How to build a chart of projection acceleration from time. Equal asked movement: Formulas, examples

§ 14. Tracks of the path and speed

Determination of the road schedule

In physics and mathematics, three methods of filing communication information between different values \u200b\u200bare used: a) as a formula, for example, S \u003d V ∙ T; b) in the form of a table; c) in the form of a graph (drawing).

The dependence of the speed of time V (T) is a speed chart is depicted using two mutually perpendicular axes. Along the horizontal axis we will postpone the time, and vertical - speed (Fig. 14.1). It is necessary to consider the scale in advance so that the drawing is not too big or too small. The end of the axis indicates the letter, which is the designation numerically equal to the area of \u200b\u200bthe shaded rectangle ABCD magnitude, which is postponed on it. Near the letters indicate a unit of measurement of this value. For example, t, С, and near the axis of the velocity V (T), MES) indicate near the time axis. Select the scale and apply divisions for each axis.

Fig. 14.1. A chart of body velocity, evenly moving at a speed of 3 m / s. The path passed by the body from the 2nd to 6 seconds,

Image of uniform movement with table and graphs

Consider the uniform body movement at a speed of 3 m / s, that is, the numerical value of the speed will be constant for the entire time of movement. Abbreviated this is written as follows: V \u003d const (constant, that is, a constant value). In our example, it is equal to three: v \u003d 3. You already know that information on the dependence of one value from another can be submitted as a table (array, as they say in computer science):

It can be seen from the table that at all these points in time, the speed is 3 m / s. Let the scale of the time axis of 2 cl. \u003d 1 C, and 2 CL speed axes. \u003d 1 m / s. A graph of speed dependence on time (abbreviated: Speed \u200b\u200bgraph) are shown in Figure 14.1.

Using the speed graph, you can find a way that the body passes for a certain time interval. To do this, you need to compare two facts: on the one hand, the path can be found, multiplying the speed for a while, and on the other - the product of speed for a while, as can be seen from the figure - this is the area of \u200b\u200bthe rectangle with the sides T and V.

For example, from the second to the sixth second, the body moved for four seconds and passed 3 m / s ∙ 4 C \u003d 12 m. It is the area of \u200b\u200bthe rectangle ABSD, the length of which is 4 C (segment AD along the time axis) and height 3 m / s ( Cut AB along the vertical). The square, however, is somewhat unusual, since it is not measured in m 2, and in the city consequently, the area under the speed chart is numerically equal to the distance traveled.

Schedule

Schedule S (T) can be depicted using the formula S \u003d V ∙ T, that is, in our case, when the speed is 3 m / s: S \u003d 3 ∙ T. Build a table:

Along the horizontal axis, the time is postponed (T, C), and along the vertical - the path. Near the axis of the path we write: S, M (Fig. 14.2).

Determination of speed on schedule

Showing now in one figure, two graphics that will correspond to movements with speeds 3 m / s (straight 2) and 6 m / s (direct 1) (Fig. 14.3). It can be seen that the larger the body speed, the steeper the line of points of the graph.

There is a feedback: having a motion schedule, you need to determine the speed and write the path equation (Fig. 14.3). Consider straight 2. From the beginning of the movement and until the time T \u003d 2, the path S \u003d 6 m passed. Consequently, its speed: v \u003d \u003d 3. The choice of another time interval will not change anything, for example, at the time T \u003d 4, the path passed by the body from the beginning of the movement is S \u003d 12 m. The ratio is again equal to 3 m / s. But it should be, since the body moves at a constant speed. Therefore, it would be the easiest way to choose the time interval 1 s, because the path passed by the body in one second is numerically equal to speed. The path passed by the first body (schedule 1) for 1 C is 6 m, that is, the speed of the first body is 6 m / s. The corresponding dependences of the distance from time in these two bodies will be:

s 1 \u003d 6 ∙ T and S 2 \u003d 3 ∙ T.

Fig. 14.2. Schedule path. The remaining points, except for the six, specified in the table set in the task that the movement of the utmost time was uniform

Fig. 14.3. Schedule path in case of different speeds

Let's summarize

Physics use three ways to submit information: graphic, analytical (by formulas) and table (array). The third method is more adapted to solve on the computer.

The path is numerically equal to the area under speed chart.

The steeper S (T) graph, the greater the speed.

Creative tasks

14.1. Draw graphs of speed and path when the body speed is evenly increased, or decreases.

Exercise 14.

1. How do the path on the speed schedule?

2. Is it possible to write a formula to depend on the time of time, having a S (T) schedule?

3. Or will the end of the slope of the path of the path change if the scale on the axes should be doubled?

4. Why is the schedule of a uniform movement depicted direct?

5. Which body (Fig. 14.4) has the highest speed?

6. Name the three ways to provide information on body movement, as well as (in your opinion) their advantages and disadvantages.

7. How can I define a speed schedule?

8. a) What is the difference between the schedules of the path for bodies moving at different speeds? b) What is in common?

9. According to the schedule (Fig. 14.1), find the way passed by the body from the beginning of the first to the end of the third second.

10. What path the body passed (Fig. 14.2) for: a) two seconds; b) Four seconds? c) Specify where the third second of the movement begins, and where it ends.

11. Position in charts of speed and path movement at speed a) 4 m / s; b) 2 m / s.

12. Record the dependence formula from time to the movements depicted in Fig. 14.3.

13. a) Find the velocities of the tel by schedules (Fig. 14.4); b) Record the corresponding path equations and speed. c) Build graphics of the speed of these bodies.

14. Build graphs of the path and velocity for bodies, the movements of which are given by the equations: S 1 \u003d 5 ∙ T and S 2 \u003d 6 ∙ T. What are the speed of tel?

15. According to graphics (Fig. 14.5), identify: a) body velocities; b) the paths passed by them in the first 5 seconds. c) write down the path equation and build the corresponding graphs for all three movements.

16. Instruct the schedule of the path for the movement of the first body relative to the second (Fig. 14.3).

Graphic representation
uniform rectilinear movement

Speed \u200b\u200bgraph Shows how the body speed changes over time. In a straight uniform movement, speed over time does not change. Therefore, the speed chart of such a movement is a straight, parallel axis of the abscissa (time axis). In fig. 6 shows the speed charts of two tel. Chart 1 refers to the case when the body moves in the positive direction of the axis of the axis (the projection of the body velocity is positive), the chart 2 is to the case when the body moves against the positive direction of the axis about x (the speed projection is negative). According to the speed graph, you can determine the body traveled (if the body does not change the directions of its movement, the path length is equal to the module of its movement).

2. Schedule of body coordinate from time which otherwise is called movement schedule

In fig. The graphs of the movement of two tel are depicted. The body whose graph is direct 1, moves in the positive direction of the axis about x, and the body, whose timeline is direct 2, moves the opposite to the positive axis direction of the x.

3. Schedule

The schedule is a straight line. This direct passes through the origin of the coordinates (Fig.). The angle of inclination of this direct to the abscissa axis is the greater, the greater the body speed. In fig. Depicted graphs 1 and 2 ways of two bodies. From this pattern it is seen that in the same time T body 1, having a greater speed than the body 2, passes the greater path (s 1\u003e s 2).

A rectilinear equilibrium movement is the easiest form of an uneven movement, in which the body moves along a straight line, and its speed is valid for any equal periods of time.

Equal asked movement is a constant acceleration movement.

The acceleration of the body in its equilibrium movement is the value equal to the ratio of the speed change by the period of time, for which this change occurred:

→ →
→ V - V 0
a \u003d ---
T.

Calculate the acceleration of the body moving straightforwardly and equally, it is possible using an equation into which the projections of the acceleration and speed vectors are included:

v x - v 0x
a x \u003d ---
T.

Acceleration unit in C: 1 m / s 2.

The speed of the straight equal to the movement.

v x \u003d v 0x + a x t

where v 0x is the projection of the initial speed, and the X is the projection of acceleration, T is time.

→
If the body rested at the initial moment, then v 0 \u003d 0. For this case, the formula takes the following form:

Move with an equal rectilinear movement S x \u003d V 0 x T + A x T ^ 2/2

Coordinate with RUPD X \u003d X 0 + V 0 x T + A x T ^ 2/2

Graphic representation
equal asked straight line

Speed \u200b\u200bgraph

Speed \u200b\u200bgraph is a straight line. If the body moves at some initial speed, this direct crosses the ordinate axis at the point V 0x. If the initial body speed is zero, the speed schedule passes through the origin of the coordinates. The graphs of the speed of the straight equilibrium movement are shown in Fig. . In this figure, graphics 1 and 2 correspond to the movement with a positive projection of the acceleration on the axis on x (speed increases), and the chart 3 corresponds to the movement with a negative acceleration projection (speed decreases). Chart 2 corresponds to the movement without the initial speed, and the graphs 1 and 3 - movement with the initial velocity V OX. The angle of inclination A graph to the axis of the abscissa depends on the acceleration of the body movement. By speed, you can determine the path passed by the body over the time t.

The path passed in a straight equilibrium movement at the initial speed is numerically equal to the area of \u200b\u200bthe trapezoid bounded by the speed graph, the axes of coordinate and the ordinate, corresponding to the value of the body velocity at time t.

Schedule of the dependence of the coordinates from time (traffic schedule)

Let the body move equally in the positive direction of the selected coordinate system. Then the equation of body movement is:

x \u003d x 0 + v 0x · t + a x T 2/2. (one)

The expression (1) corresponds to a functional dependence of y \u003d ah 2 + BX + C (square three-shred) corresponds to a math course of mathematics. In the case we consider
a \u003d | a x | / 2, b \u003d | v 0x |, c \u003d | x 0 |.

Schedule

In an equilibrium rectilinear movement, the dependence of the path of time is expressed by formulas

s \u003d V 0 T + AT 2/2, S \u003d AT 2/2 (at v 0 \u003d 0).

As can be seen from these formulas, this dependence is quadratic. From both formulas, it also follows that S \u003d 0 at T \u003d 0. Therefore, the schedule of the straight equilibrium movement is a parabola branch. In fig. Showing a path of the path at v 0 \u003d 0.

Schedule acceleration

Acceleration schedule - dependence of the projection of acceleration from time:

straightforward uniform movement. Graphic representation uniform straightforward movement. 4. Instant speed. Addition...

Lesson Subject: "Material point. System reference" Objectives: to give an idea about kinematics

Lesson

Definition uniform straightforward movement. - What is called speed uniform movement? - Name the speed unit movement In ... Projections of the velocity vector movement U (O. 2. Graphic representation movement. - At the point with ...

Tasks in physics are easy!

Do not forgetthat it is always necessary to solve the tasks in the SI system!

And now to the tasks!

Elementary tasks from the course of school physics in kinematics.

The problem of drawing up the description of the movement and drawing up the equation of movement on a given schedule of motion

Given: Schedule of body movement

To find:
1. Move the movement
2. Make a body movement equation.

Speed \u200b\u200bvector projection We define on schedule by selecting any time convenient for consideration.
Here it is convenient to take T \u003d 4C

Make up Body Movement Equation:

Write on the formula of the equation of a straight line uniform movement.

We substitute a found coefficient V x (do not forget about the minus!).
The initial coordinate of the body (x o) corresponds to the beginning of the graph, then x o \u003d 3

Make up Body Movement Description:

It is advisable to make a drawing, it will help not make a mistake!
Do not forget that all physical quantities have units of measurement, they must be specified!

The body moves straight and evenly from the starting point x O \u003d 3M at a speed of 0.75 m / s opposite to the direction of the X axis.

The task of determining the place and time of the meeting of two moving bodies (with a straight-line uniform movement)

The movement of the bodies is set by the equations of motion for each body.

Given:
1. The first body movement equation
2. The equation of the Second Body Movement

To find:
1. The coordinate of the meeting place
2. Moment time (after the start of the movement) when a meeting occurs

According to the given movement equations, we build traffic schedules for each body in one coordinate system.

Intersection pointtwo motion charts determines:

1. On the axis T - the meeting time (how much time after the start of the movement there will be a meeting)
2. On the X axis - the coordinate of the meeting place (relative to the start of the coordinates)

As a result:

Two bodies will meet at a point with a coordinate -1.75 m after 1.25 seconds after the start of the movement.

To check the answers received by the graphically, you can solve the system of equations from two specified
Traffic equations:

Everything was true!

For those who for some reason forgotHow to build a graph of a straight uniform movement:

The timing schedule is a linear dependence (straight), built by two points.
Select the two any convenient to calculate the value of T 1 and T 2.
For these values, T calculate the corresponding coordinate values \u200b\u200bof x 1 and x 2.
We lay 2 points with coordinates (T 1, x 1) and (t 2, x 2) and connect their direct - the graph is ready!

Tasks for drawing up the description of the movement of the body and the construction of graphs of movement according to a given equation of rectilinear uniform movement

Task 1.

Given: Body Movement Equation

To find:

The specified equation compare with the formula and determine the coefficients.
Do not forget to make a drawing to once again pay attention to the direction of the velocity vector.

Task 2.

Given: Body Movement Equation

To find:
1. Move the movement
2. Build a traffic schedule

Task 3.

Given: Body Movement Equation

To find:
1. Move the movement
2. Build a traffic schedule

Task 4.

Given: Body Movement Equation

To find:
1. Move the movement
2. Build a traffic schedule

Movement Description:

The body is in a state of rest at the point with the x \u003d 4m coordinate (the state of rest is a special case of movement when the body speed is zero).

Task 5.

Given:
The initial coordinate of the moving point xo \u003d -3 m
VX \u003d -2 M / s velocity vector projection

To find:
1. Write the movement equation
2. Build a traffic schedule
3. Show on drawing velocity and moving vectors
4. Find the point coordinate after 10 seconds after the start of the movement

Equal asked movement is a movement with acceleration, the vector of which does not change in the module and direction. Examples of this movement: a bicycle that rolls from a slide; Stone abandoned at an angle to the horizon.

Consider the last case in more detail. At any point of the trajectory on the stone there is an acceleration of the free fall G →, which does not change in magnitude and is always directed in one direction.

The movement of the body thrown at an angle to the horizon can be represented as the sum of movements relative to the vertical and horizontal axes.

Along the axis X, the movement is uniform and straightforward, and along the Y axis is equivalent and straightforward. We will consider projections of velocity vectors and acceleration on the axis.

The formula for speed with an equalized movement:

Here v 0 is the initial body velocity, A \u003d C O n s T - acceleration.

We show on the chart that with an equilibrium movement, the dependence V (T) has the kind of a straight line.

Acceleration can be determined by the corner of the speed chart. The figure above the acceleration module is equal to the side of the side of the ABC triangle.

a \u003d v - v 0 t \u003d b c a c

The greater the angle β, the greater the slope (steepness) of the graph with respect to the time axis. Accordingly, the greater the acceleration of the body.

For the first graph: V 0 \u003d - 2 m C; A \u003d 0, 5 m C 2.

For the second graph: V 0 \u003d 3 m C; A \u003d - 1 3 m C 2.

For this schedule, you can also calculate the movement of the body during T. How to do it?

We highlight on the chart a small period of time Δ t. We assume that it is so small that the movement during Δ t can be considered a uniform movement at a rate equal to the velocity of the body in the middle of the gap Δ t. Then, the movement Δ S for the time Δ T will be Δ s \u003d V Δ t.

We break all the time t to infinitely small gaps Δ t. Moving S for time T is equal to the area of \u200b\u200bthe trapetion O D E F.

s \u003d O D + E F 2 O F \u003d V 0 + V 2 T \u003d 2 V 0 + (V - V 0) 2 t.

We know that V - V 0 \u003d a t, so the final formula for moving the body will take the form:

s \u003d V 0 T + A T 2 2

In order to find the coordinate of finding the body at the moment, you need to add movement to the initial coordinate of the body. The change in the coordinate with an equalized movement expresses the law of an equivalent movement.

The law of an equidalist movement

The law of an equidalist movement

y \u003d Y 0 + V 0 T + A T 2 2.

Another common task that occurs when analyzing an equilibrium movement - finding the movement with the specified values \u200b\u200bof the initial and final speeds and acceleration.

Excluding from the ties recorded above and solving them, we obtain:

s \u003d V 2 - V 0 2 2 a.

At known initial speed, acceleration and movement, you can find the final body velocity:

v \u003d V 0 2 + 2 A s.

At v 0 \u003d 0 s \u003d V 2 2 A and V \u003d 2 A s

Important!

The values \u200b\u200bV, V 0, A, Y 0, S, which are included in the expressions are algebraic values. Depending on the nature of the movement and the direction of the coordinate axes in the context of a specific task, they can take both positive and negative values.

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