Let us derive a formula with which you can calculate the projection of the displacement vector of a body moving rectilinearly and uniformly accelerated for any period of time. To do this, let's turn to Figure 14. Both in Figure 14, a, and in Figure 14, b, the segment AC is a graph of the projection of the velocity vector of a body moving with constant acceleration a (at an initial speed v 0).
Rice. 14. The projection of the displacement vector of a body moving rectilinearly and uniformly accelerated is numerically equal to the area S under the graph
Let us recall that in the case of rectilinear uniform motion of a body, the projection of the displacement vector made by this body is determined by the same formula as the area of the rectangle enclosed under the graph of the projection of the velocity vector (see Fig. 6). Therefore, the projection of the displacement vector is numerically equal to the area of this rectangle.
Let us prove that in the case of rectilinear uniformly accelerated motion, the projection of the displacement vector s x can be determined by the same formula as the area of the figure enclosed between the graph AC, the Ot axis and the segments OA and BC, i.e., as in this case, the projection of the displacement vector is numerically equal to the area of the figure under the velocity graph. To do this, on the Ot axis (see Fig. 14, a) we select a small time period db. From points d and b we draw perpendiculars to the Ot axis until they intersect with the graph of the projection of the velocity vector at points a and c.
Thus, over a period of time corresponding to the segment db, the speed of the body changes from v ax to v cx.
Over a fairly short period of time, the projection of the velocity vector changes very slightly. Therefore, the motion of the body during this period of time differs little from uniform motion, that is, from motion at a constant speed.
The entire area of the OASV figure, which is a trapezoid, can be divided into such strips. Consequently, the projection of the displacement vector sx for the period of time corresponding to the segment OB is numerically equal to the area S of the trapezoid OASV and is determined by the same formula as this area.
According to the rule given in school geometry courses, the area of a trapezoid is equal to the product of half the sum of its bases and its height. From Figure 14, b it is clear that the bases of the trapezoid OASV are the segments OA = v 0x and BC = v x, and the height is the segment OB = t. Hence,
Since v x = v 0x + a x t, a S = s x, we can write:
Thus, we have obtained a formula for calculating the projection of the displacement vector during uniformly accelerated motion.
Using the same formula, the projection of the displacement vector is also calculated when the body moves with a decreasing velocity, only in this case the velocity and acceleration vectors will be directed in opposite directions, so their projections will have different signs.
Questions
- Using Figure 14, a, prove that the projection of the displacement vector during uniformly accelerated motion is numerically equal to the area of the figure OASV.
- Write down an equation to determine the projection of the displacement vector of a body during its rectilinear uniformly accelerated motion.
Exercise 7
Graphic representation of uniformly accelerated linear motion.
Moving during uniformly accelerated motion.
Ilevel.
Many physical quantities that describe the movements of bodies change over time. Therefore, for greater clarity of description, movement is often depicted graphically.
Let us show how the time dependences of kinematic quantities describing rectilinear uniformly accelerated motion are graphically depicted.
Uniformly accelerated linear motion- this is a movement in which the speed of a body changes equally over any equal periods of time, i.e. it is a movement with acceleration constant in magnitude and direction.
a=const - acceleration equation. That is, a has a numerical value that does not change over time.
By definition of acceleration
From here we have already found equations for the dependence of speed on time: v = v0 + at.
Let's see how this equation can be used to graphically represent uniformly accelerated motion.
Let us graphically depict the dependences of kinematic quantities on time for three bodies
.
1, the body moves along the 0X axis, while increasing its speed (acceleration vector a is codirectional with the velocity vector v). vx >0, akh > 0
2, the body moves along the 0X axis, while reducing its speed (the acceleration vector a is not codirectional with the velocity vector v). vx >0, ah< 0
2, the body moves against the 0X axis, while reducing its speed (the acceleration vector is not codirectional with the velocity vector v). vx< 0, ах > 0
Acceleration graph
Acceleration, by definition, is a constant value. Then, for the presented situation, the graph of acceleration versus time a(t) will look like:
From the acceleration graph, you can determine how the speed changed - increased or decreased and by what numerical value the speed changed and which body the speed changed more.
Speed graph
If we compare the dependence of the coordinate on time during uniform motion and the dependence of the velocity projection on time during uniformly accelerated motion, we can see that these dependencies are the same:
x= x0 + vx t vx = v 0 x + a X t
This means that the dependency graphs have the same appearance.
To construct this graph, the time of movement is plotted on the abscissa axis, and the speed (projection of speed) of the body is plotted on the ordinate axis. In uniformly accelerated motion, the speed of a body changes over time.
Moving during uniformly accelerated motion.
In uniformly accelerated rectilinear motion, the speed of a body is determined by the formula
vx = v 0 x + a X t
In this formula, υ0 is the speed of the body at t = 0 (starting speed ), a= const – acceleration. On the speed graph υ ( t) this dependence looks like a straight line (Fig.).
Acceleration can be determined from the slope of the velocity graph a bodies. The corresponding constructions are shown in Fig. for graph I. Acceleration is numerically equal to the ratio of the sides of the triangle ABC: MsoNormalTable">
The greater the angle β that the velocity graph forms with the time axis, i.e., the greater the slope of the graph ( steepness), the greater the acceleration of the body.
For graph I: υ0 = –2 m/s, a= 1/2 m/s2.
For graph II: υ0 = 3 m/s, a= –1/3 m/s2.
The velocity graph also allows you to determine the projection of movement s bodies for some time t. Let us select on the time axis a certain small period of time Δ t. If this period of time is small enough, then the change in speed over this period is small, i.e. the movement during this period of time can be considered uniform with a certain average speed, which is equal to the instantaneous speed υ of the body in the middle of the interval Δ t. Therefore, the displacement Δ s in time Δ t will be equal to Δ s = υΔ t. This movement is equal to the area of the shaded strip (Fig.). Breaking down the time period from 0 to some point t for small intervals Δ t, we find that the movement s for a given time t with uniformly accelerated rectilinear motion is equal to the area of the trapezoid ODEF. The corresponding constructions were made for graph II in Fig. 1.4.2. Time t taken equal to 5.5 s.
Since υ – υ0 = at s t will be written in the form:
To find the coordinates y bodies at any time t needed to the starting coordinate y 0 add movement in time t: DIV_ADBLOCK189">
Since υ – υ0 = at, the final formula for moving s body with uniformly accelerated motion over a time interval from 0 to t will be written in the form: https://pandia.ru/text/78/516/images/image009_57.gif" width="146 height=55" height="55">
When analyzing uniformly accelerated motion, sometimes the problem arises of determining the movement of a body based on the given values of the initial υ0 and final υ velocities and acceleration a. This problem can be solved using the equations written above by eliminating time from them t. The result is written in the form
If the initial speed υ0 is zero, these formulas take the form MsoNormalTable">
It should be noted once again that the quantities υ0, υ, included in the formulas for uniformly accelerated rectilinear motion s, a, y 0 are algebraic quantities. Depending on the specific type of movement, each of these quantities can take on both positive and negative values.
An example of solving a problem:
Petya slides down the mountainside from a state of rest with an acceleration of 0.5 m/s2 in 20 s and then moves along a horizontal section. Having traveled 40 m, he crashes into the gaping Vasya and falls into a snowdrift, reducing his speed to 0 m/s. With what acceleration did Petya move along the horizontal surface to the snowdrift? What is the length of the mountain slope from which Petya so unsuccessfully slid down?
Given: | |
|
a 1 = 0.5 m/s2 t 1 = 20 s s 2 = 40 m | Petit's movement consists of two stages: at the first stage, descending from the mountainside, he moves with increasing speed; at the second stage, when moving on a horizontal surface, his speed decreases to zero (collided with Vasya). We write down the values related to the first stage of movement with index 1, and those related to the second stage with index 2. |
Stage 1.
The equation for Petit's speed at the end of the descent from the mountain is:
v 1 = v 01 + a 1t 1.
In projections onto the axis X we get:
v 1x = a 1xt.
Let us write an equation connecting the projections of Petya’s velocity, acceleration and displacement at the first stage of movement:
or because Petya was driving from the very top of the hill with an initial speed of V01=0
(If I were Petya, I would be careful about driving down such high hills)
Considering that Petya’s initial speed at this 2nd stage of movement is equal to his final speed at the first stage:
v 02 x = v 1 x, v 2x = 0, where v1 is the speed with which Petya reached the foot of the hill and began to move towards Vasya. V2x - Petya's speed in a snowdrift.
2. Using this acceleration graph, tell us how the speed of the body changes. Write down the equations for the dependence of speed on time if at the moment of the start of movement (t=0) the speed of the body is v0х =0. Please note that with each subsequent section of movement, the body begins to pass at a certain speed (which was achieved during the previous time!).
3. A metro train, leaving the station, can reach a speed of 72 km/h in 20 s. Determine with what acceleration a bag, forgotten in a subway car, is moving away from you. How far will she travel?
4. A cyclist moving at a speed of 3 m/s begins to descend a mountain with an acceleration of 0.8 m/s2. Find the length of the mountain if the descent took 6 s.
5. Having started braking with an acceleration of 0.5 m/s2, the train traveled 225 m to the stop. What was its speed before braking began?
6. Having started to move, the soccer ball reached a speed of 50 m/s, covered a distance of 50 m and crashed into the window. Determine the time it took the ball to travel this path and the acceleration with which it moved.
7. Reaction time of Uncle Oleg’s neighbor = 1.5 minutes, during which time he will figure out what happened to his window and will have time to run out into the yard. Determine what speed young football players should develop so that the joyful owners of the window do not catch up with them, if they need to run 350 m to their entrance.
8. Two cyclists are riding towards each other. The first one, having a speed of 36 km/h, began to climb up the mountain with an acceleration of 0.2 m/s2, and the second, having a speed of 9 km/h, began to descend the mountain with an acceleration of 0.2 m/s2. After how long and in what place will they collide due to their absent-mindedness, if the length of the mountain is 100 m?
In previous lessons, we discussed how to determine the distance traveled during uniform linear motion. It's time to find out how to determine the coordinates of a body, the distance traveled and the displacement during rectilinear uniformly accelerated motion. This can be done if we consider rectilinear uniformly accelerated motion as a set of a large number of very small uniform displacements of the body.
The first to solve the problem of the location of a body at a certain moment in time during accelerated motion was the Italian scientist Galileo Galilei (Fig. 1).
Rice. 1. Galileo Galilei (1564-1642)
He conducted his experiments with an inclined plane. He launched a ball, a musket bullet, along the chute, and then determined the acceleration of this body. How did he do it? He knew the length of the inclined plane, and determined the time by the beat of his heart or pulse (Fig. 2).
Rice. 2. Galileo's experiment
Consider the speed dependence graph uniformly accelerated linear motion from time. You know this dependence; it is a straight line: .
Rice. 3. Determination of displacement during uniformly accelerated linear motion
We divide the speed graph into small rectangular sections (Fig. 3). Each section will correspond to a certain speed, which can be considered constant in a given period of time. It is necessary to determine the distance traveled during the first period of time. Let's write the formula: . Now let's calculate the total area of all the figures we have.
The sum of the areas during uniform motion is the total distance traveled.
Please note: the speed will change from point to point, thereby we will get the path traveled by the body precisely during rectilinear uniformly accelerated motion.
Note that during rectilinear uniformly accelerated motion of a body, when speed and acceleration are directed in the same direction (Fig. 4), the displacement module is equal to the distance traveled, therefore, when we determine the displacement module, we determine distance traveled. In this case, we can say that the displacement module will be equal to the area of the figure, limited by the graph of speed and time.
Rice. 4. The displacement module is equal to the distance traveled
Let's use mathematical formulas to calculate the area of the indicated figure.
Rice. 5 Illustration for calculating area
The area of the figure (numerically equal to the distance traveled) is equal to half the sum of the bases multiplied by the height. Please note that in the figure, one of the bases is the initial speed, and the second base of the trapezoid will be the final speed, indicated by the letter . The height of the trapezoid is equal to , this is the period of time during which the movement occurred.
We can write the final velocity, discussed in the previous lesson, as the sum of the initial velocity and the contribution due to the constant acceleration of the body. The resulting expression is:
If you open the brackets, it becomes double. We can write the following expression:
If you write each of these expressions separately, the result will be the following:
This equation was first obtained through the experiments of Galileo Galilei. Therefore, we can consider that it was this scientist who first made it possible to determine the location of a body during rectilinear uniformly accelerated motion at any time. This is the solution to the main problem of mechanics.
Now let's remember that the distance traveled, equal in our case movement module, is expressed by the difference:
If we substitute this expression into Galileo’s equation, we obtain a law according to which the coordinate of a body changes during rectilinear uniformly accelerated motion:
It should be remembered that the quantities are projections of velocity and acceleration onto the selected axis. Therefore, they can be both positive and negative.
Conclusion
The next stage of consideration of movement will be the study of movement along a curvilinear trajectory.
Bibliography
- Kikoin I.K., Kikoin A.K. Physics: textbook for 9th grade of high school. - M.: Enlightenment.
- Peryshkin A.V., Gutnik E.M., Physics. 9th grade: textbook for general education. institutions/A. V. Peryshkin, E. M. Gutnik. - 14th ed., stereotype. - M.: Bustard, 2009. - 300.
- Sokolovich Yu.A., Bogdanova G.S.. Physics: A reference book with examples of problem solving. - 2nd edition repartition. - X.: Vesta: Ranok Publishing House, 2005. - 464 p.
Additional recommended links to Internet resources
- Internet portal “class-fizika.narod.ru” ()
- Internet portal “videouroki.net” ()
- Internet portal “foxford.ru” ()
Homework
- Write down the formula that determines the projection of the displacement vector of a body during rectilinear uniformly accelerated motion.
- A cyclist, whose initial speed is 15 km/h, slides down a hill in 5 s. Determine the length of the slide if the cyclist moved with a constant acceleration of 0.5 m/s^2 .
- How do the dependences of displacement on time differ for uniform and uniformly accelerated motion?
Uniformly accelerated motion called such a movement in which the acceleration vector remains unchanged in magnitude and direction. An example of such movement is the movement of a stone thrown at a certain angle to the horizon (without taking into account air resistance). At any point in the trajectory, the acceleration of the stone is equal to the acceleration of gravity. Thus, the study of uniformly accelerated motion is reduced to the study of rectilinear uniformly accelerated motion. In the case of rectilinear motion, the velocity and acceleration vectors are directed along the straight line of motion. Therefore, speed and acceleration in projections onto the direction of motion can be considered as algebraic quantities. In uniformly accelerated rectilinear motion, the speed of the body is determined by formula (1)
In this formula, is the speed of the body at t = 0 (starting speed ), = const – acceleration. In the projection onto the selected x axis, equation (1) will be written as: (2). On the velocity projection graph υ x ( t) this dependence looks like a straight line.
Acceleration can be determined from the slope of the velocity graph a bodies. The corresponding constructions are shown in Fig. for graph I Acceleration is numerically equal to the ratio of the sides of the triangle ABC: .
The greater the angle β that the velocity graph forms with the time axis, i.e., the greater the slope of the graph ( steepness), the greater the acceleration of the body.
For graph I: υ 0 = –2 m/s, a= 1/2 m/s 2. For schedule II: υ 0 = 3 m/s, a= –1/3 m/s 2 .
The velocity graph also allows you to determine the projection of the body’s displacement s over some time t. Let us highlight a certain small time interval Δt on the time axis. If this period of time is short enough, then the change in speed over this period is small, that is, the movement during this period of time can be considered uniform with a certain average speed, which is equal to the instantaneous speed υ of the body in the middle of the interval Δt. Therefore, the displacement Δs during the time Δt will be equal to Δs = υΔt. This movement is equal to the shaded area in Fig. stripes. By dividing the time interval from 0 to a certain moment t into small intervals Δt, we can obtain that the displacement s for a given time t with uniformly accelerated rectilinear motion is equal to the area of the trapezoid ODEF. The corresponding constructions are shown in Fig. for schedule II. Time t is assumed to be 5.5 s.
(3) – the resulting formula allows you to determine the displacement during uniformly accelerated motion if the acceleration is unknown.
If we substitute the expression for speed (2) into equation (3), we obtain (4) - this formula is used to write the equation of motion of the body: (5).
If we express the time of movement (6) from equation (2) and substitute it into equality (3), then
This formula allows you to determine the movement with an unknown time of movement.
And the time of movement, you can find the distance traveled:
Substituting the expression into this formula V avg = V/2, we will find the path traveled during uniformly accelerated motion from a state of rest:
If we substitute into formula (4.1) the expression V avg = V 0 /2, then we get the path traveled during braking:
The last two formulas include speeds V 0 and V. Substituting the expression V=at into formula (4.2), and the expression V 0 =at - into formula (4.3), we get
The resulting formula is valid both for uniformly accelerated motion from a state of rest, and for motion with decreasing speed when the body stops at the end of the path. In both of these cases, the distance traveled is proportional to the square of the time of movement (and not just time, as was the case with uniform movement). The first to establish this pattern was G. Galileo.
Table 2 gives the basic formulas describing uniformly accelerated linear motion. 
Galileo did not have a chance to see his book, which outlined the theory of uniformly accelerated motion (along with many of his other discoveries). When was it published? The 74-year-old scientist was already blind. Galileo took the loss of his vision very hard. “You can imagine,” he wrote, “how I grieve when I realize that this sky, this world and the Universe, which by my observations and clear evidence have been expanded a hundred and a thousand times compared to what people thought they were sciences in all the past centuries have now become so diminished and diminished for me.”
Five years earlier, Galileo was tried by the Inquisition. His views on the structure of the world (and he adhered to the Copernican system, in which the central place was occupied by the Sun, not the Earth) had not been liked by church ministers for a long time. Back in 1614, the Dominican priest Caccini declared Galileo a heretic and mathematics an invention of the devil. And in 1616, the Inquisition officially declared that “the doctrine attributed to Copernicus that the Earth moves around the Sun, while the Sun stands at the center of the Universe, not moving from East to West, is contrary to the Holy Scriptures, and therefore it can neither be defended nor accepted for the truth." Copernicus's book outlining his system of the world was banned, and Galileo was warned that if "he did not calm down, he would be imprisoned."
But Galileo “did not calm down.” “There is no greater hatred in the world,” the scientist wrote, “than ignorance for knowledge.” And in 1632, his famous book “Dialogue on the two most important systems of the world - Ptolemaic and Copernican” was published, in which he gave numerous arguments in favor of the Copernican system. However, only 500 copies of this work were sold, since after a few months, by order of the Pope
Rimsky, the publisher of the book, received an order to suspend the sale of this work.
In the autumn of the same year, Galileo received an order from the Inquisition to appear in Rome, and after some time the sick 69-year-old scientist was taken to the capital on a stretcher. Here, in the prison of the Inquisition, Galileo was forced to renounce his views on the structure of the world, and on June 22, 1633 in a Roman monastery Minerva Galileo reads and signs the previously prepared text of renunciation
“I, Galileo Galilei, son of the late Vincenzo Galilei of Florence, 70 years of age, brought in person to the court and kneeling before Your Eminences, the most reverend gentlemen cardinals, general inquisitors against heresy throughout Christendom, having before me the sacred Gospel and offering hands on him, I swear that I have always believed, I believe now, and with God’s help I will continue to believe in everything that the Holy Catholic and Apostolic Roman Church recognizes, defines and preaches.”
According to the court decision, Galileo's book was banned, and he himself was sentenced to imprisonment for an indefinite period. However, the Pope pardoned Galileo and replaced the imprisonment with exile. Galileo moved to Arcetri and here, while under house arrest, wrote the book "Conversations and Mathematical Proofs , concerning two new branches of science related to Mechanics and Local Motion" In 1636, the manuscript of the book was sent to Holland, where it was published in 1638. With this book, Galileo summed up his many years of physical research. In the same year, Galileo became completely blind Talking about what had befallen misfortune of the great scientist, Viviani (a student of Galileo) wrote: “He suffered severe discharge from his eyes, so that after a few months he was completely left without eyes - yes, I say, without his eyes, which in a short time saw more than all human eyes over all the past centuries have been able to see and observe"
The Florentine inquisitor who visited Galileo in his letter to Rome said that he found him in a very serious condition. Based on this letter, the Pope allowed Galileo to return to his home in Florence. Here he was immediately given an order “On pain of life imprisonment in a true prison and excommunication “Don’t go out into the city and don’t talk to anyone, no matter who it is, about the damned opinion about the double movement of the Earth.”
Galileo did not stay at home for long. After a few months he was again ordered to come to Arcetri. He had about four years to live. On January 8, 1642, at four o'clock in the morning, Galileo died.
1. How does uniformly accelerated motion differ from uniform motion? 2. How does the path formula for uniformly accelerated motion differ from the path formula for uniform motion? 3. What do you know about the life and work of G. Galileo? In which year he was born?
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