Traffic schedules

Kvant 1970, No. 6 p. 39-45 Tag 5YCA

Rough automatic draft, low quality: expect mistakes and awkward wording. It will be replaced by a better translation. Original →

Mechanical motion is very convenient to depict on graphs. Their simplicity and clarity make it possible to immediately assess the nature of the movement at its various stages and solve some problems without resorting to calculations.

On university entrance exams, graphs are often used in additional questions. Therefore, you need to have a good understanding of graphs and be able to analyze them.

Speed graph

Let us first consider linear motion. The straight line along which the body moves is used as the coordinate axis $x$ . On it we will select the starting point $O$ and determine the position of the body by coordinate $x$ , counting from this zero point in one direction with the plane sign, and in the other - with the minus sign. If the direction of motion coincides with the positive direction of the $x$ axis, the body speed is considered positive, otherwise - negative (Fig. 1).

Let’s draw coordinate axes and plot the time $t$ along one of them, and the speed $V$ (Fig. 2) along the second. Any line in these coordinates will express the graphical dependence of speed on time. What can you learn from such a graph?

Prime $I$ , parallel to axis $t$ ,

characterizes movement at a constant speed $V_0$ . The fact that the speed is positive means that the direction of movement coincides with the positive direction of the $x$ axis. The path traveled by the body from moment $t_1$ to moment $t_2$ is equal in size to the area of the upper rectangle: $S = V_0(t_2 - t_1)$ .

Prima $II$ on figure 2 also corresponds to uniform motion, but only in the opposite direction (velocity is negative). The path traveled by this body in time $(t_2 - t_1)$ is a positive number equal in size to the area of the lower rectangle: $S_1 = V_1(t_2 - t_1)$ .

From the graph you can also find out the change in the body coordinate $\Delta x = x_2 - x_1 = V(t_2 - t_1)$ . In both cases, it is equal in abscillot value to the area of the figure under the graph ( $S$ or $S_1$ ), but in the first case it is positive, and in the second it is negative, since in the second case $x_2 < x_1$ and $V_1 < 0$ (Fig. 3).

With uniformly accelerated motion, the velocity graph is also straight, only moving at a certain angle to the time axis $t: V = V_0 + c t$ (Fig. 4). Here $V_0$ is the speed of the body at $t = 0$ . The tangent of the angle of inclination of the straight line $I$ to the axis $t$ is equal in magnitude to the acceleration

$a = \frac{\Delta V}{\Delta t} = tg \alpha.$

The higher the line rises in a circle, the greater the acceleration.

The motion that corresponds to straight line $II$ on figure 4 is motion with negative acceleration. Can we say that this is “slow motion”? It is possible, but only until moment $t_2$ , while the velocity and acceleration had different signs, that is, they were directed in opposite directions. At $t > t_2$ the abscilottic value of the velocity increases, since the directions of the velocity and acceleration coincide: they are both negative.

A similar movement is, for example, the flight of a stone thrown upward: its speed first decreases, then reaches zero and, finally, changing direction, returns. In this case, the acceleration is constant and equal to minus $g$ (we neglect air resistance).

The path traveled by the body from moment $t = 0$ to moment $t = t_1$ is equal in magnitude to the area of the trapezoid $A$ , and the path traveled from moment $t_1$ to moment $t_3$ is equal to the sum of the areas of triangles $B$ and $C$ . The change in the coordinate of the body during time $t_3 - t_1$ is equal to the difference in the areas of these triangles, since the body first moved in one direction ( $V > 0$ ), and then in the other ( $V < 0$ ).

If the acceleration of a body is not constant, then the velocity graph depicts

Xia curve (Fig. 5). The acceleration at any moment of time $t_0$ can be found by determining the tangent of the angle of inclination of the tangent to the graph at the point with the abscissa $t_0$ .

You can verify this by replacing the curved segment with a small straight segment. $tg \beta$ is equal to the average acceleration $a_{cp} = \frac{\Delta V}{\Delta t}$ during time $\Delta t$ . At $\Delta t \to 0$ , the segment tends to a tangent to the velocity graph, and the average acceleration for time $\Delta t$ - to the instantaneous acceleration at moment $t = t_0$ , that is, to the value $a = tg\alpha$ .

From the curve at figure 5 it is possible to determine that the acceleration first increases (the tangent of the angle of inclination of the tangent races), then decreases, passes through zero (horizontal section) and becomes negative.

We see, therefore, that the velocity graph allows us to obtain the general nature of the movement, find the distance traveled and the change in coordinates, find out the speed and acceleration for any moment in time, and also evaluate the change in these quantities in the past and future.

A velocity graph can also be constructed for curved body motion. If we know the trajectory, then by choosing one of the directions along the trajectory as positive, we can build a speed graph in the same way as we did in the case of rectilinear motion. The law will indicate the direction of movement, and the value of the ordinate will indicate the abscissa value of the speed.

So, for example, when a body moves in a circle with a constant speed $V_0$ , the velocity graph will be a straight line $I$ by figure 2. The area of the figure under the velocity graph in this case is numerically equal to the distance traveled by the body. Straight line $II$ will characterize a similar movement, but in the opposite direction ( $V_1 < 0$ ).

Path graph and coordinate graph

You need to have a good understanding of the difference between these graphs. The path traveled by the body is a positive value that increases regardless of the direction of movement of the body, therefore the path graph cannot go down and is located below $t$ .

The coordinate characterizes the position of the body relative to the selected starting point. During movement, it can increase, decrease, pass through zero, be positive and negative.

This is exactly how, for example, the coordinate changes when a body moves along the $x$ axis, first in one direction and then in the other (Fig. 6). We see that the final change in coordinates

for the entire time of movement is equal to $-2\text{ m}$ , and the distance traveled during the same time is equal to $8\text{ m}$ . In areas where the movement does not change direction (for example, $0-x_1$ or $x_1-x_2$ ), the change in coordinate is equal in abscissa to the distance traveled.

Let’s look at the coordinate graph in more detail.

The straight line $I$ (Fig. 7), going at an angle $\alpha$ to the axis $t$ , corresponds to uniform motion, since in equal intervals of time $\Delta t$ the body passes through identical segments of coordinates $\Delta x$ . The tangent of the angle of inclination of this line, as can easily be seen from figure,

numerically equal to the speed of movement:

$V = \frac{\Delta x}{\Delta t} = \frac{\Delta S}{\Delta t} = tg\alpha$

(in our example, the movement does not change direction, so the change in coordinate $\Delta x$ during time $\Delta t$ is exactly equal to the distance $\Delta S$ traveled during the same time).

The rounder the straight line rises, the greater the tangent of the angle of inclination and, therefore, the greater the speed. Point $x_0$ on line $I$ is the initial coordinate of the body. It characterizes the position of the body at the initial moment of time (at $t=0$ ). Over time, the body moves away from the origin in the positive direction ( $x>0$ and increases).

Straight line $II$ on figure 7 also corresponds to uniform motion, but with negative speed ( $tg\beta<0$ ). The body is located at $t=0$ at a distance $x_1$ from the origin. It moves towards the first body, reaches the origin at $t=t_1$ and continues to move in the negative direction.

Path graphs for both cases are shown in figure 8. These are straight lines going at angles $\alpha$ and $\beta$ to the x-axis.

Accelerated motion is depicted on a curved line graph (Fig. 9). If the acceleration is constant, the curve is a parabola. Tangent of the angle of inclination of the tangent to the graph

ku coordinates are equal in magnitude to the numerical value of the body’s speed. Pozny, if the curve is curved as in section I (Fig. 9), the tangent of the angle of inclination returns, and, consequently, the speed increases; if the curve is curved as in section II, the speed increases. The corresponding path graph is shown in figure 10. While the body coordinate is returning, the path graph coincides with the coordinate graph; when the coordinate kills, let it return so that the increase in the path is equal to the increase in the coordinate.

It should be remembered that the speed of a body cannot change abruptly and therefore the graphs of coordinates and paths cannot have kinks. Graph coordinates

dinata, for example, cannot be the same as at figure 11. At the moment of time $t=t_1$ the speed with such a movement must change abruptly from $V_1=tg\alpha$ to $V_2=tg\beta$ , which is impossible.

Let’s solve two problems.

Two bodies, being at a distance $S_0=3\text{ m}$ , begin to move towards each other with speeds $V_1=2\text{ m/сек}$ and $V_2=1\text{ m/сек}$ . Where and when will they meet?

The problem can be easily solved graphically (Fig. 12). We will assume that the first body moves from the origin of coordinates in a positive direction, that is, up on the graph, and the second body moves from point $x_1=S_0$ in a negative direction, that is, down on the graph (the coordinate increases). Both movements are depicted as inclined straight lines, moving towards each other and intersecting at the point corresponding to the coordinate and time of meeting.

The slope of the lines is determined from the condition: $tg\alpha_1=V_1=2$ , $tg\beta=V_2=-1$ . The graph shows what occurs at the point with coordinates $t=1\text{ сек}$ and $x=2\text{ м}$ .

During exams, students are often asked to comment on a movement depicted on a graph and to show (draw) what a graph of the same movement looks like in other coordinates.

Using the coordinate graph (Fig. 13), draw an approximate velocity graph.

It is convenient to build graphs one below the other so that you can compare them for the same points in time. The analysis of the graph should begin with the simplest sections (in our problem these are 1 and 5), where the speed is the same and constant in magnitude (the tangents of the slope angles are equal and do not change). Having depicted these sections on the velocity graph as straight horizontal lines, you can move on to analyzing the curved part of the coordinate graph.

It is easy to see that in section 2 the speed returns (the tangent of the tangent angle increases

changes), in section 3 it decreases, at point A the speed reaches its original value (the tangent is parallel to the initial straight line), and at point B it is zero (the tangent is zero). In section 4, the body moves in the opposite direction, and the speed is constantly negative. At point C, the speed again crosses zero and then returns to its previous level.

To successfully construct graphs, training is, of course, necessary. It is easy to implement by drawing volume graphs in one coordinate system and rowing the corresponding graphs in another.

Try solving the following problems on your own.

Using the velocity graph (Fig. 14), find the coordinate and acceleration graphs. What do negative t values mean?
Using the coordinate graphs (Fig. 15), draw velocity graphs.

What kind of motion is shown in figure 16 and what would it look like on a velocity graph? The last $t=t_1$ curve of the graph is a parabola.
Using the acceleration graph (Fig. 17), draw graphs of speed, coordinates and paths.
In the first two seconds, the body traveled 4 meters, then, with increased speed, it moved for 4 seconds in the same direction and finally returned back at its original speed. During each of the three stages, the speed is considered constant. The second body began to move a second straight from

the same point and moved with a constant speed $V = \frac{3}{4} m/sec$ . Using the path graph, determine: 1) the speed of the first body at the second stage; 2) the time and place of meetings became crazy; 3) the path traveled by the first body during the entire period of movement.

Two bicycles left in the same direction with an interval of 10 seconds. The first one moves with acceleration $a_1=2 m/sec^2$ for 5 seconds, then with a constant speed available to them, and after the second bike leaves - slowly with acceleration $a'_1=-0,25 m/sec^2$ . The second bicycle moves with acceleration $a_2=1 m/sec^2$ . As soon as the speeds of the bicycles were equal, both continued moving at a constant speed. Using the speed graph, find: 1) when and at what distance from the starting point did the speeds avel become equal; 2) what is this speed; 3) the path of the first avel until the second one leaves.
Two balls began to move simultaneously and at the same speed along a surface having the shape shown in figure 18. How will the speeds and times of movement of the balls differ when they arrive at point $B$ ?