How to determine the movement of the body according to the schedule. Determination of displacement and path according to the schedule. Graphs of uniformly accelerated motion

§ 14. GRAPHS OF PATH AND SPEED

Determination of the path according to the speed graph

In physics and mathematics, three ways of presenting information about the relationship between different quantities are used: a) in the form of a formula, for example, s = v ∙ t; b) in the form of a table; c) in the form of a graph (figure).

Velocity versus time v(t) - the velocity graph is depicted using two mutually perpendicular axes. We will plot time along the horizontal axis, and speed along the vertical axis (Fig. 14.1). It is necessary to think over the scale in advance so that the drawing is not too large or too small. At the end of the axis, a letter is indicated, which is a designation numerically equal to the area of ​​\u200b\u200bthe shaded rectangle abcd of the value that is deposited on it. Near the letter indicate the unit of measurement of this value. For example, near the time axis indicate t, s, and near the velocity axis v (t), months. Choose a scale and put divisions on each axis.

Rice. 14.1. Graph of the speed of a body moving uniformly at a speed of 3 m/s. The path traveled by the body from the 2nd to the 6th second,

Image of uniform movement by table and graphs

Consider the uniform motion of a body with a speed of 3 m/s, that is, the numerical value of the speed will be constant throughout the entire time of motion. In short, this is written as follows: v = const (constant, that is, a constant value). In our example, it is equal to three: v = 3 . You already know that information about the dependence of one quantity on another can be presented in the form of a table (an array, as they say in computer science):

It can be seen from the table that at all indicated times the speed is 3 m/s. Let the scale of the time axis be 2 cells. \u003d 1 s, and the velocity axis is 2 cells. = 1 m/sec. A graph of speed versus time (abbreviated to say: speed graph) is shown in Figure 14.1.

Using the speed graph, you can find the path that the body travels in a certain time interval. To do this, we need to compare two facts: on the one hand, the path can be found by multiplying the speed by the time, and on the other hand, the product of the speed by the time, as can be seen from the figure, is the area of ​​a rectangle with 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 s = 12 m. segment ab along the vertical). The area, however, is somewhat unusual, since it is measured not in m 2, but in g. Therefore, the area under the speed graph is numerically equal to the distance traveled.

Path chart

The graph of the path s(t) can be depicted using the formula s = v ∙ t, that is, in our case, when the speed is 3 m/s: s = 3 ∙ t. Let's build a table:

Time (t, s) is again plotted along the horizontal axis, and the path along the vertical axis. Near the axis of the path we write: s, m (Fig. 14.2).

Determination of speed according to the path schedule

Let us now depict two graphs in one figure, which will correspond to movements with speeds of 3 m / s (straight line 2) and 6 m / s (straight line 1) (Fig. 14.3). It can be seen that the greater the speed of the body, the steeper the line of points on the graph.

There is also an inverse problem: having a motion schedule, you need to determine the speed and write down the equation of the path (Fig. 14.3). Consider straight line 2. From the beginning of the movement to the moment of time t = 2 s, the body has traveled a distance s = 6 m. Therefore, its speed is: v = = 3 . Choosing another time interval will not change anything, for example, at the moment t = 4 s, the path traveled by the body from the beginning of the movement is s = 12 m. The ratio is again equal to 3 m/sec. But this is how it should be, since the body is moving at a constant speed. Therefore, it would be easiest to choose a time interval of 1 s, because the path traveled by the body in one second is numerically equal to the speed. The path traveled by the first body (graph 1) in 1 s is 6 m, that is, the speed of the first body is 6 m/s. The corresponding path-time dependencies in these two bodies will be:

s 1 \u003d 6 ∙ t and s 2 \u003d 3 ∙ t.

Rice. 14.2. Path schedule. The remaining points, except for the six indicated in the table, were set in the task that the movement was uniform throughout the entire time

Rice. 14.3. Path graph in case of different speeds

Summing up

In physics, three methods of presenting information are used: graphical, analytical (by formulas) and table (array). The third method is more suitable for solving on a computer.

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

The steeper the graph s(t), the greater the speed.

Creative tasks

14.1. Draw graphs of speed and path when the speed of the body increases or decreases uniformly.

Exercise 14

1. How is the path determined on the speed graph?

2. Is it possible to write a formula for the dependence of the path on time, having a graph of s (t)?

3. Or will the slope of the path graph change if the scale on the axes is halved?

4. Why is the graph of the path of uniform motion depicted as a straight line?

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

6. What are the three ways of presenting information about the movement of the body, and (in your opinion) their advantages and disadvantages.

7. How can you determine the path according to the speed graph?

8. a) What is the difference between path graphs for bodies moving at different speeds? b) What do they have in common?

9. According to the graph (Fig. 14.1), find the path traveled by the body from the beginning of the first to the end of the third second.

10. What is the distance traveled by the body (Fig. 14.2) in: a) two seconds; b) four seconds? c) Indicate where the third second of the movement begins and where it ends.

11. Draw on the speed and path graphs the movement at a speed of a) 4 m/s; b) 2 m/sec.

12. Write down the formula for the dependence of the path on time for the movements shown in fig. 14.3.

13. a) Find the velocities of the bodies according to the graphs (Fig. 14.4); b) write down the corresponding equations of path and velocity. c) Plot the velocity graphs of these bodies.

14. Build graphs of the path and speed for bodies whose movements are given by the equations: s 1 = 5 ∙ t and s 2 = 6 ∙ t. What are the speeds of the bodies?

15. According to the graphs (Fig. 14.5), determine: a) the speed of the body; b) the paths they traveled in the first 5 seconds. c) Write down the path equation and plot the corresponding graphs for all three movements.

16. Draw a path graph for the movement of the first body relative to the second (Fig. 14.3).

Physics problems - it's easy!

Don't forget that problems must always be solved in the SI system!

And now to the tasks!

Elementary tasks from the course of school physics in kinematics.

The task of compiling a description of the movement and compiling an equation of movement according to a given schedule of movement

Given: body movement chart

Find:
1. write a description of the movement
2. draw up an equation of motion of the body.

We determine the projection of the velocity vector according to the graph, choosing any time interval convenient for consideration.
Here it is convenient to take t=4c

Compiling body motion equation:

We write down the formula for the equation of rectilinear uniform motion.

We substitute the found coefficient V x into it (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

Compiling body movement description:

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

The body moves in a straight line and uniformly from the starting point X o = 3 m 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 rectilinear uniform motion)

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

Given:
1. equation of motion of the first body
2. equation of motion of the second body

Find:
1. meeting point coordinate
2. moment in time (after the start of movement) when the bodies meet

According to the given equations of motion, we build graphs of motion for each body in one coordinate system.

Intersection point two motion schedules defines:

1. on the t axis - the time of the meeting (how long after the start of the movement the meeting will occur)
2. on the X axis - the coordinate of the meeting place (relative to the origin)

As a result:

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

To check the answers obtained graphically, you can solve a system of equations from two given
equations of motion:

Everything was right!

For those who somehow forgot how to plot a rectilinear uniform motion graph:

The motion graph is a linear relationship (straight line), built on two points.
We choose any two values ​​t 1 and t 2 convenient for ease of calculation.
For these values ​​of t, we calculate the corresponding values ​​of the coordinates X 1 and X 2 .
Set aside 2 points with coordinates (t 1 , X 1) and (t 2 , X 2) and connect them with a straight line - the graph is ready!

Tasks for compiling a description of the motion of a body and plotting motion graphs according to a given equation of rectilinear uniform motion

Task 1

Given: body motion equation

Find:

We compare the given equation 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 motion equation

Find:
1. write a description of the movement
2. build a motion schedule

Task 3

Given: body motion equation

Find:
1. write a description of the movement
2. build a motion schedule

Task 4

Given: body motion equation

Find:
1. write a description of the movement
2. build a motion schedule

Movement description:

The body is at rest at a point with coordinate X=4m (the state of rest is a special case of motion when the speed of the body is zero).

Task 5

Given:
initial coordinate of the moving point xo=-3 m
velocity vector projection Vx=-2 m/s

Find:
1. write down the equation of motion
2. build a motion schedule
3. show the velocity and displacement vectors on the drawing
4. find the coordinate of the point 10 seconds after the start of the movement

« Physics - Grade 10 "

What is the difference between uniform motion and uniformly accelerated motion?
What is the difference between a path graph for uniformly accelerated motion and a path graph for uniform motion?
What is called the projection of a vector on any axis?

In the case of uniform rectilinear motion, you can determine the speed according to the graph of coordinates versus time.

The velocity projection is numerically equal to the tangent of the slope of the straight line x(t) to the x-axis. In this case, the greater the speed, the greater the angle of inclination.

Rectilinear uniformly accelerated motion.

Figure 1.33 shows graphs of the projection of acceleration versus time for three different values ​​of acceleration in a rectilinear uniformly accelerated motion of a point. They are straight lines parallel to the x-axis: a x = const. Graphs 1 and 2 correspond to movement when the acceleration vector is directed along the OX axis, graph 3 - when the acceleration vector is directed in the direction opposite to the OX axis.

With uniformly accelerated motion, the velocity projection depends linearly on time: υ x = υ 0x + a x t. Figure 1.34 shows the graphs of this dependence for these three cases. In this case, the initial speed of the point is the same. Let's analyze this chart.

Acceleration projection It can be seen from the graph that the greater the acceleration of the point, the greater the angle of inclination of the straight line to the t axis and, accordingly, the greater the tangent of the angle of inclination, which determines the value of acceleration.

For the same period of time at different accelerations, the speed changes by different values.

With a positive value of the acceleration projection for the same time interval, the velocity projection in case 2 increases 2 times faster than in case 1. With a negative value of the acceleration projection on the OX axis, the velocity projection modulo changes by the same value as in case 1, but the speed is decreasing.

For cases 1 and 3, the graphs of the dependence of the velocity modulus on time will coincide (Fig. 1.35).

Using the speed versus time graph (Figure 1.36), we find the change in the coordinate of the point. This change is numerically equal to the area of ​​the shaded trapezoid, in this case, the change in coordinate for 4 with Δx = 16 m.

We found a change in coordinates. If you need to find the coordinate of a point, then you need to add its initial value to the found number. Let at the initial moment of time x 0 = 2 m, then the value of the coordinate of the point at a given moment of time, equal to 4 s, is 18 m. In this case, the displacement module is equal to the path traveled by the point, or the change in its coordinates, i.e. 16 m .

If the movement is uniformly slowed down, then the point during the selected time interval can stop and start moving in the opposite direction to the initial one. Figure 1.37 shows the projection of velocity versus time for such a motion. We see that at the moment of time equal to 2 s, the direction of the velocity changes. The change in coordinate will be numerically equal to the algebraic sum of the areas of the shaded triangles.

Calculating these areas, we see that the change in coordinate is -6 m, which means that in the direction opposite to the OX axis, the point has traveled a greater distance than in the direction of this axis.

Square above we take the t axis with the plus sign, and the area under axis t, where the velocity projection is negative, with a minus sign.

If at the initial moment of time the speed of a certain point was equal to 2 m / s, then its coordinate at the moment of time equal to 6 s is equal to -4 m. The modulus of moving a point in this case is also equal to 6 m - the modulus of changing the coordinate. However, the path traveled by this point is 10 m, the sum of the areas of the shaded triangles shown in Figure 1.38.

Let's plot the dependence of the x-coordinate of a point on time. According to one of the formulas (1.14), the time dependence curve - x(t) - is a parabola.

If the point moves at a speed, the time dependence of which is shown in Figure 1.36, then the branches of the parabola are directed upwards, since a x\u003e 0 (Figure 1.39). From this graph, we can determine the coordinate of the point, as well as the speed at any given time. So, at the moment of time equal to 4 s, the coordinate of the point is 18 m.

For the initial moment of time, drawing a tangent to the curve at point A, we determine the tangent of the slope α 1, which is numerically equal to the initial speed, i.e. 2 m / s.

To determine the speed at point B, we draw a tangent to the parabola at this point and determine the tangent of the angle α 2 . It is equal to 6, therefore, the speed is 6 m/s.

The path versus time graph is the same parabola, but drawn from the origin (Fig. 1.40). We see that the path is continuously increasing with time, the movement is in one direction.

If the point moves at a speed whose projection versus time graph is shown in Figure 1.37, then the branches of the parabola are directed downwards, since a x< 0 (рис. 1.41). При этом моменту времени, равному 2 с, соответствует вершина параболы. Касательная в точке В параллельна оси t, угол наклона касательной к этой оси равен нулю, и скорость также равна нулю. До этого момента времени тангенс угла наклона касательной уменьшался, но был положителен, движение точки происходило в направлении оси ОХ.

Starting from the time t = 2 s, the tangent of the slope angle becomes negative, and its module increases, which means that the point moves in the opposite direction to the initial one, while the module of the movement speed increases.

The displacement modulus is equal to the modulus of the difference between the coordinates of the point at the final and initial moments of time and is equal to 6 m.

The graph of dependence of the path traveled by the point on time, shown in Figure 1.42, differs from the graph of the dependence of displacement on time (see Figure 1.41).

No matter how the speed is directed, the path traveled by the point continuously increases.

Let us derive the dependence of the point coordinate on the velocity projection. Velocity υx = υ 0x + a x t, hence

In the case of x 0 \u003d 0 and x\u003e 0 and υ x\u003e υ 0x, the graph of the dependence of the coordinate on the speed is a parabola (Fig. 1.43).

In this case, the greater the acceleration, the less steep the branch of the parabola will be. This is easy to explain, since the greater the acceleration, the smaller the distance that the point must travel in order for the speed to increase by the same amount as when moving with less acceleration.

In case a x< 0 и υ 0x >0 speed projection will decrease. Let us rewrite equation (1.17) in the form where a = |a x |. The graph of this dependence is a parabola with branches pointing downwards (Fig. 1.44).

Accelerated movement.

According to the graphs of dependence of the projection of velocity on time, it is possible to determine the coordinate and projection of the acceleration of a point at any moment in time for any type of movement.

Let the projection of the speed of a point depend on time as shown in Figure 1.45. It is obvious that in the time interval from 0 to t 3 the movement of the point along the X axis occurred with variable acceleration. Starting from the moment of time equal to t 3 , the motion is uniform with a constant speed υ Dx . From the graph, we see that the acceleration with which the point moved was continuously decreasing (compare the angle of inclination of the tangent at points B and C).

The change in the x coordinate of a point over time t 1 is numerically equal to the area of ​​the curvilinear trapezoid OABt 1, over time t 2 - the area OACt 2, etc. As we can see from the graph of the dependence of the velocity projection on time, you can determine the change in body coordinates for any period of time.

According to the graph of the dependence of the coordinate on time, one can determine the value of the speed at any moment of time by calculating the tangent of the slope of the tangent to the curve at the point corresponding to the given moment of time. From figure 1.46 it follows that at time t 1 the velocity projection is positive. In the time interval from t 2 to t 3 the speed is zero, the body is motionless. At time t 4 the speed is also zero (the tangent to the curve at point D is parallel to the x-axis). Then the projection of the velocity becomes negative, the direction of movement of the point changes to the opposite.

If you know the graph of the dependence of the projection of the velocity on time, you can determine the acceleration of the point, and also, knowing the initial position, determine the coordinate of the body at any time, i.e. solve the main problem of kinematics. One of the most important kinematic characteristics of movement, speed, can be determined from the graph of the dependence of coordinates on time. In addition, according to the specified graphs, you can determine the type of movement along the selected axis: uniform, with constant acceleration, or movement with variable acceleration.

Graphical representation
uniform rectilinear motion

Speed ​​Graph shows how the speed of the body changes over time. In a rectilinear uniform motion, the speed does not change over time. Therefore, the graph of the speed of such movement is a straight line parallel to the x-axis (time axis). On fig. 6 shows graphs of the speed of two bodies. Graph 1 refers to the case when the body moves in the positive direction of the O x axis (the projection of the body's velocity is positive), graph 2 - to the case when the body moves against the positive direction of the O x axis (the projection of the velocity is negative). According to the speed graph, you can determine the distance traveled by the body (If the body does not change the direction of its movement, the length of the path is equal to the modulus of its movement).

2.Graph of body coordinates versus time which is otherwise called traffic schedule

On fig. graphs of motion of two bodies are shown. The body whose graph is line 1 moves in the positive direction of the O x axis, and the body whose motion graph is line 2 moves in the opposite direction to the positive direction of the O x axis.

3.Path chart

The graph is a straight line. This straight line passes through the origin (Fig.). The angle of inclination of this straight line to the abscissa axis is the greater, the greater the speed of the body. On fig. graphs 1 and 2 of the path of two bodies are shown. From this figure it can be seen that for the same time t body 1, which has a greater speed than body 2, travels a longer distance (s 1 > s 2).

Rectilinear uniformly accelerated motion is the simplest type of non-uniform motion, in which the body moves along a straight line, and its speed changes in the same way for any equal time intervals.

Uniformly accelerated motion is motion with constant acceleration.

The acceleration of a body during its uniformly accelerated motion is a value equal to the ratio of the change in speed to the time interval during which this change occurred:

→ →
→ v – v0
a = ---
t

You can calculate the acceleration of a body moving in a straight line and uniformly accelerated using an equation that includes the projections of the acceleration and velocity vectors:

vx – v0x
x = ---
t

Unit of acceleration in SI: 1 m/s 2 .

The speed of rectilinear uniformly accelerated motion.

v x = v 0x + a x t

where v 0x is the projection of the initial velocity, a x is the projection of acceleration, t is the time.

→
If at the initial moment the body was at rest, then v 0 = 0. For this case, the formula takes the following form:

Movement with uniform rectilinear motion S x \u003d V 0 x t + a x t ^ 2/2

RAPD coordinate x=x 0 + V 0 x t + a x t^2/2

Graphical representation
uniformly accelerated rectilinear motion

Speed ​​Graph

The speed graph is a straight line. If the body moves with some initial speed, this straight line intersects the y-axis at the point v 0x . If the initial velocity of the body is zero, the velocity graph passes through the origin. Graphs of the speed of rectilinear uniformly accelerated motion are shown in fig. . In this figure, graphs 1 and 2 correspond to movement with a positive acceleration projection on the O x axis (speed increases), and graph 3 corresponds to movement with a negative acceleration projection (speed decreases). Graph 2 corresponds to movement without initial speed, and graphs 1 and 3 correspond to movement with initial speed v ox . The angle of inclination a of the graph to the x-axis depends on the acceleration of the body. According to the speed graphs, you can determine the path traveled by the body for a period of time t.

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

Graph of coordinates versus time (motion graph)

Let the body move uniformly accelerated in the positive direction O x of the chosen coordinate system. Then the equation of motion of the body has the form:

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

Expression (1) corresponds to the functional dependence known from the course of mathematics y \u003d ax 2 + bx + c (square trinomial). In our case
a=|a x |/2, b=|v 0x |, c=|x 0 |.

Path chart

In a uniformly accelerated rectilinear motion, the dependence of the path on time is expressed by the formulas

s=v 0 t+at 2/2, s= at 2/2 (for v 0 =0).

As can be seen from these formulas, this dependence is quadratic. It also follows from both formulas that s = 0 at t = 0. Therefore, the graph of the path of a uniformly accelerated rectilinear motion is a branch of a parabola. On fig. the path graph is shown for v 0 =0.

Acceleration Graph

Acceleration graph - dependence of the projection of acceleration on time:

rectilinear uniform movements. Graphic performance uniform rectilinear movements. 4. Instant speed. Addition...

Mechanical movement is represented graphically. The dependence of physical quantities is expressed using functions. designate

Graphs of uniform motion

Time dependence of acceleration. Since the acceleration is equal to zero during uniform motion, the dependence a(t) is a straight line that lies on the time axis.

Dependence of speed on time. The speed does not change with time, the graph v(t) is a straight line parallel to the time axis.

The numerical value of the displacement (path) is the area of ​​the rectangle under the speed graph.

Path versus time. Graph s(t) - sloping line.

The rule for determining the speed according to the schedule s(t): The tangent of the slope of the graph to the time axis is equal to the speed of movement.

Graphs of uniformly accelerated motion

Dependence of acceleration on time. Acceleration does not change with time, has a constant value, graph a(t) is a straight line parallel to the time axis.

Speed ​​versus time. With uniform motion, the path changes, according to a linear relationship. in coordinates. The graph is a sloping line.

The rule for determining the path according to the schedule v(t): The path of the body is the area of ​​the triangle (or trapezoid) under the velocity graph.

The rule for determining the acceleration according to the schedule v(t): The acceleration of the body is the tangent of the slope of the graph to the time axis. If the body slows down, the acceleration is negative, the angle of the graph is obtuse, so we find the tangent of the adjacent angle.

Path versus time. With uniformly accelerated movement, the path changes, according to