design, purposefulness.

Equipment:

overhead projector, screen;
cards;
computers;
table;
differentiated tasks in the form of multimedia presentations.

I. Checking homework.

1. Listen to students' reports on examples of the use of derivatives.

2. Consider examples of the use of the derivative in physics, chemistry, technology and other industries proposed by students.

II. Knowledge update.

Teacher:

Define the derivative of a function.
What operation is called differentiation?
What rules of differentiation are used to calculate the derivative? (Students are invited to the board).
- derivative of the sum;
- derivative of the work;
- derivative containing a constant factor;
- the derivative of the quotient;
- derivative of a complex function;
Give examples applied tasks, leading to the concept of a derivative.

A number of particular problems from various fields of science.

Task number 1. The body moves in a straight line according to the law x(t). Write down the formula for finding the speed and acceleration of a body at time t.

Task number 2. The circle radius R changes according to the law R = 4 + 2t 2 . Determine the rate at which its area will change. in moment t = 2 s. The radius of a circle is measured in centimeters. Answer: 603 cm 2 / s.

Task number 3. A material point with a mass of 5 kg moves in a straight line according to the law

S(t) = 2t+ , where S- distance in meters t- time in seconds. Find the force acting on the point at the moment t = 4 s.

Answer: N.

Task number 4. The flywheel held by the brake turns behind t s at an angle 3t - 0.1t 2 (rad). Find:

a) the angular velocity of the flywheel at the moment t = 7 with;
b) at what point in time the flywheel will stop.

Answer: a) 2.86; b) 150 s.

Examples of the use of the derivative can also serve as tasks for finding: specific heat substance of a given body, linear density and kinetic energy of the body, etc.

III. Fulfillment of differentiated tasks.

Those who wish to complete tasks of level “A” sit down at the computer and perform a test with a programmed answer. ( Appendix. )

1. Find the value of the derivative of the function at the point x 0 = 3.

2. Find the value of the derivative of the function y \u003d xe x at the point x 0 \u003d 1.

1) 2e;
2) e;
3) 1 + e;
4) 2 + e.

3. Solve the equation f / (x) \u003d 0 if f (x) \u003d (3x 2 + 1) (3x 2 - 1).

1) ;
2) 2;
3) ;
4) 0.

4. Calculate f / (1) if f (x) = (x 2 + 1) (x 3 - x).

5. Find the value of the derivative of the function f(t) = (t4 - 3)(t2 + 2) at the point t0 = 1.

6. The point moves rectilinearly according to the law: S(t) = t 3 - 3t 2 . Choose a formula that specifies the speed of movement of this point at time t.

1) t 2 - 2t;
2) 3t 2 – 3t;
3) 3t 2 – 6t;
4) t 3 + 6t.

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The use of the derivative in physics, technology, biology, life

Presentation for the lesson

Attention! The slide preview is for informational purposes only and may not represent the full extent of the presentation. If you are interested this work please download the full version.

Lesson type: integrated.

The purpose of the lesson: study some aspects of the application of the derivative in various fields physics, chemistry, biology.

Tasks: expanding horizons and cognitive activity students, development logical thinking and the ability to apply their knowledge.

Technical support: interactive board; computer and disk.

I. Organizational moment

II. Setting the goal of the lesson

- I would like to conduct a lesson under the motto of Alexei Nikolaevich Krylov Soviet mathematician and the shipbuilder: "Theory without practice is dead or useless; practice without theory is impossible or pernicious."

Let's review the basic concepts and answer the questions:

What is the basic definition of a derivative?
– What do you know about the derivative (properties, theorems)?
– Do you know any examples of derivative problems in physics, mathematics and biology?

Consideration of the basic definition of the derivative and its justification (answer to the first question):

Derivative is one of the fundamental concepts of mathematics. The ability to solve problems using derivatives requires good knowledge theoretical material ability to conduct research in various situations.

Therefore, today in the lesson we will consolidate and systematize the knowledge gained, consider and evaluate the work of each group and, using the example of some tasks, we will show how to use the derivative to solve other problems and non-standard tasks using a derivative.

III. Explanation of new material

1. Instantaneous power is the derivative of work with respect to time:

W = lim ∆A/∆t ∆A – job change.

2. If the body rotates around an axis, then the angle of rotation is a function of time t
Then angular velocity is equal to:

W = lim Δφ/Δt = φ׳(t) Δ t → 0

3. Current strength is a derivative Ι = lim Δg/Δt = g′, where g- positive electric charge transferred through the cross section of the conductor in time Δt.

4. Let ∆Q is the amount of heat required to change the temperature Δt time, then lim ΔQ/Δt = Q′ = C – specific heat.

5. The problem of the rate of a chemical reaction

m(t) – m(t0) – the amount of a substance that reacts over time t0 before t

V= lim ∆m/∆t = m ∆t → 0

6. Let m be the mass radioactive substance. Speed radioactive decay: V = lim ∆m/∆t = m׳(t) ∆t→0

In a differentiated form, the law of radioactive decay has the form: dN/dt = – λN, where N is the number of nuclei that have not decayed over time t.

Integrating this expression, we get: dN/N = – λdt ∫dN/N = – λ∫dt lnN = – λt + c, c = const at t = 0 number of radioactive nuclei N = N0, hence we have: log N0 = const, hence

n N = – λt + ln N0.

Potentiating this expression we get:

is the law of radioactive decay, where N0 is the number of cores at a time t0 = 0, N is the number of nuclei that have not decayed over time t.

7. According to Newton's heat transfer equation, the heat flow rate dQ/dt is directly proportional to the window area S and the temperature difference ΔT between the inner and outer glass and inversely proportional to its thickness d:

dQ/dt =A S/d ∆T

8. The phenomenon of diffusion is the process of establishing an equilibrium distribution

Within the phases of concentration. Diffusion goes to the side, equalizing concentrations.

m = D ∆c/∆x c – concentration
m = D c׳x x - coordinate, D- diffusion coefficient

9. It was known that the electric field excites either electric charges, or a magnetic field that has a single source - an electric current. James Clark Maxwell introduced one amendment to the laws of electromagnetism discovered before him: a magnetic field also arises when electric field. Small at first glance, the amendment had grandiose consequences: it appeared, albeit at the tip of the pen, a completely new physical object – electromagnetic wave. Maxwell masterfully owned, in contrast to Faraday, who seemed possible its existence, deduced the equation for the electric field:

∂E/∂x = M∂B/Mo ∂t Mo = const t

A change in the electric field causes the appearance magnetic field at any point in space, in other words, the rate of change of the electric field determines the magnitude of the magnetic field. Under the big electric shock- a larger magnetic field.

IV. Consolidation of the studied

– We studied the derivative and its properties. I would like to read Gilbert's philosophical statement: “Every person has a certain outlook. When this horizon narrows to the infinitely small, it turns into a point. Then the person says that this is his point of view.
Let's try to measure the point of view on the application of the derivative!

Plot "Leaf"(application of the derivative in biology, physics, life)

Consider the fall as uneven movement time dependent.

So: S = S(t) V = S′(t) = x′(t), a = V′(t) = S″(t)

(Theoretical survey: mechanical sense derivative).

1. Problem solving

Solve problems on your own.

2. F = ma F = mV′ F = mS″

Let's write Porton's II law, and taking into account the mechanical meaning of the derivative, we will rewrite it in the form: F = mV′ F = mS″

The plot of "Wolves, Gophers"

Let's return to the equations: Consider the differential equations of exponential growth and decrease: F = ma F = mV' F = mS"
Solving many problems of physics, technical biology and social sciences are reduced to the problem of finding functions f"(x) = kf(x), satisfying the differential equation, where k = const .

Human Formula

A man is as many times larger than an atom as he is smaller than a star:

Hence it follows that
This is the formula that determines the place of man in the universe. In accordance with it, the dimensions of a person represent the average proportional of the star and the atom.

I would like to end the lesson with the words of Lobachevsky: “There is not a single area of mathematics, no matter how abstract it may be, that will not someday be applicable to the phenomena of the real world.”

V. Solution of numbers from the collection:

Independent problem solving on the board, collective analysis of problem solutions:

№ 1 Find the speed of movement material point at the end of the 3rd second, if the movement of the point is given by the equation s = t^2 –11t + 30.

№ 2 The point moves rectilinearly according to the law s = 6t – t^2. At what point will its speed be zero?

№ 3 Two bodies move in a straight line: one according to the law s \u003d t^3 - t^2 - 27t, the other - according to the law s \u003d t^2 + 1. Determine the moment when the speeds of these bodies are equal.

№ 4 For a car moving at a speed of 30 m/s, the stopping distance is determined by the formula s(t) =30t-16t^2, where s(t) is the distance in meters, t is the braking time in seconds. How long does it take to decelerate full stop cars? Which distance will pass the car from the beginning of braking to its complete stop?

№5 A body of mass 8 kg moves in a straight line according to the law s = 2t^2+ 3t - 1. Find kinetic energy body (mv^2/2) 3 seconds after the start of movement.

Decision: Find the speed of the body at any time:
V=ds/dt=4t+3
Calculate the speed of the body at time t = 3:
V t \u003d 3 \u003d 4 * 3 + 3 \u003d 15 (m / s).
Let's determine the kinetic energy of the body at time t = 3:
mv2/2 = 8 - 15^2 /2 = 900 (J).

№6 Find the kinetic energy of the body 4 s after the start of motion, if its mass is 25 kg, and the law of motion is s = 3t^2-1.

№7 A body whose mass is 30 kg moves in a straight line according to the law s = 4t^2 + t. Prove that the motion of the body occurs under the action of a constant force.
Decision: We have s' = 8t + 1, s" = 8. Therefore, a(t) = 8 (m/s^2), i.e., under the given law of motion, the body moves with constant acceleration 8 m/s^2. Further, since the mass of the body is constant (30 kg), then, according to Newton's second law, the force acting on it F = ma = 30 * 8 = 240 (H) is also a constant value.

№8 A body with a mass of 3 kg moves in a straight line according to the law s(t) = t^3 - 3t^2 + 2. Find the force acting on the body at time t = 4s.

№9 The material point moves according to the law s = 2t^3 – 6t^2 + 4t. Find its acceleration at the end of the 3rd second.

VI. Application of the derivative in mathematics:

The derivative in mathematics shows numeric expression the degree of change in a quantity located at the same point under the influence of various conditions.

The derivative formula dates back to the 15th century. The great Italian mathematician Tartaglia, considering and developing the question - how much does the range of the projectile depend on the inclination of the gun - uses it in his writings.

The derivative formula is often found in the works of famous mathematicians of the 17th century. It is used by Newton and Leibniz.

The well-known scientist Galileo Galilei devotes a whole treatise on the role of the derivative in mathematics. Then the derivative and various presentations with its application began to be found in the works of Descartes, French mathematician Roberval and the Englishman Gregory. A great contribution to the study of the derivative was made by such minds as Lopital, Bernoulli, Langrange and others.

1. Plot and explore the function:

Solution to this problem:

A moment of relaxation

VII. Application of the derivative in physics:

When studying certain processes and phenomena, the problem often arises of determining the speed of these processes. Its solution leads to the concept of a derivative, which is the basic concept differential calculus.

The method of differential calculus was created in the 17th and 18th centuries. The names of two great mathematicians, I. Newton and G.V. Leibniz.

Newton came to the discovery of differential calculus when solving problems about the speed of a material point in this moment time (instantaneous speed).

In physics, the derivative is mainly used to calculate the largest or the smallest values any quantities.

№1 Potential energy U the field of a particle in which there is another, exactly the same particle has the form: U = a/r 2 – b/r, where a and b are positive constants, r- distance between particles. Find: a) value r0 corresponding to the equilibrium position of the particle; b) find out whether this situation is stable; in) Fmax the value of the force of attraction; d) depict approximate dependency graphs U(r) and F(r).

Solution of this problem: To determine r0 corresponding to the equilibrium position of the particle, we investigate f = U(r) to the extreme.

Using the link between potential energy fields

U and F, then F = -dU/dr, we get F = – dU/dr = – (2a/r3+ b/r2) = 0; wherein r = r0; 2a/r3 = b/r2 => r0 = 2a/b; Stable or unstable equilibrium is determined by the sign of the second derivative:
d2U/dr02= dF/dr0 = – 6a/r02 + 2b/r03 = – 6a/(2a/b)4 + 2b/(2a/b)3 = (– b4/8a3) 2 = FM / (M + µt ) 2

Consider the case when sand spills out of a filled platform.
Change in momentum over a short period of time:
Δ p = (M – µ(t + Δ t))(u+ Δ u) +Δ µtu – (M – µt)u = FΔ t
Term Δ µtu is the momentum of the amount of sand that spilled out of the platform during the time Δ t. Then:
Δ p = MΔ u-µtΔ u- Δ µtΔ u=FΔ t
Divide by Δ t and pass to the limit Δ t → 0
(M – µt)du/dt = F
Or a1= du/dt= F/(M – µt)

Answer: a = FM / (M + µt) 2 , a1= F/(M – µt)

VIII. Independent work:

Find derivatives of functions:

The line y \u003d 2x is tangent to the function: y \u003d x 3 + 5x 2 + 9x + 3. Find the abscissa of the point of contact.

IX. Summing up the lesson:

- What were the topics of the lesson?
- What did you learn in class?
What theoretical facts were summarized in the lesson?
– What were the most difficult tasks considered? Why?

Bibliography:

Amelkin V.V., Sadovsky A.P. Mathematical models and differential equations. – Minsk: graduate School, 1982. - 272p.
Amelkin V.V. Differential equations in applications. M.: Science. Main edition of physical and mathematical literature, 1987. - 160p.
Erugin N.P. Book to read by general exchange rate differential equations. - Minsk: Science and technology, 1979. - 744 p.
.Magazine "Potential" November 2007 №11
"Algebra and the Beginnings of Analysis" Grade 11 S.M. Nikolsky, M.K. Potapov and others.
"Algebra and Mathematical Analysis" N.Ya. Vilenkin and others.
"Mathematics" V.T. Lisichkin, I.L. Soloveichik, 1991

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The physical meaning of the derivative. Tasks!

physical meaning 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 problems where the law of motion is given certain point(object), expressed by the 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 are solved in one step. So:

Let the law of motion of a material point x (t) along coordinate axis, 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):

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

where x is the distance from the reference point in meters, 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.

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 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 = 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 time t the speed was equal to 3 m / s, it is necessary to solve the equation:

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!

matematikalegko.ru

Algebra and beginnings mathematical analysis, Grade 11 (S. M. Nikolsky, M. K. Potapov, N. N. Reshetnikov, A. V. Shevkin) 2009

Page number 094.

Textbook:

OCR version of the page from the tutorial (text of the page above):

As follows from the above this paragraph problems, the following statements are true:

1. If at rectilinear motion the path s traveled by the point is a function of time t, i.e. s = f(t), then the speed of the point is the derivative of the path with respect to time, i.e. v(t) =

This fact expresses the mechanical meaning of the derivative.

2. If at the point x 0 a tangent is drawn to the graph of the function y \u003d f (jc), then the number f "(xo) is the tangent of the angle a between this tangent and the positive direction of the Ox axis, i.e. /" (x 0) \u003d

Tga. This angle is called the angle of inclination of the tangent.

This fact expresses geometric meaning derivative.

EXAMPLE 3. Let's find the tangent of the slope of the tangent to the graph of the function y \u003d 0.5jc 2 - 2x + 4 at the point with the abscissa x \u003d 0.

Find the derivative of the function f(x) = 0.5jc 2 - 2x + 4 at any point x using equality (2):

0.5 2 x - 2 = jc - 2.

Let's calculate the value of this derivative at the point x = 0:

Therefore, tga = -2. The graph x of the function y \u003d / (jc) and the tangent to its graph at the point with the abscissa jc \u003d 0 are shown in Figure 95.

4.1 Let the point move rectilinearly according to the law s = t 2 . Find:

a) the increment of time Д£ on the time interval from t x \u003d 1 to £ 2 - 2;

b) increment of the path As on the time interval from t x = 1 to t 2 = 2;

in) average speed on the time interval from t x \u003d 1 to t 2 \u003d 2.

4.2 In task 4.1 find:

b) average speed over the time interval from t to t + At;

in) instantaneous speed at time t;

d) instantaneous speed at time t = 1.

4.3 Let the point move rectilinearly according to the law:

1) s = 3t + 5; 2) s \u003d t 2 - bt.

a) increment of the path As on the time interval from t to t + At;

Textbook: Algebra and beginning of mathematical analysis. Grade 11: textbook. for general education institutions: basic and profile. levels / [S. M. Nikolsky, M. K. Potapov, N. N. Reshetnikov, A. V. Shevkin]. - 8th ed. - M.: Education, 2009. - 464 p.: ill.

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:

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) = 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:

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?

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.

Decision.

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 true value 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. At curvilinear motion the speed of the point changes in magnitude and direction.

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

Increment (change) vector of velocity ΔV will

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 scalar value the direction of the vector does not change. 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.

Thus, 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 tangential 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.

Vector normal acceleration always directed towards the center of curvature, therefore it is called centripetal.