A material point moves according to a rectilinear law. A number of particular problems from various fields of science

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

Our VKontakte group
Mobile applications:

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

At t= 9 s we have:

Why are we leaving out 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 movement). Find its speed in (m/s) at the moment of 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 preferable to subtraction?

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

You did not understand the guest's answer.

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

So everything is correct, do the math for yourself.

2) multiplication/division (depends on the order in the equation; what comes first is solved first);

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

Firstly, for you - Sergei Batkovich. Secondly, did you understand what you wanted to say and to whom? I did not understand you.

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 s.

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 and systematize the material on the topic of finding the derivative;
- consolidate the rules of differentiation;
- reveal to students the polytechnic and applied significance of the topic;
developing:
- exercise control over the acquisition 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;
- To instill in students accuracy in design and determination.

Equipment:

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

I. Checking homework.

1. Listen to student reports on examples of the use of derivatives.

2. Consider examples of the use of derivatives in physics, chemistry, engineering and other fields proposed by students.

II. Updating knowledge.

Teacher:

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

A number of particular problems from various areas Sci.

Task No. 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 No. 2. The radius of the circle R varies according to the law R = 4 + 2t 2. Determine the rate at which its area changes V moment t = 2 s. The radius of a circle is measured in centimeters. Answer: 603 cm 2 /s.

Task No. 3. A material point with a mass of 5 kg moves rectilinearly 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 No. 4. The flywheel, held by the brake, turns behind t s at an angle of 3t - 0.1t 2 (rad). Find:

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

Answer: a) 2.86; b) 150 s.

Examples of using derivatives can also include problems of finding: specific heat capacity substances given body, linear density and kinetic energy of the body, etc.

III. Performance differentiated tasks.

Those who want to complete level “A” tasks sit down at the computer and complete a test with a programmed answer. ( Application. )

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 = xe x at the point x 0 = 1.

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

3. Solve the equation f / (x) = 0 if f (x) = (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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Application of derivatives in physics, technology, biology, life

Presentation for the lesson

Attention! Slide previews are for informational purposes only and may not represent all the features 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 derivatives in various fields of physics, chemistry, and biology.

Tasks: broadening one's 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 a lesson goal

– I would like to conduct a lesson under the motto of Alexei Nikolaevich Krylov, a Soviet mathematician and shipbuilder: “Theory without practice is dead or useless, practice without theory is impossible or detrimental.”

– Let’s review the basic concepts and answer the questions:

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

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

Derivative – 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 problems, we will show how to solve other problems using the derivative and non-standard tasks using derivatives.

III. Explanation of new material

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

W = lim ΔA/Δt ΔA – job change.

2. If a 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 during time Δt.

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

5. Problem about the rate of a chemical reaction

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

V= lim Δm/Δt = m Δt → 0

6. Let m be mass radioactive substance. Radioactive decay rate: V = lim Δm/Δt = m׳(t) Δt→0

In differentiated form, the law of radioactive decay has the form: dN/dt = – λN, Where N– number of nuclei that have not decayed 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, from here we have: ln N0 = const, hence

n N = – λt + ln N0.

Potentiating this expression we get:

– the law of radioactive decay, where N0– number of cores at a time t0 = 0, N– number of nuclei that have not decayed during 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 ΔT

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

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

m = D Δc/Δx 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 - electric current. James Clark Maxwell introduced one amendment to the laws of electromagnetism discovered before him: a magnetic field also arises when a change electric field. A seemingly small amendment had enormous consequences: a completely new physical object – electromagnetic wave. Maxwell masterfully, unlike Faraday, who thought its existence was possible, derived 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– greater magnetic field.

IV. Consolidation of what has been learned

– You and I studied the derivative and its properties. I'd like to read philosophical statement Gilbert: “Every person has a certain outlook. When this horizon narrows to the infinitesimal, 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!

The plot of "Leaf"(use of 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 yourself.

2. F = ma F = mV′ F = mS″

Let us write down Porton’s II law, and taking into account the mechanical meaning of the derivative, we 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 physics problems, 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 person is as many times larger than an atom as he is smaller than a star:

It follows that
This is the formula that determines man’s place in the universe. In accordance with it, the size of a person represents the average proportionality of a star and an 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 someday will not 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 motion 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 moment will its speed be equal to zero?

№ 3 Two bodies move rectilinearly: one according to the law s = t^3 – t^2 – 27t, the other according to the law s = t^2 + 1. Determine the moment when the velocities of these bodies turn out to be equal.

№ 4 For a car moving at a speed of 30 m/s, the braking 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 brake until the car comes to a complete stop? Which the distance will go the car from the beginning of braking until it comes to a complete stop?

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

Solution: Let's find the speed body movements at any time:
V = ds / dt = 4t + 3
Let's calculate the speed of the body at time t = 3:
V t=3 = 4 * 3 + 3=15 (m/s).
Let us 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 movement, if its mass is 25 kg, and the law of motion has the form s = 3t^2- 1.

№7 A body whose mass is 30 kg moves rectilinearly according to the law s = 4t^2 + t. Prove that the movement of a body occurs under the action of constant force.
Solution: We have s’ = 8t + 1, s” = 8. Therefore, a(t) = 8 (m/s^2), i.e., with this 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 weighing 3 kg moves rectilinearly according to the law s(t) = t^3 – 3t^2 + 2. Find the force acting on the body at time t = 4s.

№9 A 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 derivative in mathematics:

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

The derivative formula dates back to the 15th century. The great Italian mathematician Tartagli, considering and developing the question of how much the flight range of a projectile depends on the inclination of the gun, applies it in his works.

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

The famous scientist Galileo Galilei devotes an entire treatise on the role of derivatives in mathematics. Then the derivative and various presentations with its application began to be found in the works of Descartes, the French mathematician Roberval and the Englishman Gregory. Great contributions to the study of the derivative were made by such minds as L'Hopital, Bernoulli, Langrange and others.

1. Plot a graph and examine the function:

Solution to this problem:

A moment of relaxation

VII. Application of derivative in physics:

When studying certain processes and phenomena, the task of determining the speed of these processes often arises. Its solution leads to the concept of derivative, which is the main 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. – are associated with the emergence of this method. Leibniz.

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

In physics, the derivative is used mainly to calculate the largest or lowest 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- 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; V) Fmax the value of the force of attraction; d) depict sample graphs dependencies U(r) And F(r).

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

Using the connection between the potential energy of the field

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; Sustainable or unstable equilibrium we determine 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 impulse of the amount of sand that poured out of the platform during time Δ t. Then:
Δ p = MΔ u – µtΔ u – Δ µtΔ u = FΔ t
Divide by Δ t and move on 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 straight line y = 2x is tangent to the function: y = x 3 + 5x 2 + 9x + 3. Find the abscissa of the point of tangency.

IX. Summing up the lesson:

– What questions was the lesson devoted to?
– What did you learn in the lesson?
– What theoretical facts were summarized in the lesson?
– Which tasks considered turned out to be the most difficult? Why?

Bibliography:

Amelkin V.V., Sadovsky A.P. Mathematical models and differential equations. – Minsk: graduate School, 1982. – 272 p.
Amelkin V.V. Differential equations in applications. M.: Science. Main editorial office of physical and mathematical literature, 1987. – 160 p.
Erugin N.P. Book to read general course differential equations. – Minsk: Science and Technology, 1979. – 744 p.
.Magazine "Potential" November 2007 No. 11
“Algebra and principles of analysis” 11th grade S.M. Nikolsky, M.K. Potapov and others.
“Algebra and mathematical analysis” N.Ya. Vilenkin et al.
"Mathematics" V.T. Lisichkin, I.L. Soloveichik, 1991

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Physical meaning of derivative. Tasks!

Physical meaning 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 the equation and you need 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 coordinate axis, 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 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 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.

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.

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 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

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

Page No. 094.

Textbook:

OCR version of the page from the textbook (text of the page located above):

As follows from the problems considered at the beginning of this paragraph, the following statements are true:

1. If at straight motion the path s traversed by a 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 point x 0 a tangent is drawn to the graph of the function y = 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) =

Tga. This angle is called the tangent angle.

This fact expresses geometric meaning derivative.

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

Let's 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 x graph of the function y = /(jc) and the tangent to its graph at the point with the abscissa jc = 0 are shown in Figure 95.

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

a) time increment D£ over the time interval from t x = 1 to £ 2 - 2;

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

V) average speed over the time interval from t x = 1 to t 2 = 2.

4.2 In task 4.1 find:

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

V) 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 = t 2 - bt.

a) increment of path As over the period of time from t to t + At;

Textbook: Algebra and beginning of mathematical analysis. 11th grade: educational. 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 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. At curvilinear movement 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 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 scalar quantity 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.

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 numerical value it 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 to the side, opposite to the vector speed and the movement will be slow.

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

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:

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) = 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:

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?

Sincerely, Alexander Krutitskikh.

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