How to determine what movement is on the chart. Uniform motion in a straight line

Lesson Topic: "Material point. Reference system" Objectives: to give an idea of ​​kinematics

Definition uniform straightforward movement. - What is called speed? uniform movement? - Name the unit of speed movement in... projection of the velocity vector versus time movement U (O. 2. Graphic performance movement. - At point C...

Let's show how you can find the path traveled by a body using a graph of speed versus time.

Let's start from the very beginning simple case– uniform movement. Figure 6.1 shows a graph of v(t) – speed versus time. It represents a segment of a straight line parallel to the base of time, since with uniform motion the speed is constant.

The figure enclosed under this graph is a rectangle (it is shaded in the figure). Its area is numerically equal to the product of speed v and time of movement t. On the other hand, the product vt is equal to the path l traversed by the body. So, with uniform motion

the path is numerically equal to the area of ​​the figure enclosed under the graph of speed versus time.

Let us now show that this remarkable property It also has uneven movement.

Let, for example, the graph of speed versus time look like the curve shown in Figure 6.2.

Let us mentally divide the entire time of movement into such small intervals that during each of them the movement of the body can be considered almost uniform (this division is shown by dashed lines in Figure 6.2).

Then the path traveled during each such interval is numerically equal to the area of ​​the figure under the corresponding lump of the graph. Therefore, the entire path is equal to the area of ​​the figures contained under the entire graph. (The technique we used is the basis integral calculus, the basics of which you will study in the course “Beginnings of Mathematical Analysis”.)

2. Path and displacement during rectilinear uniformly accelerated motion

Let us now apply the method described above for finding the path to rectilinear uniformly accelerated motion.

The initial speed of the body is zero

Let's direct the x axis in the direction of body acceleration. Then a x = a, v x = v. Hence,

Figure 6.3 shows a graph of v(t).

1. Using Figure 6.3, prove that in case of rectilinear uniformly accelerated motion without initial speed, the path l is expressed in terms of the acceleration module a and the time of movement t by the formula

l = at 2 /2. (2)

Main conclusion:

in rectilinear uniformly accelerated motion without initial speed, the distance traveled by the body is proportional to the square of the time of movement.

In this way, uniformly accelerated motion differs significantly from uniform motion.

Figure 6.4 shows graphs of the path versus time for two bodies, one of which moves uniformly, and the other uniformly accelerates without an initial speed.

2. Look at Figure 6.4 and answer the questions.
a) What color is the graph for a body moving with uniform acceleration?
b) What is the acceleration of this body?
c) What are the speeds of the bodies at the moment when they have covered the same path?
d) At what point in time are the velocities of the bodies equal?

3. Having set off, the car covered a distance of 20 m in the first 4 s. Consider the car’s motion to be rectilinear and uniformly accelerated. Without calculating the acceleration of the car, determine how far the car will travel:
a) in 8 s? b) in 16 s? c) in 2 s?

Let us now find the dependence of the projection of displacement s x on time. IN in this case the projection of acceleration onto the x axis is positive, so s x = l, a x = a. Thus, from formula (2) it follows:

s x = a x t 2 /2. (3)

Formulas (2) and (3) are very similar, which sometimes leads to errors in solving simple tasks. The fact is that the displacement projection value can be negative. This will happen if the x axis is directed opposite to the displacement: then s x< 0. А путь отрицательным быть не может!

4. Figure 6.5 shows graphs of travel time and displacement projection for a certain body. What color is the displacement projection graph?

The initial speed of the body is not zero

Let us recall that in this case the dependence of the velocity projection on time is expressed by the formula

v x = v 0x + a x t, (4)

where v 0x is the projection of the initial velocity onto the x axis.

We will further consider the case when v 0x > 0, a x > 0. In this case, we can again take advantage of the fact that the path is numerically equal to the area of ​​the figure under the graph of speed versus time. (Consider other combinations of signs for the projection of initial velocity and acceleration yourself: the result will be the same general formula (5).

Figure 6.6 shows a graph of v x (t) for v 0x > 0, a x > 0.

5. Using Figure 6.6, prove that in case of rectilinear uniformly accelerated motion with an initial speed, the projection of displacement

s x = v 0x + a x t 2 /2. (5)

This formula allows you to find the dependence of the x coordinate of the body on time. Let us recall (see formula (6), § 2) that the coordinate x of a body is related to the projection of its displacement s x by the relation

s x = x – x 0 ,

where x 0 is the initial coordinate of the body. Hence,

x = x 0 + s x , (6)

From formulas (5), (6) we obtain:

x = x 0 + v 0x t + a x t 2 /2. (7)

6. The dependence of the coordinate on time for a certain body moving along the x axis is expressed in SI units by the formula x = 6 – 5t + t 2.
a) What is the initial coordinate of the body?
b) What is the projection of the initial velocity onto the x-axis?
c) What is the projection of acceleration on the x-axis?
d) Draw a graph of the x coordinate versus time.
e) Draw a graph of the projected velocity versus time.
f) At what moment is the speed of the body equal to zero?
g) Will the body return to the starting point? If so, at what point(s) in time?
h) Will the body pass through the origin? If so, at what point(s) in time?
i) Draw a graph of the displacement projection versus time.
j) Draw a graph of the distance versus time.

3. Relationship between path and speed

When solving problems, relationships between path, acceleration and speed (initial v 0, final v or both) are often used. Let us derive these relations. Let's start with movement without an initial speed. From formula (1) we obtain for the time of movement:

Let's substitute this expression into formula (2) for the path:

l = at 2 /2 = a/2(v/a) 2 = v 2 /2a. (9)

Main conclusion:

in rectilinear uniformly accelerated motion without initial speed, the distance traveled by the body is proportional to the square of the final speed.

7. Having started, the car picked up a speed of 10 m/s over a distance of 40 m. Consider the car’s motion to be linear and uniformly accelerated. Without calculating the acceleration of the car, determine how far from the start of movement the car traveled when its speed was equal to: a) 20 m/s? b) 40 m/s? c) 5 m/s?

Relationship (9) can also be obtained by remembering that the path is numerically equal to the area of ​​the figure enclosed under the graph of speed versus time (Fig. 6.7).

This consideration will help you easily cope with the next task.

8. Using Figure 6.8, prove that when braking with constant acceleration, the body travels the distance l t = v 0 2 /2a to a complete stop, where v 0 is the initial speed of the body, a is the acceleration modulus.

In case of braking vehicle(car, train) the distance traveled to a complete stop is called the braking distance. Please note: the braking distance at the initial speed v 0 and the distance traveled during acceleration from standstill to speed v 0 with the same acceleration a are the same.

9. During emergency braking on dry asphalt, the acceleration of the car is equal in absolute value to 5 m/s 2 . What is the braking distance of a car at initial speed: a) 60 km/h (maximum permitted speed in the city); b) 120 km/h? Find the braking distance at the indicated speeds during icy conditions, when the acceleration modulus is 2 m/s 2 . Compare the braking distances you found with the length of the classroom.

10. Using Figure 6.9 and the formula expressing the area of ​​a trapezoid through its height and half the sum of the bases, prove that for rectilinear uniformly accelerated motion:
a) l = (v 2 – v 0 2)/2a, if the speed of the body increases;
b) l = (v 0 2 – v 2)/2a, if the speed of the body decreases.

11. Prove that the projections of displacement, initial and final velocity, as well as acceleration are related by the relation

s x = (v x 2 – v 0x 2)/2ax (10)

12. A car on a path of 200 m accelerated from a speed of 10 m/s to 30 m/s.
a) How fast was the car moving?
b) How long did it take the car to travel the indicated distance?
c) What is it equal to average speed car?

Additional questions and tasks

13. The last car is uncoupled from a moving train, after which the train moves uniformly, and the car moves with constant acceleration until it comes to a complete stop.
a) Draw on one drawing graphs of speed versus time for a train and a carriage.
b) How many times the distance traveled by the car to the stop? less way traveled by the train in the same time?

14. Having left the station, the train drove uniformly accelerated for some time, then for 1 minute – uniformly at a speed of 60 km/h, and then again uniformly accelerated until it stopped at the next station. The acceleration modules during acceleration and braking were different. The train covered the distance between stations in 2 minutes.
a) Draw a schematic graph of the projection of the speed of the train as a function of time.
b) Using this graph, find the distance between the stations.
c) How far would the train travel if it accelerated on the first section of the route and slowed down on the second? What would be its maximum speed?

15. A body moves uniformly accelerated along the x axis. At the initial moment it was at the origin of coordinates, and the projection of its speed was equal to 8 m/s. After 2 s, the coordinate of the body became 12 m.
a) What is the projection of the acceleration of the body?
b) Plot a graph of v x (t).
c) Write a formula expressing the dependence x(t) in SI units.
d) Will the speed of the body be zero? If yes, at what point in time?
e) Will the body visit the point with coordinate 12 m a second time? If yes, at what point in time?
f) Will the body return to the starting point? If yes, then at what point in time, and what will be the distance traveled?

16. After the push, the ball rolls up inclined plane, after which it returns to the starting point. At a distance b from starting point the ball visited twice at intervals t 1 and t 2 after the push. The ball moved up and down along the inclined plane with the same magnitude of acceleration.
a) Direct the x-axis up along the inclined plane, select the origin at the point initial position ball and write a formula expressing the dependence x(t), which includes the modulus of the initial velocity of the ball v0 and the modulus of acceleration of the ball a.
b) Using this formula and the fact that the ball was at a distance b from the starting point at times t 1 and t 2, create a system of two equations with two unknowns v 0 and a.
c) Having solved this system of equations, express v 0 and a in terms of b, t 1 and t 2.
d) Express the entire path l traveled by the ball in terms of b, t 1 and t 2.
e) Find numeric values v 0 , a and l at b = 30 cm, t 1 = 1 s, t 2 = 2 s.
f) Plot graphs of v x (t), s x (t), l(t).
g) Using the graph of sx(t), determine the moment when the ball’s modulus of displacement was maximum.

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. If a body moves along the O x axis, the dependence of its speed on time is expressed by the formulas
v x =v 0x +a x t and v x =at (for v 0x = 0).

From these formulas it is clear that the dependence of v x on t is linear, therefore, the speed graph is a straight line. If the body moves with a certain initial speed, this straight line intersects the ordinate axis at point v 0x. If the initial velocity of the body is zero, the velocity graph passes through the origin.

The velocity graphs of rectilinear uniformly accelerated motion are shown in Fig. 9. In this figure, graphs 1 and 2 correspond to movement with a positive projection of acceleration on the O x axis (speed increases), and graph 3 corresponds to movement with a negative projection of acceleration (speed decreases). Graph 2 corresponds to movement without an initial speed, and graphs 1 and 3 to movement with an initial speed v ox. The angle of inclination a of the graph to the abscissa axis depends on the acceleration of the body. As can be seen from Fig. 10 and formulas (1.10),

tg=(v x -v 0x)/t=a x .

Using velocity graphs, you can determine the distance traveled by a body during a period of time t. To do this, we determine the area of ​​the trapezoid and the triangle shaded in Fig. eleven.

On the selected scale, one base of the trapezoid is numerically equal to the modulus of the projection of the initial velocity v 0x of the body, and its other base is equal to the modulus of the projection of its velocity v x at time t. The height of the trapezoid is numerically equal to the duration of the time interval t. Area of ​​trapezoid

S=(v 0x +v x)/2t.

Using formula (1.11), after transformations we find that the area of ​​the trapezoid

S=v 0x t+at 2 /2.

the path traveled in rectilinear uniformly accelerated motion with an initial speed is numerically equal to the area of ​​the trapezoid limited by the velocity graph, coordinate axes and ordinate corresponding to the value of the body’s speed at time t.

On the chosen scale, the height of the triangle (Fig. 11, b) is numerically equal to the modulus of the projection of the velocity v x of the body at time t, and the base of the triangle is numerically equal to the duration of the time interval t. Area of ​​the triangle S=v x t/2.

Using formula 1.12, after transformations we find that the area of ​​the triangle

Right part The last equality is an expression that determines the path traveled by the body. Hence, the path covered in rectilinear uniformly accelerated motion without initial speed is numerically equal to the area of ​​the triangle, limited by schedule speed, x-axis and ordinate corresponding to the speed of the body at time t.

3.1. Equally alternating motion in a straight line.

3.1.1. Uniform motion in a straight line- movement in a straight line with acceleration constant in magnitude and direction:

3.1.2. Acceleration()- physical vector quantity, showing how much the speed will change in 1 s.

IN vector form:

where is the initial speed of the body, is the speed of the body at the moment of time t.

In projection onto the axis Ox:

where is the projection of the initial velocity onto the axis Ox, - projection of the body velocity onto the axis Ox at a point in time t.

The signs of the projections depend on the direction of the vectors and the axis Ox.

3.1.3. Projection graph of acceleration versus time.

With uniformly alternating motion, the acceleration is constant, therefore it will appear as straight lines parallel to the time axis (see figure):

3.1.4. Speed ​​during uniform motion.

In vector form:

In projection onto the axis Ox:

For uniformly accelerated motion:

For uniform slow motion:

3.1.5. Projection graph of speed versus time.

The graph of the projection of speed versus time is a straight line.

Direction of movement: if the graph (or part of it) is above the time axis, then the body is moving in the positive direction of the axis Ox.

Acceleration value: the greater the tangent of the angle of inclination (the steeper it goes up or down), the greater the acceleration module; where is the change in speed over time

Intersection with the time axis: if the graph intersects the time axis, then before the intersection point the body slowed down (uniformly slow motion), and after the intersection point it began to accelerate in the opposite side(uniformly accelerated motion).

3.1.6. Geometric meaning area under the graph in axes

Area under the graph when on the axis Oy the speed is delayed, and on the axis Ox- time is the path traveled by the body.

In Fig. 3.5 shows the case of uniformly accelerated motion. The path in this case will be equal to the area of ​​the trapezoid: (3.9)

3.1.7. Formulas for calculating path

All formulas presented in the table work only when the direction of movement is maintained, that is, until the straight line intersects with the time axis on the graph of the velocity projection versus time.

If the intersection has occurred, then the movement is easier to divide into two stages:

before crossing (braking):

After the intersection (acceleration, movement in reverse side)

In the formulas above - the time from the beginning of movement to the intersection with the time axis (time before stopping), - the path that the body has traveled from the beginning of movement to the intersection with the time axis, - the time elapsed from the moment of intersection of the time axis to at this moment t, - the path that the body has traveled in the opposite direction during the time elapsed from the moment of crossing the time axis to this moment t, - the module of the displacement vector for the entire time of movement, L- the path traveled by the body during the entire movement.

3.1.8. Movement in the th second.

Over time the body will go the way:

During this time the body will travel the following distance:

Then during the th interval the body will travel the following distance:

Any period of time can be taken as an interval. Most often with.

Then in 1 second the body travels the following distance:

In 2 seconds:

In 3 seconds:

If we look carefully, we will see that, etc.

Thus, we arrive at the formula:

In words: the paths traversed by a body over successive periods of time are related to each other as a series of odd numbers, and this does not depend on the acceleration with which the body moves. We emphasize that this relation is valid for

3.1.9. Equation of body coordinates for uniform motion

Coordinate equation

The signs of the initial velocity and acceleration projections depend on relative position corresponding vectors and axis Ox.

To solve problems, it is necessary to add to the equation the equation for changing the velocity projection onto the axis:

3.2. Graphs of kinematic quantities for rectilinear motion

3.3. Free fall body

By free fall we mean the following physical model:

1) The fall occurs under the influence of gravity:

2) There is no air resistance (in problems they sometimes write “neglect air resistance”);

3) All bodies, regardless of mass, fall with the same acceleration (sometimes they add “regardless of the shape of the body,” but we consider the movement only material point, so body shape is no longer taken into account);

4) The acceleration of gravity is directed strictly downwards and is equal on the surface of the Earth (in problems we often assume for convenience of calculations);

3.3.1. Equations of motion in projection onto the axis Oy

Unlike movement along a horizontal straight line, when not all tasks involve a change in direction of movement, when free fall it is best to immediately use the equations written in projections onto the axis Oy.

Body coordinate equation:

Velocity projection equation:

As a rule, in problems it is convenient to select the axis Oy in the following way:

Axis Oy directed vertically upward;

The origin coincides with the level of the Earth or the lowest point of the trajectory.

With this choice, the equations and will be rewritten in the following form:

3.4. Movement in a plane Oxy.

We considered the motion of a body with acceleration along a straight line. However, the uniformly variable motion is not limited to this. For example, a body thrown at an angle to the horizontal. In such problems, it is necessary to take into account movement along two axes at once:

Or in vector form:

And changing the projection of speed on both axes:

3.5. Application of the concept of derivative and integral

We will not give here detailed definition derivative and integral. To solve problems we need only a small set of formulas.

Derivative:

Where A, B and that is, constant values.

Integral:

Now let's see how the concept of derivative and integral applies to physical quantities. In mathematics, the derivative is denoted by """, in physics, the derivative with respect to time is denoted by "∙" above the function.

Speed:

that is, the speed is a derivative of the radius vector.

For velocity projection:

Acceleration:

that is, acceleration is a derivative of speed.

For acceleration projection:

Thus, if the law of motion is known, then we can easily find both the speed and acceleration of the body.

Now let's use the concept of integral.

Speed:

that is, the speed can be found as the time integral of the acceleration.

Radius vector:

that is, the radius vector can be found by taking the integral of the velocity function.

Thus, if the function is known, we can easily find both the speed and the law of motion of the body.

The constants in the formulas are determined from initial conditions- values ​​and at time

3.6. Velocity triangle and displacement triangle

3.6.1. Speed ​​triangle

In vector form at constant acceleration the law of speed change has the form (3.5):

This formula means that a vector is equal to the vector sum of vectors and the vector sum can always be depicted in a figure (see figure).

In each problem, depending on the conditions, the velocity triangle will have its own form. This representation allows the use of geometric considerations in the solution, which often simplifies the solution of the problem.

3.6.2. Triangle of movements

In vector form, the law of motion with constant acceleration has the form:

When solving a problem, you can choose the reference system in the most convenient way, therefore, without losing generality, we can choose the reference system in such a way that, that is, we place the origin of the coordinate system at the point where the body is located at the initial moment. Then

that is, the vector is equal to the vector sum of the vectors and Let us depict it in the figure (see figure).

As in the previous case, depending on the conditions, the displacement triangle will have its own shape. This representation allows the use of geometric considerations in the solution, which often simplifies the solution of the problem.