Formula for terminal velocity during uniformly accelerated motion. Uniformly accelerated linear motion

The most important characteristic when moving a body is its speed. Knowing it, as well as some other parameters, we can always determine the time of movement, distance traveled, initial and final speed and acceleration. Uniformly accelerated motion is only one type of motion. It is usually found in physics problems from the kinematics section. In such problems, the body is taken as a material point, which significantly simplifies all calculations.

Speed. Acceleration

First of all, I would like to draw the reader’s attention to the fact that these two physical quantities are not scalar, but vector. This means that when solving certain types of problems, it is necessary to pay attention to what acceleration the body has in terms of sign, as well as what the vector of the body’s velocity itself is. In general, in problems of a purely mathematical nature, such moments are omitted, but in problems in physics this is quite important, since in kinematics, due to one incorrect sign, the answer may turn out to be erroneous.

Examples

An example is uniformly accelerated and uniformly decelerated motion. Uniformly accelerated motion is characterized, as is known, by acceleration of the body. The acceleration remains constant, but the speed continuously increases at each individual moment. And with uniformly slow motion, the acceleration has a negative value, the speed of the body continuously decreases. These two types of acceleration form the basis of many physical problems and are quite often found in problems in the first part of physics tests.

Example of uniformly accelerated motion

We encounter uniformly accelerated motion everywhere every day. No car moves uniformly in real life. Even if the speedometer needle shows exactly 6 kilometers per hour, you should understand that this is actually not entirely true. Firstly, if we analyze this issue from a technical point of view, then the first parameter that will give inaccuracy will be the device. Or rather, its error.

We find them in all control and measuring instruments. The same lines. Take about ten rulers, at least identical (15 centimeters, for example), or different (15, 30, 45, 50 centimeters). Put them next to each other and you will notice that there are slight inaccuracies and their scales do not quite line up. This is an error. In this case, it will be equal to half the division value, as with other devices that produce certain values.

The second factor that will cause inaccuracy is the scale of the device. The speedometer does not take into account values ​​such as half a kilometer, one-half kilometer, and so on. It is quite difficult to notice this on the device with the eye. Almost impossible. But there is a change in speed. Albeit by such a small amount, but still. Thus, it will be uniformly accelerated motion, not uniform. The same can be said about a regular step. Let’s say we’re walking, and someone says: our speed is 5 kilometers per hour. But this is not entirely true, and why was explained a little higher.

Body acceleration

Acceleration can be positive or negative. This was discussed earlier. Let us add that acceleration is a vector quantity, which is numerically equal to the change in speed over a certain period of time. That is, through the formula it can be denoted as follows: a = dV/dt, where dV is the change in speed, dt is the time interval (change in time).

Nuances

The question may immediately arise as to how acceleration in this situation can be negative. Those people who ask a similar question motivate this by the fact that even speed cannot be negative, let alone time. In fact, time really cannot be negative. But very often they forget that the speed can easily take negative values. This is a vector quantity, we should not forget about it! It's probably all about stereotypes and incorrect thinking.

So, to solve problems, it is enough to understand one thing: the acceleration will be positive if the body accelerates. And it will be negative if the body slows down. That's it, quite simple. The simplest logical thinking or the ability to see between the lines will, in fact, be part of the solution to a physical problem related to speed and acceleration. A special case is the acceleration of gravity, and it cannot be negative.

Formulas. Problem solving

It should be understood that problems related to speed and acceleration are not only practical, but also theoretical. Therefore, we will analyze them and, if possible, try to explain why this or that answer is correct or, conversely, incorrect.

Theoretical problem

Very often in physics exams in grades 9 and 11 you can come across similar questions: “How will a body behave if the sum of all forces acting on it is zero?” In fact, the wording of the question can be very different, but the answer is still the same. Here, the first thing you need to do is to use superficial buildings and ordinary logical thinking.

The student is given 4 answers to choose from. First: “the speed will be zero.” Second: “the speed of the body decreases over a certain period of time.” Third: “the speed of the body is constant, but it is definitely not zero.” Fourth: “the speed can have any value, but at each moment of time it will be constant.”

The correct answer here is, of course, the fourth. Now let's figure out why this is so. Let's try to consider all the options in turn. As is known, the sum of all forces acting on a body is the product of mass and acceleration. But our mass remains a constant value, we will discard it. That is, if the sum of all forces is zero, the acceleration will also be zero.

So, let's assume that the speed will be zero. But this cannot be, since our acceleration is equal to zero. Purely physically this is permissible, but not in this case, since now we are talking about something else. Let the speed of the body decrease over a period of time. But how can it decrease if the acceleration is constant and equal to zero? There are no reasons or prerequisites for a decrease or increase in speed. Therefore, we reject the second option.

Let us assume that the speed of the body is constant, but it is definitely not zero. It will indeed be constant due to the fact that there is simply no acceleration. But it cannot be said unequivocally that the speed will be different from zero. But the fourth option is right on target. The speed can be any, but since there is no acceleration, it will be constant over time.

Practical problem

Determine which path was traveled by the body in a certain period of time t1-t2 (t1 = 0 seconds, t2 = 2 seconds) if the following data are available. The initial speed of the body in the interval from 0 to 1 second is 0 meters per second, the final speed is 2 meters per second. The speed of the body at the time of 2 seconds is also 2 meters per second.

Solving such a problem is quite simple, you just need to grasp its essence. So, we need to find a way. Well, let's start looking for it, having previously identified two areas. As is easy to see, the body passes through the first section of the path (from 0 to 1 second) with uniform acceleration, as evidenced by the increase in its speed. Then we will find this acceleration. It can be expressed as the difference in speed divided by the time of movement. The acceleration will be (2-0)/1 = 2 meters per second squared.

Accordingly, the distance traveled on the first section of the path S will be equal to: S = V0t + at^2/2 = 0*1 + 2*1^2/2 = 0 + 1 = 1 meter. On the second section of the path, in the period from 1 second to 2 seconds, the body moves uniformly. This means that the distance will be equal to V*t = 2*1 = 2 meters. Now we sum up the distances, we get 3 meters. This is the answer.

And the time of movement, you can find the distance traveled:

Substituting the expression into this formula V avg = V/2, we will find the path traveled during uniformly accelerated motion from a state of rest:

If we substitute into formula (4.1) the expression V avg = V 0 /2, then we get the path traveled during braking:

The last two formulas include speeds V 0 and V. Substituting the expression V=at into formula (4.2), and the expression V 0 =at - into formula (4.3), we get

The resulting formula is valid both for uniformly accelerated motion from a state of rest, and for motion with decreasing speed when the body stops at the end of the path. In both of these cases, the distance traveled is proportional to the square of the time of movement (and not just time, as was the case with uniform movement). The first to establish this pattern was G. Galileo.

Table 2 gives the basic formulas describing uniformly accelerated linear motion.

Galileo did not have a chance to see his book, which outlined the theory of uniformly accelerated motion (along with many of his other discoveries). When was it published? The 74-year-old scientist was already blind. Galileo took the loss of his vision very hard. “You can imagine,” he wrote, “how I grieve when I realize that this sky, this world and the Universe, which by my observations and clear evidence have been expanded a hundred and a thousand times compared to what people thought they were sciences in all the past centuries have now become so diminished and diminished for me.”

Five years earlier, Galileo was tried by the Inquisition. His views on the structure of the world (and he adhered to the Copernican system, in which the central place was occupied by the Sun, not the Earth) had not been liked by church ministers for a long time. Back in 1614, the Dominican priest Caccini declared Galileo a heretic and mathematics an invention of the devil. And in 1616, the Inquisition officially declared that “the doctrine attributed to Copernicus that the Earth moves around the Sun, while the Sun stands at the center of the Universe, not moving from East to West, is contrary to the Holy Scriptures, and therefore it can neither be defended nor accepted for the truth." Copernicus's book outlining his system of the world was banned, and Galileo was warned that if "he did not calm down, he would be imprisoned."

But Galileo “did not calm down.” “There is no greater hatred in the world,” the scientist wrote, “than ignorance for knowledge.” And in 1632, his famous book “Dialogue on the two most important systems of the world - Ptolemaic and Copernican” was published, in which he gave numerous arguments in favor of the Copernican system. However, only 500 copies of this work were sold, since after a few months, by order of the Pope
Rimsky, the publisher of the book, received an order to suspend the sale of this work.

In the autumn of the same year, Galileo received an order from the Inquisition to appear in Rome, and after some time the sick 69-year-old scientist was taken to the capital on a stretcher. Here, in the prison of the Inquisition, Galileo was forced to renounce his views on the structure of the world, and on June 22, 1633 in a Roman monastery Minerva Galileo reads and signs the previously prepared text of renunciation

“I, Galileo Galilei, son of the late Vincenzo Galilei of Florence, 70 years of age, brought in person to the court and kneeling before Your Eminences, the most reverend gentlemen cardinals, general inquisitors against heresy throughout Christendom, having before me the sacred Gospel and offering hands on him, I swear that I have always believed, I believe now, and with God’s help I will continue to believe in everything that the Holy Catholic and Apostolic Roman Church recognizes, defines and preaches.”

According to the court decision, Galileo's book was banned, and he himself was sentenced to imprisonment for an indefinite period. However, the Pope pardoned Galileo and replaced the imprisonment with exile. Galileo moved to Arcetri and here, while under house arrest, wrote the book "Conversations and Mathematical Proofs , concerning two new branches of science related to Mechanics and Local Motion" In 1636, the manuscript of the book was sent to Holland, where it was published in 1638. With this book, Galileo summed up his many years of physical research. In the same year, Galileo became completely blind Talking about what had befallen misfortune of the great scientist, Viviani (a student of Galileo) wrote: “He suffered severe discharge from his eyes, so that after a few months he was completely left without eyes - yes, I say, without his eyes, which in a short time saw more than all human eyes over all the past centuries have been able to see and observe"

The Florentine inquisitor who visited Galileo in his letter to Rome said that he found him in a very serious condition. Based on this letter, the Pope allowed Galileo to return to his home in Florence. Here he was immediately given an order “On pain of life imprisonment in a true prison and excommunication “Don’t go out into the city and don’t talk to anyone, no matter who it is, about the damned opinion about the double movement of the Earth.”

Galileo did not stay at home for long. After a few months he was again ordered to come to Arcetri. He had about four years to live. On January 8, 1642, at four o'clock in the morning, Galileo died.

1. How does uniformly accelerated motion differ from uniform motion? 2. How does the path formula for uniformly accelerated motion differ from the path formula for uniform motion? 3. What do you know about the life and work of G. Galileo? In which year he was born?

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Materials from physics 8th grade, assignments and answers from physics by grade, notes for preparing for physics lessons, plans for lesson notes on physics 8th grade

Topics of the Unified State Examination codifier: types of mechanical motion, speed, acceleration, equations of rectilinear uniformly accelerated motion, free fall.

Uniformly accelerated motion - this is movement with a constant acceleration vector. Thus, with uniformly accelerated motion, the direction and absolute magnitude of the acceleration remain unchanged.

Dependence of speed on time.

When studying uniform rectilinear motion, the question of the dependence of speed on time did not arise: the speed was constant during the movement. However, with uniformly accelerated motion, the speed changes over time, and we have to find out this dependence.

Let's practice some basic integration again. We proceed from the fact that the derivative of the velocity vector is the acceleration vector:

. (1)

In our case we have . What needs to be differentiated to get a constant vector? Of course, the function. But not only that: you can add an arbitrary constant vector to it (after all, the derivative of a constant vector is zero). Thus,

. (2)

What is the meaning of the constant? At the initial moment of time, the speed is equal to its initial value: . Therefore, assuming in formula (2) we get:

So, the constant is the initial speed of the body. Now relation (2) takes its final form:

. (3)

In specific problems, we choose a coordinate system and move on to projections onto coordinate axes. Often two axes and a rectangular Cartesian coordinate system are enough, and vector formula (3) gives two scalar equalities:

, (4)

. (5)

The formula for the third velocity component, if needed, is similar.)

Law of motion.

Now we can find the law of motion, that is, the dependence of the radius vector on time. We recall that the derivative of the radius vector is the speed of the body:

We substitute here the expression for speed given by formula (3):

(6)

Now we have to integrate equality (6). It is not difficult. To get , you need to differentiate the function. To obtain, you need to differentiate. Let's not forget to add an arbitrary constant:

It is clear that is the initial value of the radius vector at time . As a result, we obtain the desired law of uniformly accelerated motion:

. (7)

Moving on to projections onto coordinate axes, instead of one vector equality (7), we obtain three scalar equalities:

. (8)

. (9)

. (10)

Formulas (8) - (10) give the dependence of the coordinates of the body on time and therefore serve as a solution to the main problem of mechanics for uniformly accelerated motion.

Let's return again to the law of motion (7). Note that - movement of the body. Then
we get the dependence of displacement on time:

Rectilinear uniformly accelerated motion.

If uniformly accelerated motion is rectilinear, then it is convenient to choose a coordinate axis along the straight line along which the body moves. Let, for example, this be the axis. Then to solve problems we will only need three formulas:

where is the projection of displacement onto the axis.

But very often another formula that is a consequence of them helps. Let us express time from the first formula:

and substitute it into the formula for moving:

After algebraic transformations (be sure to do them!) we arrive at the relation:

This formula does not contain time and allows you to quickly come to an answer in those problems where time does not appear.

Free fall.

An important special case of uniformly accelerated motion is free fall. This is the name given to the movement of a body near the surface of the Earth without taking into account air resistance.

The free fall of a body, regardless of its mass, occurs with a constant free fall acceleration directed vertically downward. In almost all problems, m/s is assumed in calculations.

Let's look at several problems and see how the formulas we derived for uniformly accelerated motion work.

Task. Find the landing speed of a raindrop if the height of the cloud is km.

Solution. Let's direct the axis vertically downward, placing the origin at the point of separation of the drop. Let's use the formula

We have: - the required landing speed, . We get: , from . We calculate: m/s. This is 720 km/h, about the speed of a bullet.

In fact, raindrops fall at speeds of the order of several meters per second. Why is there such a discrepancy? Windage!

Task. A body is thrown vertically upward with a speed of m/s. Find its speed in c.

Here, so. We calculate: m/s. This means the speed will be 20 m/s. The projection sign indicates that the body will fly down.

Task. From a balcony located at a height of m, a stone was thrown vertically upward at a speed of m/s. How long will it take for the stone to fall to the ground?

Solution. Let's direct the axis vertically upward, placing the origin on the surface of the Earth. We use the formula

We have: so , or . Solving the quadratic equation, we get c.

Horizontal throw.

Uniformly accelerated motion is not necessarily linear. Consider the motion of a body thrown horizontally.

Suppose that a body is thrown horizontally with a speed from a height. Let's find the time and flight range, and also find out what trajectory the movement takes.

Let us choose a coordinate system as shown in Fig.

1 .

We use the formulas:

. (11)

In our case . We get:

We find the flight time from the condition that at the moment of fall the coordinate of the body becomes zero:

Flight range is the coordinate value at the moment of time:

We obtain the trajectory equation by excluding time from equations (11). We express from the first equation and substitute it into the second:

We obtained a dependence on , which is the equation of a parabola. Consequently, the body flies in a parabola.

Throw at an angle to the horizontal.

Let's consider a slightly more complex case of uniformly accelerated motion: the flight of a body thrown at an angle to the horizon.

Let us assume that a body is thrown from the surface of the Earth with a speed directed at an angle to the horizon. Let's find the time and flight range, and also find out what trajectory the body is moving along.

Let us choose a coordinate system as shown in Fig.

2.

Trajectory
called a mental line that a material point describes during its movement. According to the shape of the trajectory, the movement is divided into: A)
rectilinear - the trajectory is a straight line segment; b)

curvilinear- the trajectory is a segment of a curve.
Path is a vector connecting the initial position of a material point with its final position (see figure).

It is very important to understand how a path differs from a movement. The most important difference is that movement is a vector with a beginning at the point of departure and an end at the destination (it does not matter at all what route this movement took). And the path is, on the contrary, a scalar quantity that reflects the length of the trajectory traveled.

Uniform linear movement is a movement in which a material point makes equal movements in any equal intervals of time
Speed ​​of uniform linear motion is called the ratio of movement to the time during which this movement occurred:

For uneven motion they use the concept average speed. Average speed is often introduced as a scalar quantity. This is the speed of such uniform motion in which the body travels the same path in the same time as during uneven motion:

Instant speed call the speed of a body at a given point in the trajectory or at a given moment in time.
Uniformly accelerated linear motion- this is a rectilinear movement in which the instantaneous speed for any equal periods of time changes by the same amount

The dependence of the body coordinates on time in uniform rectilinear motion has the form: x = x 0 + V x t, where x 0 is the initial coordinate of the body, V x is the speed of movement.
Free fall called uniformly accelerated motion with constant acceleration g = 9.8 m/s 2, independent of the mass of the falling body. It occurs only under the influence of gravity.

Free fall speed is calculated using the formula:

Vertical movement is calculated using the formula:

One type of motion of a material point is motion in a circle. With such movement, the speed of the body is directed along a tangent drawn to the circle at the point where the body is located (linear speed). You can describe the position of a body on a circle using a radius drawn from the center of the circle to the body. The displacement of a body when moving in a circle is described by turning the radius of the circle connecting the center of the circle with the body. The ratio of the angle of rotation of the radius to the period of time during which this rotation occurred characterizes the speed of movement of the body in a circle and is called angular velocity ω:

Angular velocity is related to linear velocity by the relation

where r is the radius of the circle.
The time it takes a body to complete a complete revolution is called circulation period. The reciprocal of the period is the circulation frequency - ν

Since during uniform motion in a circle the velocity module does not change, but the direction of the velocity changes, with such motion there is acceleration. He is called centripetal acceleration, it is directed radially towards the center of the circle:

Basic concepts and laws of dynamics

The part of mechanics that studies the reasons that caused the acceleration of bodies is called dynamics

Newton's first law:
There are reference systems relative to which a body maintains its speed constant or is at rest if other bodies do not act on it or the action of other bodies is compensated.
The property of a body to maintain a state of rest or uniform linear motion with balanced external forces acting on it is called inertia. The phenomenon of maintaining the speed of a body under balanced external forces is called inertia. Inertial reference systems are systems in which Newton's first law is satisfied.

Galileo's principle of relativity:
in all inertial reference systems under the same initial conditions, all mechanical phenomena proceed in the same way, i.e. subject to the same laws
Weight is a measure of body inertia
Force is a quantitative measure of the interaction of bodies.

Newton's second law:
The force acting on a body is equal to the product of the mass of the body and the acceleration imparted by this force:
$F↖(→) = m⋅a↖(→)$

The addition of forces consists of finding the resultant of several forces, which produces the same effect as several simultaneously acting forces.

Newton's third law:
The forces with which two bodies act on each other are located on the same straight line, equal in magnitude and opposite in direction:
$F_1↖(→) = -F_2↖(→) $

Newton's III law emphasizes that the action of bodies on each other is in the nature of interaction. If body A acts on body B, then body B acts on body A (see figure).

Or in short, the force of action is equal to the force of reaction. The question often arises: why does a horse pull a sled if these bodies interact with equal forces? This is possible only through interaction with the third body - the Earth. The force with which the hooves press into the ground must be greater than the frictional force of the sled on the ground. Otherwise, the hooves will slip and the horse will not move.
If a body is subjected to deformation, forces arise that prevent this deformation. Such forces are called elastic forces.

Hooke's law written in the form

where k is the spring stiffness, x is the deformation of the body. The “−” sign indicates that the force and deformation are directed in different directions.

When bodies move relative to each other, forces arise that impede the movement. These forces are called friction forces. A distinction is made between static friction and sliding friction. Sliding friction force calculated by the formula

where N is the support reaction force, µ is the friction coefficient.
This force does not depend on the area of ​​the rubbing bodies. The friction coefficient depends on the material from which the bodies are made and the quality of their surface treatment.

Static friction occurs if the bodies do not move relative to each other. The static friction force can vary from zero to a certain maximum value

By gravitational forces are the forces with which any two bodies are attracted to each other.

Here R is the distance between the bodies. The law of universal gravitation in this form is valid either for material points or for spherical bodies.

Body weight called the force with which the body presses on a horizontal support or stretches the suspension.

Gravity- this is the force with which all bodies are attracted to the Earth:

With a stationary support, the weight of the body is equal in magnitude to the force of gravity:

If a body moves vertically with acceleration, then its weight will change.
When a body moves with upward acceleration, its weight

It can be seen that the weight of the body is greater than the weight of the body at rest.

When a body moves with downward acceleration, its weight

In this case, the weight of the body is less than the weight of the body at rest.

Weightlessness is the movement of a body in which its acceleration is equal to the acceleration of gravity, i.e. a = g. This is possible if only one force acts on the body - gravity.
Artificial Earth satellite- this is a body that has a speed V1 sufficient to move in a circle around the Earth
There is only one force acting on the Earth's satellite - the force of gravity directed towards the center of the Earth
First escape velocity- this is the speed that must be imparted to the body so that it revolves around the planet in a circular orbit.

where R is the distance from the center of the planet to the satellite.
For the Earth, near its surface, the first escape velocity is equal to

1.3. Basic concepts and laws of statics and hydrostatics

Lever equilibrium condition:
the algebraic sum of the moments of all forces rotating the body is equal to zero.
Pressure is a physical quantity equal to the ratio of the force acting on the platform, perpendicular to this force, to the area of ​​the platform:

Valid for liquids and gases Pascal's law:
pressure spreads in all directions without changes.
If a liquid or gas is in a gravity field, then each layer above presses on the layers below, and as the liquid or gas is immersed inside, the pressure increases. For liquids

where ρ is the density of the liquid, h is the depth of penetration into the liquid.

A homogeneous liquid in communicating vessels is established at the same level. If liquid with different densities is poured into the elbows of communicating vessels, then the liquid with a higher density is installed at a lower height. In this case

The heights of liquid columns are inversely proportional to densities:

Hydraulic Press is a vessel filled with oil or other liquid, in which two holes are cut, closed by pistons. The pistons have different areas. If a certain force is applied to one piston, then the force applied to the second piston turns out to be different.
Thus, the hydraulic press serves to convert the magnitude of the force. Since the pressure under the pistons must be the same, then

Then A1 = A2.
A body immersed in a liquid or gas is acted upon by an upward buoyant force from the side of this liquid or gas, which is called by the power of Archimedes
The magnitude of the buoyancy force is determined by Archimedes' law: a body immersed in a liquid or gas is acted upon by a buoyant force directed vertically upward and equal to the weight of the liquid or gas displaced by the body:

where ρ liquid is the density of the liquid in which the body is immersed; V submergence is the volume of the submerged part of the body.

Body floating condition- a body floats in a liquid or gas when the buoyant force acting on the body is equal to the force of gravity acting on the body.

1.4. Conservation laws

Momentum is a vector quantity. [p] = kg m/s. Along with body impulse, they often use impulse of power. This is the product of force and the duration of its action
The change in the momentum of a body is equal to the momentum of the force acting on this body. For an isolated system of bodies (a system whose bodies interact only with each other) law of conservation of momentum: the sum of the impulses of the bodies of an isolated system before interaction is equal to the sum of the impulses of the same bodies after the interaction.
Mechanical work called a physical quantity that is equal to the product of the force acting on the body, the displacement of the body and the cosine of the angle between the direction of the force and the displacement:

Power is the work done per unit of time:

The ability of a body to do work is characterized by a quantity called energy. Mechanical energy is divided into kinetic and potential. If a body can do work due to its motion, it is said to have kinetic energy. The kinetic energy of the translational motion of a material point is calculated by the formula

If a body can do work by changing its position relative to other bodies or by changing the position of parts of the body, it has potential energy. An example of potential energy: a body raised above the ground, its energy is calculated using the formula

where h is the lift height

Compressed spring energy:

where k is the spring stiffness coefficient, x is the absolute deformation of the spring.

The sum of potential and kinetic energy is mechanical energy. For an isolated system of bodies in mechanics, law of conservation of mechanical energy: if there are no frictional forces between the bodies of an isolated system (or other forces leading to energy dissipation), then the sum of the mechanical energies of the bodies of this system does not change (the law of conservation of energy in mechanics). If there are friction forces between the bodies of an isolated system, then during interaction part of the mechanical energy of the bodies turns into internal energy.

1.5. Mechanical vibrations and waves

The quantity A equal to the largest absolute value of the fluctuating physical quantity x is called amplitude of oscillations. The expression α = ωt + ϕ determines the value of x at a given time and is called the oscillation phase. Period T is the time it takes for an oscillating body to complete one complete oscillation. Frequency of periodic oscillations is the number of complete oscillations completed per unit of time:

Frequency is measured in s -1. This unit is called hertz (Hz).

Mathematical pendulum is a material point of mass m suspended on a weightless inextensible thread and oscillating in a vertical plane.
If one end of the spring is fixed motionless, and a body of mass m is attached to its other end, then when the body is removed from the equilibrium position, the spring will stretch and oscillations of the body on the spring will occur in the horizontal or vertical plane. Such a pendulum is called a spring pendulum.

Period of oscillation of a mathematical pendulum determined by the formula

where l is the length of the pendulum.

Period of oscillation of a load on a spring determined by the formula

where k is the spring stiffness, m is the mass of the load.

Propagation of vibrations in elastic media.
A medium is called elastic if there are interaction forces between its particles. Waves are the process of propagation of vibrations in elastic media.
The wave is called transverse, if the particles of the medium oscillate in directions perpendicular to the direction of propagation of the wave. The wave is called longitudinal, if the vibrations of the particles of the medium occur in the direction of wave propagation.
Wavelength is the distance between two closest points oscillating in the same phase:

where v is the speed of wave propagation.

Sound waves are called waves in which oscillations occur with frequencies from 20 to 20,000 Hz.
The speed of sound varies in different environments. The speed of sound in air is 340 m/s.
Ultrasonic waves are called waves whose oscillation frequency exceeds 20,000 Hz. Ultrasonic waves are not perceived by the human ear.

Page 8 of 12

§ 7. Movement under uniform acceleration
straight motion

1. Using a graph of speed versus time, you can obtain a formula for the displacement of a body during uniform rectilinear motion.

Figure 30 shows a graph of the projection of the speed of uniform motion onto the axis X from time. If we restore the perpendicular to the time axis at some point C, then we get a rectangle OABC. The area of ​​this rectangle is equal to the product of the sides O.A. And O.C.. But side length O.A. equal to v x, and the side length O.C. - t, from here S = v x t. Product of the projection of velocity onto an axis X and time is equal to the projection of displacement, i.e. s x = v x t.

Thus, the projection of displacement during uniform rectilinear motion is numerically equal to the area of ​​the rectangle bounded by the coordinate axes, the velocity graph and the perpendicular to the time axis.

2. We obtain in a similar way the formula for the projection of displacement in rectilinear uniformly accelerated motion. To do this, we will use the graph of the velocity projection onto the axis X from time to time (Fig. 31). Let's select a small area on the graph ab and drop the perpendiculars from the points a And b on the time axis. If time interval D t, corresponding to the site CD on the time axis is small, then we can assume that the speed does not change during this period of time and the body moves uniformly. In this case the figure cabd differs little from a rectangle and its area is numerically equal to the projection of the movement of the body over the time corresponding to the segment CD.

The whole figure can be divided into such strips OABC, and its area will be equal to the sum of the areas of all strips. Therefore, the projection of the movement of the body over time t numerically equal to the area of ​​the trapezoid OABC. From your geometry course you know that the area of ​​a trapezoid is equal to the product of half the sum of its bases and height: S= (O.A. + B.C.)O.C..

As can be seen from Figure 31, O.A. = v 0x , B.C. = v x, O.C. = t. It follows that the displacement projection is expressed by the formula: s x= (v x + v 0x)t.

With uniformly accelerated rectilinear motion, the speed of the body at any moment of time is equal to v x = v 0x + a x t, hence, s x = (2v 0x + a x t)t.

From here:

To obtain the equation of motion of a body, we substitute its expression in terms of the difference in coordinates into the displacement projection formula s x = x – x 0 .

We get: x – x 0 = v 0x t+ , or

Using the equation of motion, you can determine the coordinate of a body at any time if the initial coordinate, initial velocity and acceleration of the body are known.

3. In practice, there are often problems in which it is necessary to find the displacement of a body during uniformly accelerated rectilinear motion, but the time of motion is unknown. In these cases, a different displacement projection formula is used. Let's get it.

From the formula for the projection of the velocity of uniformly accelerated rectilinear motion v x = v 0x + a x t Let's express time:

t = .

Substituting this expression into the displacement projection formula, we obtain:

s x = v 0x + .

From here:

s x = , or
–= 2a x s x.

If the initial speed of the body is zero, then:

2a x s x.

4. Example of problem solution

A skier slides down a mountain slope from a state of rest with an acceleration of 0.5 m/s 2 in 20 s and then moves along a horizontal section, having traveled 40 m to a stop. With what acceleration did the skier move along a horizontal surface? What is the length of the mountain slope?

We connect the reference system with the Earth, the axis X let's direct the skier in the direction of speed at each stage of his movement (Fig. 32).

Let's write the equation for the skier's speed at the end of the descent from the mountain:

v 1 = v 01 + a 1 t 1 .

In projections onto the axis X we get: v 1x = a 1x t. Since the projections of velocity and acceleration onto the axis X are positive, the speed modulus of the skier is equal to: v 1 = a 1 t 1 .

Let us write an equation connecting the projections of speed, acceleration and displacement of the skier at the second stage of movement:

–= 2a 2x s 2x .

Considering that the initial speed of the skier at this stage of movement is equal to his final speed at the first stage

v 02 = v 1 , v 2x= 0 we get

– = –2a 2 s 2 ; (a 1 t 1) 2 = 2a 2 s 2 .

From here a 2 = ;

a 2 == 0.125 m/s 2 .

The module of movement of the skier at the first stage of movement is equal to the length of the mountain slope. Let's write the equation for displacement:

s 1x = v 01x t + .

Hence the length of the mountain slope is s 1 = ;

s 1 == 100 m.

Answer: a 2 = 0.125 m/s 2 ; s 1 = 100 m.

Self-test questions

1. As in the graph of the projection of the speed of uniform rectilinear motion onto the axis X

2. As in the graph of the projection of the speed of uniformly accelerated rectilinear motion onto the axis X determine the projection of body movement from time to time?

3. What formula is used to calculate the projection of the displacement of a body during uniformly accelerated linear motion?

4. What formula is used to calculate the projection of the displacement of a body moving uniformly accelerated and rectilinearly if the initial speed of the body is zero?

Task 7

1. What is the module of movement of the car in 2 minutes, if during this time its speed changed from 0 to 72 km/h? What is the coordinate of the car at the moment of time t= 2 min? The initial coordinate is considered equal to zero.

2. The train moves with an initial speed of 36 km/h and an acceleration of 0.5 m/s 2 . What is the displacement of the train in 20 s and its coordinate at the moment of time? t= 20 s if the initial coordinate of the train is 20 m?

3. What is the cyclist’s displacement in 5 s after the start of braking, if his initial speed during braking is 10 m/s and the acceleration is 1.2 m/s 2? What is the coordinate of the cyclist at the moment of time? t= 5 s, if at the initial moment of time it was at the origin?

4. A car moving at a speed of 54 km/h stops when braking for 15 s. What is the modulus of movement of a car during braking?

5. Two cars are moving towards each other from two settlements located at a distance of 2 km from each other. The initial speed of one car is 10 m/s and the acceleration is 0.2 m/s 2 , the initial speed of the other is 15 m/s and the acceleration is 0.2 m/s 2 . Determine the time and coordinates of the meeting place of the cars.

Laboratory work No. 1

Study of uniformly accelerated
rectilinear motion

Goal of the work:

learn to measure acceleration during uniformly accelerated linear motion; to experimentally establish the ratio of the paths traversed by a body during uniformly accelerated rectilinear motion in successive equal intervals of time.

Devices and materials:

trench, tripod, metal ball, stopwatch, measuring tape, metal cylinder.

Work order

1. Secure one end of the chute in the tripod leg so that it makes a small angle with the surface of the table. At the other end of the chute, place a metal cylinder in it.

2. Measure the paths traveled by the ball in 3 consecutive periods of time equal to 1 s each. This can be done in different ways. You can put chalk marks on the gutter that record the positions of the ball at times equal to 1 s, 2 s, 3 s, and measure the distances s_ between these marks. You can, by releasing the ball from the same height each time, measure the path s, traveled by it first in 1 s, then in 2 s and in 3 s, and then calculate the path traveled by the ball in the second and third seconds. Record the measurement results in table 1.

3. Find the ratio of the path traveled in the second second to the path traveled in the first second, and the path traveled in the third second to the path traveled in the first second. Draw a conclusion.

4. Measure the time the ball moves along the chute and the distance it travels. Calculate the acceleration of its motion using the formula s = .

5. Using the experimentally obtained acceleration value, calculate the distances that the ball must travel in the first, second and third seconds of its movement. Draw a conclusion.

Table 1