How to write the equation of uniformly accelerated motion using a graph. Uniform linear motion

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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 meanings acceleration during 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 in this case coordinate change 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 it to the found number initial value. 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 equal to zero, and the speed is also zero. Until this point in time, the tangent of the tangent angle decreased, but was positive, the point moved in the direction of the OX axis.

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 the coordinate versus time, you can determine the value of the speed at any point in time by calculating the tangent of the tangent to the curve at the point corresponding to at this moment 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.

If the trajectory of a point’s movement is known, then the dependence of the path traversed by the point on the elapsed time interval gives Full description this movement. We have seen that for uniform motion such a dependence can be given in the form of formula (9.2). The relationship between and for individual points in time can also be specified in the form of a table containing the corresponding values ​​of the time period and the distance traveled. Let us be given that the speed of some uniform motion is 2 m/s. Formula (9.2) in this case has the form . Let's make a table of the path and time of such movement:

The dependence of one quantity on another is often convenient to depict not with formulas or tables, but with graphs, which more clearly show the picture of changes in variable quantities and can facilitate calculations. Let us construct a graph of the distance traveled versus time for the movement in question. To do this, take two mutually perpendicular straight lines - coordinate axes; We will call one of them (the abscissa axis) the time axis, and the other (the ordinate axis) the path axis. Let's choose scales for depicting time intervals and paths and take the point of intersection of the axes as the initial moment and as the starting point on the trajectory. Let us plot on the axes the values ​​of time and distance traveled for the movement under consideration (Fig. 18). To “bind” the values ​​of the distance traveled to moments in time, we draw perpendiculars to the axes from the corresponding points on the axes (for example, points 3 s and 6 m). The point of intersection of the perpendiculars corresponds simultaneously to both quantities: path and moment, and in this way the “binding” is achieved. The same construction can be performed for any other points in time and corresponding paths, obtaining for each such pair of time - path values ​​one point on the graph. In Fig. 18 such a construction is made, replacing both rows of the table with one row of points. If such a construction were carried out for all points in time, then instead of individual points, a solid line would be obtained (also shown in the figure). This line is called a path versus time graph or, in short, a path graph.

Rice. 18. Graph of the path of uniform motion at a speed of 2 m/s

Rice. 19. For exercise 12.1

In our case, the path graph turned out to be a straight line. It can be shown that the graph of the path of uniform motion is always a straight line; and vice versa: if the graph of the path versus time is a straight line, then the movement is uniform.

Repeating the construction for a different speed, we find that the graph points for higher speeds lie higher than the corresponding graph points for lower speeds (Fig. 20). Thus, the greater the speed of uniform motion, the steeper the rectilinear graph of the path, i.e., the larger angle it composes with the time axis.

Rice. 20. Graphs of the path of uniform movements with speeds of 2 and 3 m/s

Rice. 21. Graph of the same movement as in Fig. 18, drawn on a different scale

The slope of the graph depends, of course, not only on the numerical value of the speed, but also on the choice of time and length scales. For example, the graph shown in Fig. 21 gives the path versus time for the same movement as the graph in Fig. 18, although it has a different slope. From here it is clear that it is possible to compare movements by the slope of graphs only if they are drawn on the same scale.

Using path graphs, you can easily solve various motion problems. For example in Fig. 18 dashed lines show the constructions necessary to solve the following problems for a given movement: a) find the path traveled in 3.5 s; b) find the time it takes to travel 9 m. In the figure, the answers are found graphically (dashed lines): a) 7 m; b) 4.5 s.

On graphs describing uniform rectilinear movement, you can plot the coordinate of the moving point along the ordinate instead of the path. This description reveals great opportunities. In particular, it makes it possible to distinguish the direction of movement relative to the axis. In addition, by taking the origin of time to be zero, it is possible to show the movement of the point at earlier moments of time, which should be considered negative.

Rice. 22. Graphs of movements with the same speed, but at different initial positions of the moving point

Rice. 23. Graphs of several movements with negative speeds

For example, in Fig. 22 straight line I is a graph of motion occurring at a positive speed of 4 m/s (i.e. in the direction of the axis), and at the initial moment the moving point was at a point with coordinate m. For comparison, the same figure shows a graph of the motion that occurs with the same speed, but at which at the initial moment the moving point is at the point with the coordinate (line II). Straight. III corresponds to the case when at the moment the moving point was at a point with coordinate m. Finally, straight line IV describes the movement in the case when the moving point had a coordinate at the moment c.

We see that the slopes of all four graphs are the same: the slope depends only on the speed of the moving point, and not on its initial position. When changing the initial position, the entire graph is simply transferred parallel to itself along the axis up or down at the appropriate distance.

Graphs of movements occurring at negative speeds (i.e. in the direction opposite to the direction of the axis) are shown in Fig. 23. They are straight, inclined downwards. For such movements, the coordinate of the point decreases over time., had coordinates

Path graphs can also be constructed for cases in which a body moves uniformly for a certain period of time, then moves uniformly but at a different speed for another period of time, then changes speed again, etc. For example, in Fig. 26 shows a motion graph in which the body moved during the first hour at a speed of 20 km/h, during the second hour at a speed of 40 km/h and during the third hour at a speed of 15 km/h.

Exercise: 12.8. Construct a graph of the path for movement in which, over successive hourly intervals, the body had speeds of 10, -5, 0, 2, -7 km/h. What is the total displacement of the body?

Graphing is used to show the dependence of one quantity on another. In this case, the change in one quantity is plotted on one axis, and the change in another quantity is plotted on the other axis. In rectilinear uniform motion, the speed of the body remains constant, only time and the distance traveled, which depends on it, change. Therefore, the greatest interest for such movement is the graph showing the dependence of the path on time.

When constructing such a graph, a change in time (t) is noted on one of the axes of the coordinate plane. For example, 1s, 2s, 3s, etc. Let this be the x-axis. The other axis (in this case y) marks the change in the distance traveled. For example, 10m, 20m, 30m, etc.

The origin of the coordinate system is taken as the origin of motion. This is the starting point at which the time spent moving is zero, and the distance traveled is also zero. This is the first point on the path versus time graph.

Next, the second point of the graph is found on the coordinate plane. To do this, for a given time, the paths are found to be the path traveled during this time. If the speed of the body is 30 m/s, then it can be a point with coordinates (1; 30) or (2; 60) and so on.

After the second point is marked, draw a ray through two points (the first is the origin). The origin of the ray is the origin of coordinates. This ray is a graph of the path versus time for rectilinear uniform motion. The beam has no end, which means that the longer the time spent on the path, the longer the distance traveled.

In general, they say that the graph of the path versus time is a straight line passing through the origin of coordinates.

To prove that the graph is a straight line, and, say, not a broken line, you can construct a series of points on the coordinate plane. For example, if the speed is 5 km/h, then points (1; 5), (2; 10), (3; 15), (4; 20) can be marked on the coordinate plane. Then connect them in series with each other. You will see that it will be straight.

The greater the speed of the body, the faster the distance traveled increases. If on the same coordinate plane we draw the dependence of the path on time for two bodies moving with at different speeds, then the graph of a body that moves faster will have a larger angle with the positive direction of the time axis.

For example, if one body moves at a speed of 10 km/h, and the second - 20 km/h, then on the coordinate plane you can mark points (1; 10) for one body and (1; 20) for the other. It is clear that the second point is further from the time axis, and the straight line through it forms a larger angle than the straight line through the point marked for the first body.

Graphs of the path versus time for rectilinear uniform motion can be used to quickly find the elapsed time by known value the path traveled or the path over a known time. To do this you need to carry out perpendicular line from the value of the coordinate axis, which is known, to the intersection with the graph. Next, from the resulting intersection point, draw a perpendicular to the other axis, thereby obtaining the desired value.

In addition to graphs of path versus time, you can plot graphs of path versus speed and speed versus time. However, since in rectilinear uniform motion the speed is constant, these graphs are straight lines parallel to the axes of the path or time and passing at the level of the declared speed.

Instructions

Consider the function f(x) = |x|. To begin with, this is an unsigned modulus, that is, the graph of the function g(x) = x. This graph is a straight line passing through the origin and the angle between this straight line and the positive direction of the x-axis is 45 degrees.

Since the modulus is a non-negative quantity, the part that is below the abscissa axis must be mirrored relative to it. For the function g(x) = x, we find that the graph after such a mapping will look like V. This new schedule and will be a graphical interpretation of the function f(x) = |x|.

Video on the topic

note

The graph of the modulus of a function will never be in the 3rd and 4th quarters, since the modulus cannot take negative values.

Helpful advice

If a function contains several modules, then they need to be expanded sequentially and then stacked on top of each other. The result will be the desired graph.

Sources:

• how to graph a function with modules

Kinematics problems in which you need to calculate speed, time or the path of uniformly and rectilinearly moving bodies that meet in school course algebra and physics. To solve them, find in the condition quantities that can be equalized. If the condition requires defining time at a known speed, use the following instructions.

You will need

• - pen;
• - paper for notes.

Instructions

The simplest case is the movement of one body with a given uniform speed Yu. The distance that the body has traveled is known. Find on the way: t = S/v, hour, where S is the distance, v is the average speed bodies.

The second is for oncoming movement of bodies. A car moves from point A to point B speed 50 km/h. A moped with a speed 30 km/h. The distance between points A and B is 100 km. Need to find time through which they will meet.

Label the meeting point K. Let the distance AK of the car be x km. Then the motorcyclist’s path will be 100 km. From the problem conditions it follows that time On the road, a car and a moped have the same experience. Make up the equation: x/v = (S-x)/v’, where v, v’ – and the moped. Substituting the data, solve the equation: x = 62.5 km. Now time: t = 62.5/50 = 1.25 hours or 1 hour 15 minutes.

Create an equation similar to the previous one. But in this case time the journey of a moped will be 20 minutes longer than that of a car. To equalize the parts, subtract one third of an hour from the right side of the expression: x/v = (S-x)/v’-1/3. Find x – 56.25. Calculate time: t = 56.25/50 = 1.125 hours or 1 hour 7 minutes 30 seconds.

The fourth example is a problem involving the movement of bodies in one direction. A car and a moped are moving from point A at the same speeds. It is known that the car left half an hour later. After what time will he catch up with the moped?

In this case, the distance traveled by the vehicles will be the same. Let time the car will travel x hours, then time the moped's journey will be x+0.5 hours. You have the equation: vx = v’(x+0.5). Solve the equation by substituting , and find x – 0.75 hours or 45 minutes.

Fifth example – a car and a moped are moving at the same speeds in the same direction, but the moped left point B, located 10 km from point A, half an hour earlier. Calculate after what time After the start, the car will catch up with the moped.

The distance traveled by the car is 10 km more. Add this difference to the motorcyclist’s path and equalize the parts of the expression: vx = v’(x+0.5)-10. Substituting the speed values ​​and solving it, you get: t = 1.25 hours or 1 hour 15 minutes.

Sources:

• what is the speed of the time machine

Instructions

Calculate the average of a body moving uniformly along a section of path. Such speed is the easiest to calculate, since it does not change over the entire segment movement and equals the average. This can be expressed in the form: Vрд = Vср, where Vрд – speed uniform movement, and Vav – average speed.

Calculate the average speed uniformly slow (uniformly accelerated) movement in this area, for which it is necessary to add the initial and final speed. Divide the result by two, which is the average speed Yu. This can be written more clearly as a formula: Vср = (Vн + Vк)/2, where Vн represents

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

Let's start with the simplest case - uniform motion. 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 uneven motion also has this remarkable property.

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 of 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 case of 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 started, the car covered a distance of 20 m in the first 4 s. Consider the car’s motion to be linear 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 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, the 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 beginning of the 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 the average speed of the 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 is the distance covered by the carriage to the stop less than the distance covered by the train in the same time?

14. Having left the station, the train traveled at a uniform acceleration for some time, then for 1 minute at a uniform speed of 60 km/h, and then again at a uniform acceleration 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) What distance 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 so, at what point in time, and what will be the distance traveled?

16. After the push, the ball rolls up an inclined plane, after which it returns to the starting point. The ball was at a distance b from the initial point twice at time intervals t 1 and t 2 after the push. The ball moved up and down along the inclined plane with the same acceleration.
a) Direct the x-axis upward along the inclined plane, select the origin at the initial position of the 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 the 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.