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 the length of the trajectory that a material point describes over a given period of time. This is a scalar quantity.
Moving 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 called a movement in which a material point makes the same movements over any equal periods 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 rotating 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, 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 a 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 The number of complete oscillations completed per unit of time is called:

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.

In general uniformly accelerated motion called such a movement in which the acceleration vector remains unchanged in magnitude and direction. An example of such movement is the movement of a stone thrown at a certain angle to the horizon (without taking into account air resistance). At any point in the trajectory, the acceleration of the stone is equal to the acceleration of gravity. For a kinematic description of the movement of a stone, it is convenient to choose a coordinate system so that one of the axes, for example the axis OY, was directed parallel to the acceleration vector. Then the curvilinear movement of the stone can be represented as the sum of two movements - rectilinear uniformly accelerated motion along the axis OY And uniform rectilinear motion in the perpendicular direction, i.e. along the axis OX(Fig. 1.4.1).

Thus, the study of uniformly accelerated motion is reduced to the study of rectilinear uniformly accelerated motion. In the case of rectilinear motion, the velocity and acceleration vectors are directed along the straight line of motion. Therefore, the speed υ and acceleration a in projections onto the direction of movement can be considered as algebraic quantities.

In uniformly accelerated rectilinear motion, the speed of a body is determined by the formula

(*)

In this formula, υ 0 is the speed of the body at t = 0 (starting speed ), a= const - acceleration. On the speed graph υ ( t) this dependence looks like a straight line (Fig. 1.4.2).

Acceleration can be determined from the slope of the velocity graph a bodies. The corresponding constructions are shown in Fig. 1.4.2 for graph I. Acceleration is numerically equal to the ratio of the sides of the triangle ABC:

The greater the angle β that the velocity graph forms with the time axis, i.e., the greater the slope of the graph ( steepness), the greater the acceleration of the body.

For graph I: υ 0 = -2 m/s, a= 1/2 m/s 2.

For schedule II: υ 0 = 3 m/s, a= -1/3 m/s 2

The velocity graph also allows you to determine the projection of movement s bodies for some time t. Let us select on the time axis a certain small period of time Δ t. If this period of time is small enough, then the change in speed over this period is small, i.e. the movement during this period of time can be considered uniform with a certain average speed, which is equal to the instantaneous speed υ of the body in the middle of the interval Δ t. Therefore, the displacement Δ s in time Δ t will be equal to Δ s = υΔ t. This movement is equal to the area of ​​the shaded strip (Fig. 1.4.2). Breaking down the time period from 0 to some point t for small intervals Δ t, we find that the movement s for a given time t with uniformly accelerated rectilinear motion is equal to the area of ​​the trapezoid ODEF. The corresponding constructions were made for graph II in Fig. 1.4.2. Time t taken equal to 5.5 s.

Since υ - υ 0 = at, the final formula for moving s body with uniformly accelerated motion over a time interval from 0 to t will be written in the form:

(**)

To find the coordinates y bodies at any time t needed to the starting coordinate y 0 add movement in time t:

(***)

This expression is called law of uniformly accelerated motion .

When analyzing uniformly accelerated motion, sometimes the problem arises of determining the movement of a body based on the given values ​​of the initial υ 0 and final υ velocities and acceleration a. This problem can be solved using the equations written above by eliminating time from them t. The result is written in the form

From this formula we can obtain an expression for determining the final speed υ of a body if the initial speed υ 0 and acceleration are known a and moving s:

If the initial speed υ 0 is zero, these formulas take the form

It should be noted once again that the quantities υ 0, υ, included in the formulas for uniformly accelerated rectilinear motion s, a, y 0 are algebraic quantities. Depending on the specific type of movement, each of these quantities can take on both positive and negative values.

In previous lessons, we discussed how to determine the distance traveled during uniform linear motion. It's time to find out how to determine the coordinates of a body, the distance traveled and the displacement during rectilinear uniformly accelerated motion. This can be done if we consider rectilinear uniformly accelerated motion as a set of a large number of very small uniform displacements of the body.

The first to solve the problem of the location of a body at a certain point in time during accelerated motion was the Italian scientist Galileo Galilei (Fig. 1).

Rice. 1. Galileo Galilei (1564-1642)

He conducted his experiments with an inclined plane. He launched a ball, a musket bullet, along the chute, and then determined the acceleration of this body. How did he do it? He knew the length of the inclined plane, and determined the time by the beat of his heart or pulse (Fig. 2).

Rice. 2. Galileo's experiment

Consider the speed dependence graph uniformly accelerated linear motion from time. You know this dependence; it is a straight line: .

Rice. 3. Determination of displacement during uniformly accelerated linear motion

We divide the speed graph into small rectangular sections (Fig. 3). Each section will correspond to a certain speed, which can be considered constant in a given period of time. It is necessary to determine the distance traveled during the first period of time. Let's write the formula: . Now let's calculate the total area of ​​all the figures we have.

The sum of the areas during uniform motion is the total distance traveled.

Please note: the speed will change from point to point, thereby we will get the path traveled by the body precisely during rectilinear uniformly accelerated motion.

Note that during rectilinear uniformly accelerated motion of a body, when speed and acceleration are directed in the same direction (Fig. 4), the displacement module is equal to the distance traveled, therefore, when we determine the displacement module, we determine distance traveled. In this case, we can say that the displacement module will be equal to the area of ​​the figure, limited by the graph of speed and time.

Rice. 4. The displacement module is equal to the distance traveled

Let's use mathematical formulas to calculate the area of ​​the indicated figure.

Rice. 5 Illustration for calculating area

The area of ​​the figure (numerically equal to the distance traveled) is equal to half the sum of the bases multiplied by the height. Please note that in the figure, one of the bases is the initial speed, and the second base of the trapezoid will be the final speed, indicated by the letter . The height of the trapezoid is equal to , this is the period of time during which the movement occurred.

We can write the final velocity, discussed in the previous lesson, as the sum of the initial velocity and the contribution due to the constant acceleration of the body. The resulting expression is:

If you open the brackets, it becomes double. We can write the following expression:

If you write each of these expressions separately, the result will be the following:

This equation was first obtained through the experiments of Galileo Galilei. Therefore, we can consider that it was this scientist who first made it possible to determine the location of a body during rectilinear uniformly accelerated motion at any time. This is the solution to the main problem of mechanics.

Now let's remember that the distance traveled, equal in our case movement module, is expressed by the difference:

If we substitute this expression into Galileo’s equation, we obtain a law according to which the coordinate of a body changes during rectilinear uniformly accelerated motion:

It should be remembered that the quantities are projections of velocity and acceleration onto the selected axis. Therefore, they can be both positive and negative.

Conclusion

The next stage of consideration of movement will be the study of movement along a curvilinear trajectory.

Bibliography

1. Kikoin I.K., Kikoin A.K. Physics: textbook for 9th grade of high school. - M.: Enlightenment.
2. Peryshkin A.V., Gutnik E.M., Physics. 9th grade: textbook for general education. institutions/A. V. Peryshkin, E. M. Gutnik. - 14th ed., stereotype. - M.: Bustard, 2009. - 300.
3. Sokolovich Yu.A., Bogdanova G.S.. Physics: A reference book with examples of problem solving. - 2nd edition repartition. - X.: Vesta: Ranok Publishing House, 2005. - 464 p.

Additional recommended links to Internet resources

1. Internet portal “class-fizika.narod.ru” ()
2. Internet portal “videouroki.net” ()
3. Internet portal “foxford.ru” ()

Homework

1. Write down the formula that determines the projection of the displacement vector of a body during rectilinear uniformly accelerated motion.
2. A cyclist, whose initial speed is 15 km/h, slides down a hill in 5 s. Determine the length of the slide if the cyclist moved with a constant acceleration of 0.5 m/s^2 .
3. How do the dependences of displacement on time differ for uniform and uniformly accelerated motion?

In this topic we will look at a very special type of irregular motion. Based on the opposition to uniform movement, uneven movement is movement at unequal speed along any trajectory. What is the peculiarity of uniformly accelerated motion? This is an uneven movement, but which "equally accelerated". We associate acceleration with increasing speed. Let's remember the word "equal", we get an equal increase in speed. How do we understand “equal increase in speed”, how can we evaluate whether the speed is increasing equally or not? To do this, we need to record time and estimate the speed over the same time interval. For example, a car starts to move, in the first two seconds it develops a speed of up to 10 m/s, in the next two seconds it reaches 20 m/s, and after another two seconds it already moves at a speed of 30 m/s. Every two seconds the speed increases and each time by 10 m/s. This is uniformly accelerated motion.

The physical quantity that characterizes how much the speed increases each time is called acceleration.

Can the movement of a cyclist be considered uniformly accelerated if, after stopping, in the first minute his speed is 7 km/h, in the second - 9 km/h, in the third - 12 km/h? It is forbidden! The cyclist accelerates, but not equally, first he accelerated by 7 km/h (7-0), then by 2 km/h (9-7), then by 3 km/h (12-9).

Typically, movement with increasing speed is called accelerated movement. Movement with decreasing speed is slow motion. But physicists call any movement with changing speed accelerated movement. Whether the car starts moving (the speed increases!) or brakes (the speed decreases!), in any case it moves with acceleration.

Uniformly accelerated motion- this is the movement of a body in which its speed for any equal intervals of time changes(can increase or decrease) the same

Body acceleration

Acceleration characterizes the rate of change in speed. This is the number by which the speed changes every second. If the acceleration of a body is large in magnitude, this means that the body quickly gains speed (when it accelerates) or quickly loses it (when braking). Acceleration is a physical vector quantity, numerically equal to the ratio of the change in speed to the period of time during which this change occurred.

Let's determine the acceleration in the next problem. At the initial moment of time, the speed of the ship was 3 m/s, at the end of the first second the speed of the ship became 5 m/s, at the end of the second - 7 m/s, at the end of the third 9 m/s, etc. Obviously, . But how did we determine? We are looking at the speed difference over one second. In the first second 5-3=2, in the second second 7-5=2, in the third 9-7=2. But what if the speeds are not given for every second? Such a problem: the initial speed of the ship is 3 m/s, at the end of the second second - 7 m/s, at the end of the fourth 11 m/s. In this case, you need 11-7 = 4, then 4/2 = 2. We divide the speed difference by the time period.

This formula is most often used in a modified form when solving problems:

The formula is not written in vector form, so we write the “+” sign when the body is accelerating, the “-” sign when it is slowing down.

Acceleration vector direction

The direction of the acceleration vector is shown in the figures

In this figure, the car moves in a positive direction along the Ox axis, the velocity vector always coincides with the direction of movement (directed to the right). When the acceleration vector coincides with the direction of the speed, this means that the car is accelerating. Acceleration is positive.

During acceleration, the direction of acceleration coincides with the direction of speed. Acceleration is positive.

In this picture, the car is moving in the positive direction along the Ox axis, the velocity vector coincides with the direction of movement (directed to the right), the acceleration does NOT coincide with the direction of the speed, this means that the car is braking. Acceleration is negative.

When braking, the direction of acceleration is opposite to the direction of speed. Acceleration is negative.

Let's figure out why the acceleration is negative when braking. For example, in the first second the ship slowed down from 9 m/s to 7 m/s, in the second second to 5 m/s, in the third to 3 m/s. The speed changes to "-2m/s". 3-5=-2; 5-7=-2; 7-9=-2m/s. This is where the negative acceleration value comes from.

When solving problems, if the body slows down, acceleration is substituted into the formulas with a minus sign!!!

Moving during uniformly accelerated motion

An additional formula called timeless

Formula in coordinates

Medium speed communication

With uniformly accelerated motion, the average speed can be calculated as the arithmetic mean of the initial and final speeds

From this rule follows a formula that is very convenient to use when solving many problems

Path ratio

If a body moves uniformly accelerated, the initial speed is zero, then the paths traversed in successive equal intervals of time are related as a successive series of odd numbers.

The main thing to remember

1) What is uniformly accelerated motion;
2) What characterizes acceleration;
3) Acceleration is a vector. If a body accelerates, the acceleration is positive, if it slows down, the acceleration is negative;
3) Direction of the acceleration vector;
4) Formulas, units of measurement in SI

Exercises

Two trains are moving towards each other: one is heading north at an accelerated rate, the other is moving slowly to the south. How are train accelerations directed?

Equally to the north. Because the first train's acceleration coincides in direction with the movement, and the second train's acceleration is opposite to the movement (it slows down).

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 get:

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 rectilinear motion?

4. What formula is used to calculate the projection of 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 movement

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 slight angle with the table surface. 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