B11. Using the graphs of the dependence of the coordinates of bodies on time (Fig. 1), determine for each body:

a) initial coordinate;

b) coordinate after 4 s;

c) velocity projection;

d) equation of coordinates (equation of motion);

e) when will the coordinate be equal to 20 m?

Solution

a) Determine the initial coordinate for each body.

Graphic method. Using the graph, we find the coordinate values ​​of the points of intersection of the graphs with the axis 0x(in Fig. 2a these points are highlighted):

x 01 = 30 m; x 02 = 10 m; x 03 = –10 m.

b) Determine the coordinate for each body after 4 s.

Graphic method. Using the graph, we find the coordinate values ​​of the points of intersection of the graphs with the perpendicular drawn to the axis 0t at point t = 4 s (in Fig. 2 b these points are highlighted): x 1 (4 s) = 0; x 2 (4 s) = 10 m; x 3 (4 s) ≈ 20 m.

Analytical method. Create an equation of motion and use it to determine the value of the coordinate at t= 4 s (see point d).

c) Determine the velocity projection for each body.

Graphic method. Projection of velocity \(~\upsilon_x = \tan \alpha = \frac(\Delta x)(\Delta t) = \frac(x_2 - x_1)(t_2-t_1)\) , where α is the angle of inclination of the graph to the axis 0t; Δ t = t 2 – t 1 – arbitrary period of time; Δ υ = υ 2 – υ 1 – speed interval corresponding to the time interval Δ t = t 2 – t 1 .

For graph 1: let t 2 = 4 s, t 1 = 0 then x 2 = 0, x 1 = 30 m and υ 1x= (0 - 30 m)/(4 s - 0) = –7.5 m/s (Fig. 3 a).

For graph 2: let t 2 = 6 s, t 1 = 0 then x 2 = 10 m, x 1 = 10 m and υ 2x= (10 m - 10 m)/(6 s - 0) = 0 (Fig. 3 b).

For graph 3: let t 2 = 5 s, t 1 = 0 then x 2 = 30 m, x 1 = –10 m and υ 3x= (30 - (-10 m))/(5 s - 0) = 8 m/s (Fig. 3 c).

Analytical method. Let us write the coordinate equation for uniform rectilinear motion in general form x = x 0 + υ x · t. Using the values ​​of the initial coordinate (see point a) and the coordinates at t = 4 s (see point b), we find the value of the velocity projection\[~\upsilon_x = \frac(x - x_0)(t)\] .

d) Determine the coordinate equation for each body.

The coordinate equation for uniform rectilinear motion in the general form "x = x 0 + υ x · t .

For schedule 1: because x 01 = 30 m, υ 1x= –7.5 m/s, then x 1 = 30 – 7,5t. Let's check point b: x 1 (4 s) = 30 – 7.5 4 = 0, which corresponds to the answer.

For schedule 2: because x 02 = 10 m, υ 2x= 0, then x 2 = 10. Let's check point b: x 2 (4 s) = 10 (m), which corresponds to the answer.

For schedule 3: because x 03 = –10 m, υ 3x= 8 m/s, then x 3 = –10 + 8t. Let's check point b: x 3 (4 s) = –10 + 8 4 = 22 (m), which approximately corresponds to the answer.

e) Determine when the coordinate of the body will be 20 m?

Graphic method. Using the graph, we find the time values ​​of the points of intersection of the graphs with the perpendicular drawn to the axis 0x at the point x= 20 m (in Fig. 4 these points are highlighted): t 1 (20 m) ≈ 1.5 s; t 3 (20 m) ≈ 3.5 s.

Graph 2 is parallel to the perpendicular, therefore, the coordinate of body 2 will never be equal to 20 m.

Analytical method. Write down the coordinate equation for each body and find at what value of time t, the coordinate becomes equal to 20 m.

« Physics - 10th grade"

How does uniform motion differ from uniformly accelerated motion?
How does the path graph for uniformly accelerated motion differ from the path graph for uniform motion?
What is the projection of a vector onto any axis?

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

The velocity projection is numerically equal to the tangent of the angle of inclination of the straight line x(t) to the abscissa axis. Moreover, the higher 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 for rectilinear uniformly accelerated motion of a point. They are straight lines parallel to the abscissa 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 opposite direction 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 graphs of this dependence for these three cases. In this case, the initial speed of the point is the same. Let's analyze this graph.

Projection of acceleration From the graph it is clear that the greater the acceleration of a 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 the acceleration.

Over the same period of time, with different accelerations, the speed changes to different values.

With a positive value of the acceleration projection for the same period of time, 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 to the same value as in case 1, but the speed decreases.

For cases 1 and 3, the graphs of the velocity modulus versus time will be the same (Fig. 1.35).

Using the graph of speed versus time (Figure 1.36), we find the change in the coordinates of the point. This change is numerically equal to the area of ​​the shaded trapezoid, in this case the change in coordinate in 4 s 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 equal to 18 m. In this case, the displacement module is equal to the path traveled by the point, or the change in its coordinate, i.e. 16 m .

If the movement is uniformly slow, then the point during the selected time interval can stop and begin to move in the direction opposite to the initial one. Figure 1.37 shows the dependence of the velocity projection on time for such a movement. We see that at a 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 a plus sign, and the area under the t axis, 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 displacement of the point in this case is also equal to 6 m - the modulus of change in coordinates. However, the path traveled by this point is equal to 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 curve of coordinate versus time - x(t) - is a parabola.

If the point moves at a speed, the graph of which versus time is shown in Figure 1.36, then the branches of the parabola are directed upward, since a x > 0 (Figure 1.39). From this graph we can determine the coordinate of the point, as well as the speed at any time. So, at a 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 angle of inclination α 1, which is numerically equal to the initial speed, i.e. 2 m/s.

To determine the speed at point B, 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 graph of the path versus time is the same parabola, but drawn from the origin (Fig. 1.40). We see that the path continuously increases over time, the movement occurs in one direction.

If the point moves at a speed, the graph of the projection of which versus time is shown in Figure 1.37, then the branches of the parabola are directed downward, 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 moment of time t = 2 s, the tangent of the angle of inclination becomes negative, and its module increases, this means that the point moves in the direction opposite to the initial one, while the module of the movement speed increases.

The displacement module is equal to the module 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 the distance traveled by a point versus time, shown in Figure 1.42, differs from the graph of displacement versus time (see Figure 1.41).

Regardless of the direction of the speed, the path traveled by the point continuously increases.

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

In the case of x 0 = 0 and x > 0 and υ x > υ 0x, the graph of the coordinate versus 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 less the distance that the point must travel for the speed to increase by the same amount as when moving with less acceleration.

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

Accelerated movement.

Using graphs of the velocity projection versus time, you can determine the coordinate and acceleration projection of a point at any time for any type of movement.

Let the projection of the point's velocity 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 movement is uniform with a constant speed υ Dx. According to 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 during time t 1 is numerically equal to the area of ​​the curvilinear trapezoid OABt 1, during time t 2 - the area OACt 2, etc. As we can see from the graph of the velocity projection versus time, we can determine the change in the coordinate of the body over any period of time.

From a graph of coordinates versus time, you can determine the value of speed at any point in time by calculating the tangent of the tangent to the curve at the point corresponding to a given point in 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 velocity projection becomes negative, the direction of motion of the point changes to the opposite.

If the graph of the velocity projection versus time is known, 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. From the graph of coordinates versus time, one can determine one of the most important kinematic characteristics of movement - speed. In addition, using these graphs, you can determine the type of movement along the selected axis: uniform, with constant acceleration, or movement with variable acceleration.

As according to the coordinate dependence graph

from time x = x(t) build a graph

path versus time s = s(t)?

Let us note the following features of the graph s = s(t):

1) schedule s = s(t) always starts from the origin, since at the initial moment the distance traveled is always zero;

2) schedule s = s(t) always does not decrease: it either increases if the body is moving, or does not change if the body is standing;

3) function s = s(t) cannot take a negative value.

From the above it follows that the graph X = X (t) coincides with the schedule s = s(t) only if X(0) = 0 and x(t) does not decrease all the time, i.e. the body moves only in a positive direction or stands still.

Here are some examples of plotting charts: s = s(t) according to these graphs X = X(t).

Example 4.2. On schedule X = = X(t) in Fig. 4.4, A build a graph s = s(t).

Schedule X = X(t) increases, but begins not at the origin, but at the point (0, X 0). To get the schedule s = s(t) it is necessary to omit the graph X = X(t) on x 0 down (Fig. 4.4, b).

Example 4.3. On schedule X = X(t) in Fig. 4.5, A build a graph s = s(t).

In this case X(0) = 0, but the body moves in the negative direction of the axis X. In this case it is true s(t) = |x(t)|, and to plot s = s(t) just display the graph X = X(t) mirrored onto the upper half-plane (Fig. 4.5, b).

Rice. 4.5

Example 4.4. On schedule X = X(t) in Fig. 4.6, A build a graph s = s(t).

First let's lower the graph X = X(t) on X 0 down to X(0) = 0, as we did in example 4.2, and then straight line 2 (Fig. 4.6, b) will be mirrored onto the upper half-plane, as we did in Example 4.3.

Rice. 4.6

Example 4.5. On schedule X = X(t) in Fig. 4.7, A build a graph s = s(t).

Rice. 4.7

Schedule X = X(t) consists of two sections: in the first section X(t) increases, and in the second section it decreases, i.e. the body moves in the negative direction of the axis X. Therefore, to plot a graph s = s(t) first part of the graph X = X(t) we leave unchanged, and mirror the second part relative to the straight line passing through the turning point (2t, 2 X 0) parallel to the axis t(Fig. 4.7,b).

STOP! Solve for yourself: C2 (a, b, c).

Statement. Let the dependence graph be given υ x(t), X(t 1) = x 0 (Fig. 4.8). Area values ​​above the graph s+ and below the chart s– , expressed taking into account scales in units of length, are known. Then the path traveled during the period of time [ t 1 , t 2 ], is equal to:

s = s – + s + . (4.2)

Coordinate at time t 2 is equal to:

X(t 2) = x 0 – s – + s + . (4.3)

Problem 4.2. According to the graph of coordinates versus time (Fig. 4.9, A) build dependency graphs υ x = υ x(t) And υ = υ (t).

Solution. Let's consider a period of time. On this interval D X= = 1 m, D t= 1 s, hence = 1 m/s, υ = = |υ x| = 1 m/s.

Let's consider a period of time. On this interval D X= 0, which means υ x = υ = 0.

Let's consider a period of time. On this interval D X= (–2) – 1 = = –3 m, D t= 1 s, which means = –3 m/s, υ = |υ x| = 3 m/s.

Let's consider a period of time. On this interval D X= 0, therefore, υ x = υ = 0.

The graphs are shown in Fig. 4.9, b and 4.9, V.

STOP! Solve for yourself: Q3 (a, b, c).

Problem 4.3. According to the dependency graph υ x = υ x(t) (Fig. 4.10) find the values ​​of the traveled path and coordinates at times 1s, 2 s, 3 s, 4 s, 5 s, if X(0) = 2.0 m.

Solution.

1. Consider a period of time. In this interval υ x(t) decreased from 1 m/s to 0, i.e. the body moved along the axis X slowly and in the moment t= 1 s stopped. The distance traveled is equal to the area under the graph on the section: m. Coordinate at moment t= 1 s is equal to X(1) = X(0) + s 01 = 2.0 m + 0.5 m = 2.5 m.

2. Consider a period of time. In this interval υ x decreased from 0 to –1 m/s, i.e. the body accelerates from rest in the direction opposite to the direction of the axis X. The path traveled during this period of time is equal to the area above the graph υ x = υ x(t) on the interval: m. Therefore, the total path traveled by the body at the moment t= 2 s, equal s(2) = s(1) + s 12 = 0.5 m + 0.5 m = 1.0 m. Coordinate at moment t= 1 s is equal to X(2) = X(1) – s 12 = 2.5 m – 0.5 m = 2.0 m.

3. Consider a period of time. During this interval the body moves uniformly in the negative direction of the axis X with ground speed υ = 1 m/s. The distance traveled is s 23 = (1 m/s)´ ´(1 s) = 1.0 m. Therefore, the path traveled to the moment t= 3 s, equal s(3) = s(2) + s 23 = 1.0 m + 1.0 m = 2.0 m.

The coordinate during this period of time decreased by the amount of the distance traveled, since the body moved in the opposite direction: X(3) = X(2) – s 23 = 2.0 m – 1.0 m = 1.0 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:

V cp = v

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:

X = x 0 + vt

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:

X = x 0 - vt

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 greater the angle α, the faster the body moves, that is, the greater its speed (the greater 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:

Tg α = v

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

Tg α 1 > tg α 2

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

Tg 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

X = x 0

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