In this lesson, we will consider an important characteristic of uneven motion - acceleration. In addition, we will consider uneven motion with constant acceleration. Such a movement is also called uniformly accelerated or equally slowed down. Finally, we will talk about how to graphically depict the dependence of the speed of a body on time for uniformly accelerated motion.
Homework
Having solved the problems to this lesson, you will be able to prepare for questions 1 GIA and questions A1, A2 of the exam.
1. Problems 48, 50, 52, 54 sb. tasks of A.P. Rymkevich, ed. ten.
2. Write down the dependences of speed on time and draw graphs of the dependence of the body's speed on time for the cases shown in fig. 1, cases b) and d). Mark pivot points on the charts, if any.
3. Consider the following questions and their answers:
Question. Is gravitational acceleration acceleration as defined above?
Answer.Of course it is. Free fall acceleration is the acceleration of a body that freely falls from a certain height (air resistance must be neglected).
Question. What happens if the acceleration of the body is directed perpendicular to the speed of the body?
Answer.The body will move evenly around the circumference.
Question. Can I calculate the tangent of the slope using a protractor and calculator?
Answer.No! Because the acceleration obtained in this way will be dimensionless, and the dimension of the acceleration, as we showed earlier, must have the dimension of m / s 2.
Question.What about driving if the plot of speed versus time is not straight?
Answer.We can say that the acceleration of this body changes over time. Such movement will not be uniformly accelerated.
In general uniformly accelerated motion is called a movement in which the acceleration vector remains unchanged in magnitude and direction. An example of such a movement is the movement of a stone thrown at an angle to the horizon (excluding air resistance). At any point on the trajectory, the acceleration of the stone is equal to the acceleration of gravity. For a kinematic description of the movement of a stone, it is convenient to choose a coordinate system so that one of the axes, for example, the axis OY, was directed parallel to the acceleration vector. Then the curvilinear movement of the stone can be represented as the sum of two movements - rectilinear uniformly accelerated motion along the axis OY and uniform straight motion in the perpendicular direction, i.e. along the axis OX (fig. 1.4.1).
Thus, the study of uniformly accelerated motion is reduced to the study of rectilinear uniformly accelerated motion. In the case of rectilinear motion, the vectors of speed and acceleration are directed along the straight line of motion. Therefore, the speed υ and acceleration a in projections on the direction of motion can be considered as algebraic quantities.
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Figure 1.4.1. Projections of vectors of speed and acceleration on the coordinate axes. ax = 0, ay = –g |
With uniformly accelerated rectilinear motion, the speed of the body is determined by the formula
(*)
In this formula υ 0 is the speed of the body at t = 0 (starting speed ), a \u003d const - acceleration. On the graph of the speed υ ( t) this dependence has the form of a straight line (Fig. 1.4.2).
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Figure 1.4.2. Coordinate speed graphs |
Acceleration can be determined from the slope of the speed graph a body. The corresponding constructions are shown in Fig. 1.4.2 for graph I. Acceleration is numerically equal to the ratio of the sides of the triangle ABC:
The greater the angle β, which forms the velocity graph 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 \u003d –2 m / s, a \u003d 1/2 m / s 2.
For graph II: υ 0 \u003d 3 m / s, a \u003d –1/3 m / s 2
The speed graph also allows you to determine the projection of the movement s bodies for a while t... Let us select on the time axis a certain small time interval Δ t... If this time interval is small enough, then the change in speed during this interval is small, i.e., the motion during this time interval can be considered uniform with some average speed, which is equal to the instantaneous velocity 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 \u200b\u200bthe shaded strip (Fig. 1.4.2). Breaking down the time span from 0 to some point t for small intervals Δ t, we get that the displacement s for a given time t with uniformly accelerated rectilinear motion is equal to the area of \u200b\u200bthe trapezoid ODEF... The corresponding constructions are made for graph II in Fig. 1.4.2. Time t taken equal to 5.5 s.
Since υ - υ 0 \u003d at, the final formula for moving s body with uniformly accelerated motion over a time interval from 0 to t will be written as:
(**)
To find the coordinate y bodies at any given time t you need to start coordinate y 0 add movement over time t:
(***)
This expression is called uniformly accelerated motion law .
When analyzing uniformly accelerated motion, sometimes the problem arises of determining the displacement of a body according to the given values \u200b\u200bof the initial υ 0 and final υ velocities and accelerations a... This task can be solved using the equations written above by excluding the time t... The result is written as
From this formula, one can obtain an expression for determining the final velocity υ of the body, if the initial velocity υ 0, the acceleration a and moving s:
If the initial velocity υ 0 is equal to zero, these formulas take the form
It should be noted once again that the quantities υ 0, υ, included in the formulas of uniformly accelerated rectilinear motion s, a, y 0 are algebraic quantities. Depending on the specific type of movement, each of these values \u200b\u200bcan take both positive and negative values.
1) Analytical method.We consider the highway to be straightforward. Let's write the equation of motion of the cyclist. Since the cyclist was moving uniformly, his equation of motion is:
(we place the origin at the starting point, so the initial coordinate of the cyclist is zero).
The motorcyclist was moving at uniform acceleration. He also began to move from the starting point, so his initial coordinate is zero, the initial speed of the motorcyclist is also zero (the motorcyclist began to move from rest).
Considering that the motorcyclist started moving later, the equation of motion for the motorcyclist is:
In this case, the speed of the motorcyclist changed according to the law:
At the moment when the motorcyclist caught up with the cyclist, their coordinates are equal, i.e. or:
Solving this equation for, we find the meeting time:
This is a quadratic equation. Determine the discriminant:
We define the roots:
Substitute in the formulas numerical values and calculate:
Discard the second root as inappropriate physical conditions Objectives: the motorcyclist could not catch up with the cyclist in 0.37 s after the cyclist started to move, since he himself left the starting point only 2 s after the cyclist started.
Thus, the time when the motorcyclist caught up with the cyclist:
Substitute this value of time into the formula for the law of change in the speed of a motorcyclist and find the value of his speed at this moment:
2) Graphic method.
On one coordinate plane, we build graphs of the change over time of the coordinates of the cyclist and the motorcyclist (the graph for the coordinate of the cyclist is red, for the motorcyclist - green) It can be seen that the dependence of the coordinate on time for a cyclist is a linear function, and the graph of this function is a straight line (the case of uniform rectilinear movement). The motorcyclist was moving uniformly, so the dependence of the motorcyclist's coordinate on time is a quadratic function, the graph of which is a parabola.
The most important characteristic when moving a body is its speed. Knowing it, as well as some other parameters, we can always determine the movement time, distance traveled, initial, final speed and acceleration. Equally accelerated movement is only one type of movement. It is usually found in physics problems from the kinematics section. In such problems, the body is taken as a material point, which greatly 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. And this means that when solving a certain kind of problems, it is necessary to pay attention to what acceleration the body has in terms of the sign, and also what is the vector of the body's velocity itself. In general, in problems of an exclusively mathematical plan, such moments are omitted, but in problems in physics this is quite important, since in kinematics, due to one incorrectly set sign, the answer can turn out to be erroneous.
Examples of
An example is uniformly accelerated and equally slow motion. Equally accelerated movement is characterized, as is known, by the acceleration of the body. The acceleration remains constant, but the speed increases continuously at every single moment in time. And with equally slow motion, the acceleration has a negative value, the speed of the body is continuously decreasing. These two types of acceleration are the basis of many physical problems and are quite often encountered in the problems of the first part of physics tests.
An example of uniformly accelerated motion
We meet uniformly accelerated movement everywhere every day. No car moves evenly in real life. Even if the speedometer needle shows exactly 6 kilometers per hour, it should be understood that this is actually not entirely true. First, if we analyze this issue from a technical point of view, then the device will become the first parameter that will give inaccuracy. Rather, its error.
We meet them in all control and measuring devices. The same rulers. Take ten pieces of at least the same (15 centimeters, for example) rulers, even different (15, 30, 45, 50 centimeters). Attach them to each other, and you will notice that there are small inaccuracies, and their scales do not quite coincide. This is the error. In this case, it will be equal to half of the scale division, like other devices that give out certain values.
The second factor that will give inaccuracy is the scale of the instrument. The speedometer does not take into account such values \u200b\u200bas half a kilometer, one second 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. Let it be so small, but still. Thus, it will be a uniformly accelerated motion, not a uniform one. The same can be said for the normal step. Let's go, let's say we are on foot, and someone says: our speed is 5 kilometers per hour. But this is not entirely true, and why, it was told a little above.
Body acceleration
Acceleration can be positive or negative. This was discussed earlier. We add that acceleration is a vector quantity that is numerically equal to the change in speed over a certain period of time. That is, through the formula it can be designated as follows: a \u003d dV / dt, where dV is the change in speed, dt is the time interval (change in time).
Nuances
The question immediately arises as to how the acceleration in this situation can be negative. Those people who ask such a question motivate it by the fact that even the speed cannot be negative, let alone time. In fact, time cannot really be negative. But it is very often forgotten that the speed can take negative values. This is a vector quantity, do not forget about it! The whole point is probably in stereotypes and incorrect thinking.
So, to solve problems, it is enough to understand one thing: the acceleration will be positive if the body is accelerating. And it will be negative if the body slows down. That's all, simple enough. The simplest logical thinking or the ability to see between the lines will already be, in fact, 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. Solving problems
It should be understood that tasks 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 on exams in physics in the 9th and 11th grades one can come across similar questions: "How will the body behave if the sum of all forces acting on it is equal to zero?" In fact, the wording of the question may be very different, but the answer is still the same. Here, the first thing to do is to use surface buildings and ordinary logical thinking.
There are 4 answers to the student's choice. First: "the speed will be zero." Second: "the speed of the body decreases over a 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 every moment of time it will be constant”.
The correct answer here is, of course, the fourth. Now let's figure out why this is. Let's try to consider all the options in turn. As you know, the sum of all forces acting on a body is the product of mass and acceleration. But our mass remains constant, we will discard it. That is, if the sum of all forces is zero, the acceleration will also be zero.
So, suppose the speed is zero. But this cannot be, since acceleration is equal to zero. Physically, this is permissible, but not in this case, since now we are talking about something else. Let the body speed decrease over a period of time. But how can it decrease if the acceleration is constant and it is equal to zero? There are no reasons and prerequisites for a decrease or increase in speed. Therefore, we reject the second option.
Suppose the body's speed is constant, but it is definitely not zero. It really will be constant due to the fact that acceleration is simply absent. But it cannot be said unequivocally that the speed will be different from zero. But the fourth option is right in the bull's-eye. The speed can be anything, but since there is no acceleration, it will be constant over time.
Practical task
Determine which path was covered by the body in a certain period of time t1-t2 (t1 \u003d 0 seconds, t2 \u003d 2 seconds) if the following data is available. The initial speed of the body in the interval from 0 to 1 second is equal to 0 meters per second, the final speed is 2 meters per second. The speed of the body as of 2 seconds is also equal to 2 meters per second.
To solve such a problem is quite simple, you just need to grasp its essence. So, you need to find a way. Well, let's start looking for it, having previously selected two areas. It is easy to see that the body passes the first part of the path (from 0 to 1 second) with uniform acceleration, as evidenced by an increase in its speed. Then we will find this acceleration. It can be expressed as the difference in speed divided by the travel time. The acceleration will be (2-0) / 1 \u003d 2 meters per second squared.
Accordingly, the distance traveled on the first section of the path S will be equal to: S \u003d V0t + at ^ 2/2 \u003d 0 * 1 + 2 * 1 ^ 2/2 \u003d 0 + 1 \u003d 1 meter. On the second section of the path, the body moves uniformly in the period from 1 second to 2 seconds. This means that the distance will be equal to V * t \u003d 2 * 1 \u003d 2 meters. Now we sum up the distances, we get 3 meters. This is the answer.
Topics uSE codifier: types of mechanical motion, speed, acceleration, equations of rectilinear uniformly accelerated motion, free fall.
Equally accelerated movement is a motion with a constant acceleration vector. Thus, during uniformly accelerated motion, the direction and absolute value of acceleration remain unchanged.
Speed \u200b\u200bversus time.
When studying uniform rectilinear motion, the question of the dependence of speed on time did not arise: the speed was constant during the motion. However, with uniformly accelerated motion, the speed changes over time, and we have to find out this dependence.
Let's practice some basic integration again. We proceed from the fact that the derivative of the velocity vector is the acceleration vector:
. (1)
In our case, we have. What needs to be differentiated to get a constant vector? Function, of course. But not only: you can add an arbitrary constant vector to it (after all, the derivative of a constant vector is equal to zero). In this way,
. (2)
What is the meaning of the constant? At the initial moment of time, the speed is equal to its initial value:. Therefore, assuming in formula (2), we get:
So, the constant is the initial velocity of the body. Now relation (2) takes its final form:
. (3)
In specific tasks, we choose a coordinate system and move on to projections onto the coordinate axes. Often enough two axes and a rectangular cartesian system coordinates, and vector formula (3) gives two scalar equalities:
, (4)
. (5)
The formula for the third velocity component, if needed, looks similar.)
The law of motion.
Now we can find the law of motion, that is, the dependence of the radius vector on time. We recall that the derivative of the radius vector is the speed of the body:
We substitute here the expression for the speed given by the formula (3):
(6)
Now we have to integrate equality (6). This is not difficult. To get it, you need to differentiate the function. To get it, you need to differentiate. Let's not forget to add an arbitrary constant:
It is clear that is the initial value of the radius vector at the moment of time. As a result, we obtain the required law of uniformly accelerated motion:
. (7)
Passing to projections onto the coordinate axes, instead of one vector equality (7), we obtain three scalar equalities:
. (8)
. (9)
. (10)
Formulas (8) - (10) give the dependence of the coordinates of the body on time and therefore serve as a solution to the main problem of mechanics for uniformly accelerated motion.
Let's go back to the law of motion (7). Note that - body movement. Then
we get the dependence of displacement on time:
Rectilinear uniformly accelerated motion.
If the uniformly accelerated motion is rectilinear, then it is convenient to choose the coordinate axis along the straight line along which the body moves. For example, let it be an axis. Then three formulas will be enough for us to solve problems:
where is the projection of the displacement onto the axis.
But very often one more formula helps, which is their consequence. Let us express the time from the first formula:
and substitute in the formula for moving:
After algebraic transformations (make sure to do them!) We come to the relation:
This formula does not contain time and allows you to come to an answer faster in those problems where time does not appear.
Free fall.
Free fall is an important particular case of uniformly accelerated motion. This is the name for the movement of a body near the surface of the Earth without taking into account air resistance.
The free fall of a body, regardless of its mass, occurs with a constant acceleration of free fall directed vertically downward. In almost all problems, m / s is assumed for calculations.
Let's analyze several problems and see how the formulas we derived for uniformly accelerated motion work.
A task... Find the landing speed of a raindrop if the cloud height is km.
Decision. Let us direct the axis vertically downward, placing the origin at the drop separation point. Let's use the formula
We have: - the required landing speed,. We get: from where. We calculate: m / s. That's 720 km / h, of the order of the speed of a bullet.
In fact, raindrops fall at a speed of several meters per second. Why is there such a discrepancy? Windage!
A task... The body is thrown vertically upward at a speed of m / s. Find its speed in c.
Here, so what. We calculate: m / s. This means that the speed will be equal to 20 m / s. The projection sign indicates that the body will fly down.
A task. A stone was thrown vertically upwards from a balcony at a height of m at a speed of m / s. How long does it take for the stone to fall to the ground?
Decision. Let us direct the axis vertically up, placing the origin on the surface of the Earth. We use the formula
We have: so, or. Solving the quadratic equation, we get c.
Horizontal throw.
Equally accelerated motion is not necessarily linear. Consider the movement of a body thrown horizontally.
Suppose the body is thrown horizontally at a speed from a height. We will find the time and distance of the flight, as well as find out along which trajectory the movement takes place.
Let's choose a coordinate system as shown in Fig. 1 .
We use the formulas:
In our case . We get:
. (11)
We find the flight time from the condition that at the moment of falling, the coordinate of the body vanishes:
The flight range is the coordinate value at the moment in time:
We obtain the trajectory equation by excluding time from equations (11). We express from the first equation and substitute into the second:
Received dependence on, which is the equation of a parabola. Consequently, the body flies in a parabola.
Throw at an angle to the horizon.
Consider a slightly more complex case of uniformly accelerated motion: the flight of a body thrown at an angle to the horizon.
Suppose that the body is thrown from the surface of the Earth with a velocity directed at an angle to the horizon. Let's find the time and distance of the flight, and also find out along which trajectory the body is moving.
Let's choose a coordinate system as shown in Fig. 2.
We start with the equations:
(Be sure to do these calculations yourself!) As you can see, the dependence on is again the equation of the parabola. Try also to show that the maximum lift is determined by the formula.