In this lesson, we will look at an important characteristic of uneven motion - acceleration. In addition, we will consider uneven movement with constant acceleration. Such movement is also called uniformly accelerated or uniformly decelerated. Finally, we will talk about how to graphically depict the dependence of the speed of a body on time at uniformly accelerated motion.
Homework
Having solved the problems for this lesson, you can prepare for questions 1 of the GIA and questions A1, A2 of the Unified State Exam.
1. Problems 48, 50, 52, 54 sb. problems A.P. Rymkevich, ed. 10.
2. Write down the dependence of speed on time and draw graphs of the dependence of the speed of the body on time for the cases shown in Fig. 1, cases b) and d). Mark turning points on the graphs, if any.
3. Consider the following questions and their answers:
Question. Is acceleration free fall acceleration, according to the definition given above?
Answer. Of course it is. The acceleration of gravity is the acceleration of a body that is freely falling from a certain height (air resistance must be neglected).
Question. What will happen if the acceleration of the body is directed perpendicular to the speed of the body?
Answer. The body will move uniformly around the circle.
Question. Is it possible to calculate the tangent of an angle using a protractor and a calculator?
Answer. No! Because the acceleration obtained in this way will be dimensionless, and the dimension of acceleration, as we showed earlier, should have the dimension m/s 2.
Question. What can be said about motion if the graph of speed versus time is not straight?
Answer. We can say that the acceleration of this body changes with time. Such a movement will not be uniformly accelerated.
The most important characteristic when moving a body is its speed. Knowing it, as well as some other parameters, we can always determine the time of movement, distance traveled, initial and final speed and acceleration. Uniformly accelerated motion is only one type of motion. It is usually found in physics problems from the kinematics section. In such problems, the body is taken as a material point, which significantly simplifies all calculations.
Speed. Acceleration
First of all, I would like to draw the reader’s attention to the fact that these two physical quantities are not scalar, but vector. This means that when solving certain types of problems, it is necessary to pay attention to what acceleration the body has in terms of sign, as well as what the vector of the body’s velocity itself is. In general, in problems of a purely mathematical nature, such moments are omitted, but in problems in physics this is quite important, since in kinematics, due to one incorrect sign, the answer may turn out to be erroneous.
Examples
An example is uniformly accelerated and uniformly decelerated motion. Uniformly accelerated motion is characterized, as is known, by acceleration of the body. The acceleration remains constant, but the speed continuously increases at each individual moment. And with uniformly slow motion, the acceleration has a negative value, the speed of the body continuously decreases. These two types of acceleration form the basis of many physical problems and are quite often found in problems in the first part of physics tests.
Example of uniformly accelerated motion
We encounter uniformly accelerated motion everywhere every day. No car is moving in real life evenly. Even if the speedometer needle shows exactly 6 kilometers per hour, you should understand that this is actually not entirely true. Firstly, if we analyze this issue from a technical point of view, then the first parameter that will give inaccuracy will be the device. Or rather, its error.
We find them in all control and measuring instruments. The same lines. Take about ten rulers, at least identical (15 centimeters, for example), or different (15, 30, 45, 50 centimeters). Put them next to each other and you will notice that there are slight inaccuracies and their scales do not quite line up. This is an error. IN in this case it will be equal to half the division value, as with other devices that produce certain values.
The second factor that will cause inaccuracy is the scale of the device. The speedometer does not take into account values such as half a kilometer, one-half kilometer, and so on. It is quite difficult to notice this on the device with the eye. Almost impossible. But there is a change in speed. Albeit by such a small amount, but still. Thus, it will be uniformly accelerated motion, not uniform. The same can be said about a regular step. Let’s say we’re walking, and someone says: our speed is 5 kilometers per hour. But this is not entirely true, and why was explained a little higher.
Body acceleration
Acceleration can be positive or negative. This was discussed earlier. Let us add that acceleration is a vector quantity, which is numerically equal to the change in speed over a certain period of time. That is, through the formula it can be denoted as follows: a = dV/dt, where dV is the change in speed, dt is the time interval (change in time).
Nuances
The question may immediately arise as to how acceleration in this situation can be negative. Those people who ask a similar question motivate this by the fact that even speed cannot be negative, let alone time. In fact, time really cannot be negative. But very often they forget that speed is taken negative values quite possibly. This is a vector quantity, we should not forget about it! It's probably all about stereotypes and incorrect thinking.
So, to solve problems, it is enough to understand one thing: the acceleration will be positive if the body accelerates. And it will be negative if the body slows down. That's it, quite simple. The simplest logical thinking or the ability to see between the lines will, in fact, be part of the solution to a physical problem related to speed and acceleration. A special case is the acceleration of gravity, and it cannot be negative.
Formulas. Problem solving
It should be understood that problems related to speed and acceleration are not only practical, but also theoretical. Therefore, we will analyze them and, if possible, try to explain why this or that answer is correct or, conversely, incorrect.
Theoretical problem
Very often in physics exams in grades 9 and 11 you can come across similar questions: “How will a body behave if the sum of all forces acting on it is zero?” In fact, the wording of the question can be very different, but the answer is still the same. Here, the first thing you need to do is to use superficial buildings and ordinary logical thinking.
The student is given 4 answers to choose from. First: “the speed will be zero.” Second: “the speed of the body decreases over a certain period of time.” Third: “the speed of the body is constant, but it is definitely not zero.” Fourth: “the speed can have any value, but at each moment of time it will be constant.”
The correct answer here is, of course, the fourth. Now let's figure out why this is so. Let's try to consider all the options in turn. As is known, the sum of all forces acting on a body is the product of mass and acceleration. But our mass remains a constant value, we will discard it. That is, if the sum of all forces is zero, the acceleration will also be zero.
So, let's assume that the speed will be zero. But this cannot be, since our acceleration is equal to zero. Purely physically this is permissible, but not in this case, since now we are talking about something else. Let the speed of the body decrease over a period of time. But how can it decrease if the acceleration is constant and equal to zero? There are no reasons or prerequisites for a decrease or increase in speed. Therefore, we reject the second option.
Let us assume that the speed of the body is constant, but it is definitely not zero. It will indeed be constant due to the fact that there is simply no acceleration. But it cannot be said unequivocally that the speed will be different from zero. But the fourth option is right on target. The speed can be any, but since there is no acceleration, it will be constant over time.
Practical problem
Determine which path was traveled by the body in a certain period of time t1-t2 (t1 = 0 seconds, t2 = 2 seconds) if the following data are available. The initial speed of the body in the interval from 0 to 1 second is 0 meters per second, the final speed is 2 meters per second. The speed of the body at the time of 2 seconds is also 2 meters per second.
Solving such a problem is quite simple, you just need to grasp its essence. So, we need to find a way. Well, let's start looking for it, having previously identified two areas. As is easy to see, the body passes through the first section of the path (from 0 to 1 second) with uniform acceleration, as evidenced by the increase in its speed. Then we will find this acceleration. It can be expressed as the difference in speed divided by the time of movement. The acceleration will be (2-0)/1 = 2 meters per second squared.
Accordingly, the distance traveled on the first section of the path S will be equal to: S = V0t + at^2/2 = 0*1 + 2*1^2/2 = 0 + 1 = 1 meter. On the second section of the path, in the period from 1 second to 2 seconds, the body moves uniformly. This means that the distance will be equal to V*t = 2*1 = 2 meters. Now we sum up the distances, we get 3 meters. This is the answer.
How, knowing the braking distance, determine the initial speed of the car and how, knowing the characteristics of movement, such as initial speed, acceleration, time, determine the movement of the car? We will get the answers after we get acquainted with the topic of today's lesson: “Movement during uniformly accelerated motion, dependence of coordinates on time during uniformly accelerated motion”
With uniformly accelerated motion, the graph looks like a straight line going upward, since its projection of acceleration is greater than zero.
With uniform rectilinear motion, the area will be numerically equal to the module of the projection of the movement of the body. It turns out that this fact can be generalized for the case of not only uniform motion, but also for any motion, that is, it can be shown that the area under the graph is numerically equal to the modulus of the displacement projection. This is done strictly mathematically, but we will use a graphical method.
Rice. 2. Graph of speed versus time for uniformly accelerated motion ()
Let us divide the graph of the projection of velocity versus time for uniformly accelerated motion into small time intervals Δt. Let us assume that they are so small that the speed practically did not change throughout them, that is, the graph linear dependence in the figure we will conditionally turn it into a ladder. At each step we believe that the speed has practically not changed. Let's imagine that we make the time intervals Δt infinitesimal. In mathematics they say: we make the transition to the limit. In this case, the area of such a ladder will coincide indefinitely closely with the area of the trapezoid, which is limited by the graph V x (t). This means that for the case of uniformly accelerated motion we can say that the module of the displacement projection is numerically equal to the area limited by the graph V x (t): the abscissa and ordinate axes and the perpendicular lowered to the abscissa, that is, the area of the trapezoid OABC that we see in Figure 2.
The problem turns from a physical one into a mathematical problem - finding the area of a trapezoid. This is a standard situation when physicists create a model that describes a particular phenomenon, and then mathematics comes into play, enriching this model with equations, laws - something that turns the model into a theory.
We find the area of the trapezoid: the trapezoid is rectangular, since the angle between the axes is 90 0, we divide the trapezoid into two figures - a rectangle and a triangle. Obviously, the total area will be equal to the sum of the areas of these figures (Fig. 3). Let's find their areas: the area of the rectangle is equal to the product of the sides, that is, V 0x t, the area of the right triangle will be equal to half the product of the legs - 1/2AD BD, substituting the values of the projections, we get: 1/2t (V x - V 0x), and, remembering the law of changes in speed over time during uniformly accelerated motion: V x (t) = V 0x + a x t, it is quite obvious that the difference in velocity projections is equal to the product of the acceleration projection a x by time t, that is, V x - V 0x = a x t.
Rice. 3. Determination of the area of the trapezoid ( Source)
Taking into account the fact that the area of the trapezoid is numerically equal to the module of the displacement projection, we obtain:
S x(t) = V 0 x t + a x t 2 /2
We have obtained the law of the dependence of the projection of displacement on time during uniformly accelerated motion in scalar form; in vector form it will look like this:
(t) = t + t 2 / 2
Let us derive another formula for the displacement projection, which will not include time as a variable. Let's solve the system of equations, eliminating time from it:
S x (t) = V 0 x + a x t 2 /2
V x (t) = V 0 x + a x t
Let's imagine that time is unknown to us, then we will express time from the second equation:
t = V x - V 0x / a x
Let's substitute the resulting value into the first equation:
Let's get this cumbersome expression, square it and give similar ones:
We have obtained a very convenient expression for the projection of movement for the case when we do not know the time of movement.
Let our initial speed of the car, when braking began, be V 0 = 72 km/h, final speed V = 0, acceleration a = 4 m/s 2 . Find out the length of the braking distance. Converting kilometers to meters and substituting the values in the formula, we find that the braking distance will be:
S x = 0 - 400(m/s) 2 / -2 · 4 m/s 2 = 50 m
Let's analyze the following formula:
S x = (V 0 x + V x) / 2 t
The displacement projection is the half-sum of the projections of the initial and final velocities, multiplied by the time of movement. Let us recall the displacement formula for average speed
S x = V av · t
In the case of uniformly accelerated motion, the average speed will be:
V av = (V 0 + V k) / 2
We have come close to solving the main problem of the mechanics of uniformly accelerated motion, that is, obtaining the law according to which the coordinate changes with time:
x(t) = x 0 + V 0 x t + a x t 2 /2
In order to learn how to use this law, let's analyze a typical problem.
A car, moving from rest, acquires an acceleration of 2 m/s 2 . Find the distance traveled by the car in 3 seconds and in a third second.
Given: V 0 x = 0
Let us write down the law according to which displacement changes with time at
uniformly accelerated motion: S x = V 0 x t + a x t 2 /2. 2 s< Δt 2 < 3.
We can answer the first question of the problem by plugging in the data:
t 1 = 3 c S 1x = a x t 2 /2 = 2 3 2 / 2 = 9 (m) - this is the path traveled
c car in 3 seconds.
Let's find out how far he traveled in 2 seconds:
S x (2 s) = a x t 2 /2 = 2 2 2 / 2 = 4 (m)
So, you and I know that in two seconds the car traveled 4 meters.
Now, knowing these two distances, we can find the path he traveled in the third second:
S 2x = S 1x + S x (2 s) = 9 - 4 = 5 (m)
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 rectilinear movement.
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 great scientist's misfortune, Viviani (Galileo's student) wrote: “Severe discharges occurred from his eyes, so that after a few months he was completely left without eyes - yes, I say, without his eyes, which short time saw more in this world than everyone else human eyes over all the past centuries we were 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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