Physics lectures. Graphic representation of uniformly accelerated linear motion. Moving during uniformly accelerated motion

Questions.

1. Write down the formula by which you can calculate the projection of the instantaneous velocity vector of rectilinear uniformly accelerated motion if you know: a) the projection of the initial velocity vector and the projection of the acceleration vector; b) projection of the acceleration vector given that the initial speed is zero.

2. What is the projection graph of the velocity vector of uniformly accelerated motion at an initial speed: a) equal to zero; b) not equal to zero?

3. How are the movements, the graphs of which are presented in Figures 11 and 12, similar and different from each other?

In both cases, the movement occurs with acceleration, but in the first case the acceleration is positive, and in the second case it is negative.

Exercises.

1. A hockey player lightly hit the puck with his stick, giving it a speed of 2 m/s. What will be the speed of the puck 4 s after impact if, as a result of friction with ice, it moves with an acceleration of 0.25 m/s 2?

2. A skier slides down a mountain from a state of rest with an acceleration equal to 0.2 m/s 2 . After what period of time will its speed increase to 2 m/s?

3. In the same coordinate axes, construct graphs of the projection of the velocity vector (on the X axis, codirectional with the initial velocity vector) for a rectilinear uniformly accelerated motion for cases: a) v ox = 1 m/s, a x = 0.5 m/s 2 ; b) v ox = 1 m/s, a x = 1 m/s 2; c) v ox = 2 m/s, a x = 1 m/s 2.
The scale is the same in all cases: 1 cm - 1 m/s; 1cm - 1s.

4. In the same coordinate axes, construct graphs of the projection of the velocity vector (on the X axis, codirectional with the initial velocity vector) for rectilinear uniformly accelerated motion for the cases: a) v ox = 4.5 m/s, a x = -1.5 m/s 2 ; b) v ox = 3 m/s, a x = -1 m/s 2
Choose the scale yourself.

5. Figure 13 shows graphs of the velocity vector modulus versus time for the rectilinear motion of two bodies. With what absolute acceleration does body I move? body II?

Graphical representation of uniformly accelerated rectilinear movement.

Moving during uniformly accelerated motion.

Ilevel.

Many physical quantities that describe the movements of bodies change over time. Therefore, for greater clarity of description, movement is often depicted graphically.

Let us show how the time dependences of kinematic quantities describing rectilinear uniformly accelerated motion are graphically depicted.

Uniformly accelerated linear motion- this is a movement in which the speed of a body changes equally over any equal periods of time, i.e. it is a movement with acceleration constant in magnitude and direction.

a=const - acceleration equation. That is, a has a numerical value that does not change over time.

By definition of acceleration

From here we have already found equations for the dependence of speed on time: v = v0 + at.

Let's see how this equation can be used to graphically represent uniformly accelerated motion.

Let us graphically depict the dependences of kinematic quantities on time for three bodies

.

1, the body moves along the 0X axis, while increasing its speed (acceleration vector a is codirectional with the velocity vector v). vx >0, akh > 0

2, the body moves along the 0X axis, while reducing its speed (the acceleration vector a is not codirectional with the velocity vector v). vx >0, ah< 0

2, the body moves against the 0X axis, while reducing its speed (the acceleration vector is not codirectional with the velocity vector v). vx< 0, ах > 0

Acceleration graph

Acceleration, by definition, is a constant value. Then, for the presented situation, the graph of acceleration versus time a(t) will look like:

From the acceleration graph, you can determine how the speed changed - increased or decreased and by what numerical value the speed changed and which body the speed changed more.

Speed ​​graph

If we compare the dependence of the coordinate on time during uniform motion and the dependence of the velocity projection on time during uniformly accelerated motion, we can see that these dependencies are the same:

x= x0 + vx t vx = v 0 x + a X t

This means that the dependency graphs have the same appearance.

To construct this graph, the time of movement is plotted on the abscissa axis, and the speed (projection of speed) of the body is plotted on the ordinate axis. In uniformly accelerated motion, the speed of a body changes over time.

Moving during uniformly accelerated motion.

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

vx = v 0 x + a X t

In this formula, υ0 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.).

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

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: υ0 = –2 m/s, a= 1/2 m/s2.

For graph II: υ0 = 3 m/s, a= –1/3 m/s2.

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 short 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.). 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 – υ0 = at s 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: DIV_ADBLOCK189">

Since υ – υ0 = 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: https://pandia.ru/text/78/516/images/image009_57.gif" width="146 height=55" height="55">

When analyzing uniformly accelerated motion, sometimes the problem arises of determining the movement of a body based on the given values ​​of the initial υ0 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

If the initial speed υ0 is zero, these formulas take the form MsoNormalTable">

It should be noted once again that the quantities υ0, 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.

An example of solving a problem:

Petya slides down the mountainside from a state of rest with an acceleration of 0.5 m/s2 in 20 s and then moves along a horizontal section. Having traveled 40 m, he crashes into the gaping Vasya and falls into a snowdrift, reducing his speed to 0 m/s. With what acceleration did Petya move along the horizontal surface to the snowdrift? What is the length of the mountain slope from which Petya so unsuccessfully slid down?

Stage 1.

The equation for Petit's speed at the end of the descent from the mountain is:

v 1 = v 01 + a 1t 1.

In projections onto the axis X we get:

v 1x = a 1xt.

Let us write an equation connecting the projections of Petya’s velocity, acceleration and displacement at the first stage of movement:

or because Petya was driving from the very top of the hill with an initial speed of V01=0

(If I were Petya, I would be careful about driving down such high hills)

Considering that Petya’s initial speed at this 2nd stage of movement is equal to his final speed at the first stage:

v 02 x = v 1 x, v 2x = 0, where v1 is the speed with which Petya reached the foot of the hill and began to move towards Vasya. V2x - Petya's speed in a snowdrift.

2. By this schedule acceleration, tell us how the speed of the body changes. Write down the equations for the dependence of speed on time if at the moment of the start of movement (t=0) the speed of the body is v0х =0. Please note that with each subsequent section of movement, the body begins to pass at a certain speed (which was achieved during the previous time!).

3. A metro train, leaving the station, can reach a speed of 72 km/h in 20 s. Determine with what acceleration a bag, forgotten in a subway car, is moving away from you. How far will she travel?

4. A cyclist moving at a speed of 3 m/s begins to descend a mountain with an acceleration of 0.8 m/s2. Find the length of the mountain if the descent took 6 s.

5. Having started braking with an acceleration of 0.5 m/s2, the train traveled 225 m to the stop. What was its speed before braking began?

6. Having started to move, the soccer ball reached a speed of 50 m/s, covered a distance of 50 m and crashed into the window. Determine the time it took the ball to travel this path and the acceleration with which it moved.

7. Reaction time of Uncle Oleg’s neighbor = 1.5 minutes, during which time he will figure out what happened to his window and will have time to run out into the yard. Determine what speed young football players should develop so that the joyful owners of the window do not catch up with them, if they need to run 350 m to their entrance.

8. Two cyclists are riding towards each other. The first one, having a speed of 36 km/h, began to climb up the mountain with an acceleration of 0.2 m/s2, and the second, having a speed of 9 km/h, began to descend the mountain with an acceleration of 0.2 m/s2. After how long and in what place will they collide due to their absent-mindedness, if the length of the mountain is 100 m?

Uniform movement– this is movement at a constant speed, that is, when the speed does not change (v = const) and acceleration or deceleration does not occur (a = 0).

Straight-line movement- this is movement in a straight line, that is, the trajectory of rectilinear movement is a straight line.

Uniform linear movement- this is a movement in which a body makes equal movements at any equal intervals of time. For example, if we divide a certain time interval into one-second intervals, then with uniform motion the body will move the same distance for each of these time intervals.

The speed of uniform rectilinear motion does not depend on time and at each point of the trajectory is directed in the same way as the movement of the body. That is, the displacement vector coincides in direction with the velocity vector. In this case, the average speed for any period of time is equal to the instantaneous speed:

Speed ​​of uniform rectilinear motion is a physical vector quantity equal to the ratio of the movement of a body over any period of time to the value of this interval t:

Thus, the speed of uniform rectilinear motion shows how much movement a material point makes per unit time.

Moving with uniform linear motion is determined by the formula:

Distance traveled in linear motion is equal to the displacement module. If the positive direction of the OX axis coincides with the direction of movement, then the projection of the velocity onto the OX axis is equal to the magnitude of the velocity and is positive:

v x = v, that is v > 0

The projection of displacement onto the OX axis is equal to:

s = vt = x – x 0

where x 0 is the initial coordinate of the body, x is the final coordinate of the body (or the coordinate of the body at any time)

Equation of motion, that is, the dependence of the body coordinates on time x = x(t), takes the form:

If the positive direction of the OX axis is opposite to the direction of motion of the body, then the projection of the body’s velocity onto the OX axis is negative, the speed is less than zero (v< 0), и тогда уравнение движения принимает вид:

Dependence of speed, coordinates and path on time

The dependence of the projection of the body velocity on time is shown in Fig. 1.11. Since the speed is constant (v = const), the speed graph is a straight line parallel to the time axis Ot.

Rice. 1.11. Dependence of the projection of body velocity on time for uniform rectilinear motion.

The projection of movement onto the coordinate axis is numerically equal to the area of ​​the rectangle OABC (Fig. 1.12), since the magnitude of the movement vector is equal to the product of the velocity vector and the time during which the movement was made.

Rice. 1.12. Dependence of the projection of body displacement on time for uniform rectilinear motion.

A graph of displacement versus time is shown in Fig. 1.13. The graph shows that the projection of the velocity is equal to

v = s 1 / t 1 = tan α

where α is the angle of inclination of the graph to the time axis.

The larger the angle α, the faster the body moves, that is, the greater its speed (the longer the distance the body travels in less time). The tangent of the tangent to the graph of the coordinate versus time is equal to the speed:

Rice. 1.13. Dependence of the projection of body displacement on time for uniform rectilinear motion.

The dependence of the coordinate on time is shown in Fig. 1.14. From the figure it is clear that

tan α 1 > tan α 2

therefore, the speed of body 1 is higher than the speed of body 2 (v 1 > v 2).

tan α 3 = v 3< 0

If the body is at rest, then the coordinate graph is a straight line parallel to the time axis, that is

Rice. 1.14. Dependence of body coordinates on time for uniform rectilinear motion.

Relationship between angular and linear quantities

Individual points of a rotating body have different linear velocities. The speed of each point, being directed tangentially to the corresponding circle, continuously changes its direction. The magnitude of the speed is determined by the speed of rotation of the body and the distance R of the point in question from the axis of rotation. Let the body turn through an angle in a short period of time (Figure 2.4). A point located at a distance R from the axis travels a path equal to

Linear speed of a point by definition.

Tangential acceleration

Using the same relation (2.6) we obtain

Thus, both normal and tangential accelerations increase linearly with the distance of the point from the axis of rotation.

Basic concepts.

Periodic oscillation is a process in which a system (for example, a mechanical one) returns to the same state after a certain period of time. This period of time is called the oscillation period.

restoring force- the force under the influence of which the oscillatory process occurs. This force tends to return a body or a material point, deviated from its position of rest, to its original position.

Depending on the nature of the impact on the oscillating body, a distinction is made between free (or natural) vibrations and forced vibrations.

Free vibrations occur when only a restoring force acts on the oscillating body. In the event that no energy dissipation occurs, free oscillations are undamped. However, real oscillatory processes are damped, because the oscillating body is subject to motion resistance forces (mainly friction forces).

Forced vibrations are performed under the influence of an external periodically changing force, which is called forcing. In many cases, systems undergo oscillations that can be considered harmonic.

Harmonic vibrations are called oscillatory movements in which the displacement of a body from the equilibrium position occurs according to the law of sine or cosine:

To illustrate the physical meaning, consider a circle and rotate the radius OK with angular velocity ω counterclockwise (7.1) counterclockwise. If at the initial moment of time the OK lay in the horizontal plane, then after time t it will shift by an angle. If the starting angle is non-zero and equal to φ 0 , then the angle of rotation will be equal to The projection onto the XO 1 axis is equal to . As the radius OK rotates, the magnitude of the projection changes, and the point will oscillate relative to the point - up, down, etc. In this case, the maximum value of x is equal to A and is called the amplitude of oscillations; ω - circular or cyclic frequency; - oscillation phase; – initial phase. For one revolution of point K around the circle, its projection will make one complete oscillation and return to the starting point.

Period T is called the time of one complete oscillation. After time T, the values ​​of all physical quantities characterizing the oscillations are repeated. In one period, the oscillating point travels a path numerically equal to four amplitudes.

Angular velocity is determined from the condition that during the period T the radius OK will make one revolution, i.e. will rotate by an angle of 2π radians:

Oscillation frequency- the number of oscillations of a point per second, i.e. the oscillation frequency is defined as the reciprocal of the oscillation period:

Spring pendulum elastic forces.

A spring pendulum consists of a spring and a massive ball mounted on a horizontal rod along which it can slide. Let a ball with a hole be attached to a spring and slide along a guide axis (rod). In Fig. 7.2a shows the position of the ball at rest; in Fig. 7.2, b - maximum compression and in Fig. 7.2,c - arbitrary position of the ball.

Under the influence of a restoring force equal to the compression force, the ball will oscillate. Compression force F = -kx, where k is the spring stiffness coefficient. The minus sign indicates that the direction of the force F and the displacement x are opposite. Potential energy of a compressed spring

kinetic

To derive the equation of motion of the ball, it is necessary to relate x and t. The conclusion is based on the law of conservation of energy. The total mechanical energy is equal to the sum of the kinetic and potential energy of the system. In this case:

. In position b): .

Since the law of conservation of mechanical energy is satisfied in the movement under consideration, we can write:

. Let's determine the speed from here:

But in turn and therefore . Let's separate the variables . Integrating this expression, we get: ,

where is the integration constant. From the latter it follows that

Thus, under the action of elastic force, the body performs harmonic oscillations. Forces of a different nature than elastic, but in which the condition F = -kx is satisfied, are called quasi-elastic. Under the influence of these forces, bodies also perform harmonic vibrations. Wherein:

Mathematical pendulum.

A mathematical pendulum is a material point suspended on an inextensible weightless thread, performing oscillatory motion in one vertical plane under the influence of gravity.

Such a pendulum can be considered a heavy ball of mass m, suspended on a thin thread, the length l of which is much greater than the size of the ball. If it is deflected by an angle α (Fig. 7.3.) from the vertical line, then under the influence of force F, one of the components of weight P, it will oscillate. The other component, directed along the thread, is not taken into account, because is balanced by the tension of the thread. At small displacement angles, then the x coordinate can be measured in the horizontal direction. From Fig. 7.3 it is clear that the weight component perpendicular to the thread is equal to

The minus sign on the right side means that the force F is directed towards decreasing angle α. Taking into account the smallness of the angle α

To derive the law of motion of mathematical and physical pendulums, we use the basic equation of the dynamics of rotational motion

Moment of force relative to point O: , and moment of inertia: M=FL. Moment of inertia J in this case Angular acceleration:

Taking these values ​​into account, we have:

His decision ,

As we can see, the period of oscillation of a mathematical pendulum depends on its length and the acceleration of gravity and does not depend on the amplitude of the oscillations.

Damped oscillations.

All real oscillatory systems are dissipative. The energy of mechanical vibrations of such a system is gradually spent on work against frictional forces, therefore free vibrations always fade - their amplitude gradually decreases. In many cases, when there is no dry friction, as a first approximation we can assume that at low speeds of movement the forces causing attenuation of mechanical vibrations are proportional to the speed. These forces, regardless of their origin, are called resistance forces.

Let's rewrite this equation as follows:

and denote:

where represents the frequency with which free oscillations of the system would occur in the absence of environmental resistance, i.e. at r = 0. This frequency is called the natural frequency of oscillation of the system; β is the attenuation coefficient. Then

We will look for a solution to equation (7.19) in the form where U is some function of t.

Let us differentiate this expression twice with respect to time t and, substituting the values ​​of the first and second derivatives into equation (7.19), we obtain

The solution to this equation significantly depends on the sign of the coefficient at U. Let us consider the case when this coefficient is positive. Let us introduce the notation then With a real ω, the solution to this equation, as we know, is the function

Thus, in the case of low resistance of the medium, the solution to equation (7.19) will be the function

The graph of this function is shown in Fig. 7.8. The dotted lines show the limits within which the displacement of the oscillating point lies. The quantity is called the natural cyclic frequency of oscillations of the dissipative system. Damped oscillations are non-periodic oscillations, because they never repeat, for example, the maximum values ​​of displacement, speed and acceleration. The quantity is usually called the period of damped oscillations, or more correctly, the conditional period of damped oscillations,

The natural logarithm of the ratio of displacement amplitudes following each other through a time interval equal to the period T is called the logarithmic attenuation decrement.

Let us denote by τ the period of time during which the amplitude of the oscillations decreases by e times. Then

Consequently, the attenuation coefficient is a physical quantity inverse to the time period τ during which the amplitude decreases by a factor of e. The quantity τ is called the relaxation time.

Let N be the number of oscillations after which the amplitude decreases by a factor of e, Then

Therefore, the logarithmic damping decrement δ is physical quantity, reciprocal to the number of oscillations N, after which the amplitude decreases by e times

Forced vibrations.

In the case of forced oscillations, the system oscillates under the influence of an external (forcing) force, and due to the work of this force, the energy losses of the system are periodically compensated. The frequency of forced oscillations (forcing frequency) depends on the frequency of change of the external force. Let us determine the amplitude of forced oscillations of a body of mass m, considering the oscillations undamped due to a constantly acting force.

Let this force change with time according to the law where is the amplitude of the driving force. Restoring force and resistance force Then Newton's second law can be written in the following form.

If the trajectory of a point’s movement is known, then the dependence of the path traversed by the point on the elapsed time interval gives Full description this movement. We have seen that for uniform motion such a dependence can be given in the form of formula (9.2). The relationship between and for individual points in time can also be specified in the form of a table containing the corresponding values ​​of the time period and the distance traveled. Let us be given that the speed of some uniform motion is 2 m/s. Formula (9.2) in this case has the form . Let's make a table of the path and time of such movement:

The dependence of one quantity on another is often convenient to depict not with formulas or tables, but with graphs, which more clearly show the picture of changes in variable quantities and can facilitate calculations. Let us construct a graph of the distance traveled versus time for the movement in question. To do this, take two mutually perpendicular straight lines - coordinate axes; We will call one of them (the abscissa axis) the time axis, and the other (the ordinate axis) the path axis. Let's choose scales for depicting time intervals and paths and take the point of intersection of the axes as the initial moment and as the starting point on the trajectory. Let us plot on the axes the values ​​of time and distance traveled for the movement under consideration (Fig. 18). To “bind” the values ​​of the distance traveled to moments in time, we draw perpendiculars to the axes from the corresponding points on the axes (for example, points 3 s and 6 m). The point of intersection of the perpendiculars corresponds simultaneously to both quantities: path and moment, and in this way the “binding” is achieved. The same construction can be performed for any other points in time and corresponding paths, obtaining for each such pair of time - path values ​​one point on the graph. In Fig. 18 such a construction is made, replacing both rows of the table with one row of points. If such a construction were carried out for all points in time, then instead of individual points, a solid line would be obtained (also shown in the figure). This line is called a path versus time graph or, in short, a path graph.

Rice. 18. Graph of the path of uniform motion at a speed of 2 m/s

Rice. 19. For exercise 12.1

In our case, the path graph turned out to be a straight line. It can be shown that the graph of the path of uniform motion is always a straight line; and vice versa: if the graph of the path versus time is a straight line, then the movement is uniform.

Repeating the construction for a different speed, we find that the graph points for higher speeds lie higher than the corresponding graph points for lower speeds (Fig. 20). Thus, the greater the speed of uniform motion, the steeper the rectilinear path graph, i.e., the greater the angle it makes with the time axis.

Rice. 20. Graphs of the path of uniform movements with speeds of 2 and 3 m/s

Rice. 21. Graph of the same movement as in Fig. 18, drawn on a different scale

The slope of the graph depends, of course, not only on the numerical value of the speed, but also on the choice of time and length scales. For example, the graph shown in Fig. 21 gives the path versus time for the same movement as the graph in Fig. 18, although it has a different slope. From here it is clear that it is possible to compare movements by the slope of graphs only if they are drawn on the same scale.

Using path graphs, you can easily solve various motion problems. For example in Fig. 18 dashed lines show the constructions necessary to solve the following problems for a given movement: a) find the path traveled in 3.5 s; b) find the time it takes to travel 9 m. In the figure, the answers are found graphically (dashed lines): a) 7 m; b) 4.5 s.

On graphs describing uniform rectilinear motion, the coordinate of the moving point can be plotted along the ordinate axis instead of the path. This description opens up great possibilities. In particular, it makes it possible to distinguish the direction of movement relative to the axis. In addition, by taking the origin of time to be zero, it is possible to show the movement of the point at earlier moments of time, which should be considered negative.

Rice. 22. Graphs of movements with the same speed, but at different initial positions of the moving point

Rice. 23. Graphs of several movements with negative speeds

For example, in Fig. 22 straight line I is a graph of motion occurring at a positive speed of 4 m/s (i.e. in the direction of the axis), and at the initial moment the moving point was at a point with coordinate m. For comparison, the same figure shows a graph of the motion that occurs with the same speed, but at which at the initial moment the moving point is at the point with the coordinate (line II). Straight. III corresponds to the case when at the moment the moving point was at a point with coordinate m. Finally, straight line IV describes the movement in the case when the moving point had a coordinate at the moment c.

We see that the slopes of all four graphs are the same: the slope depends only on the speed of the moving point, and not on its initial position. When changing the initial position, the entire graph is simply transferred parallel to itself along the axis up or down at the appropriate distance.

Graphs of movements occurring at negative speeds (i.e. in the direction opposite to the direction of the axis) are shown in Fig. 23. They are straight, inclined downwards. For such movements, the coordinate of the point decreases over time., had coordinates

Path graphs can also be constructed for cases in which a body moves uniformly for a certain period of time, then moves uniformly but at a different speed for another period of time, then changes speed again, etc. For example, in Fig. 26 shows a motion graph in which the body moved during the first hour at a speed of 20 km/h, during the second hour at a speed of 40 km/h and during the third hour at a speed of 15 km/h.

Exercise: 12.8. Construct a graph of the path for movement in which, over successive hourly intervals, the body had speeds of 10, -5, 0, 2, -7 km/h. What is the total displacement of the body?