Acceleration speed when moving with constant acceleration. Rectilinear motion with constant acceleration is called uniformly accelerated if the velocity module increases with time, or uniformly decelerated if it decreases. Example of problem solution

In this lesson, the topic of which is: “Equation of motion with constant acceleration. Forward movement,” we will remember what movement is, what it happens. Let’s also remember what acceleration is, consider the equation of motion with constant acceleration and how to use it to determine the coordinates of a moving body. Let's consider an example of a task for consolidating material.

The main task of kinematics is to determine the position of the body at any time. The body can be at rest, then its position will not change (see Fig. 1).

Rice. 1. Body at rest

A body can move in a straight line at a constant speed. Then its movement will change uniformly, that is, equally over equal periods of time (see Fig. 2).

Rice. 2. Movement of a body when moving at a constant speed

Movement, speed multiplied by time, we have been able to do this for a long time. A body can move with constant acceleration; consider such a case (see Fig. 3).

Rice. 3. Body motion with constant acceleration

Acceleration

Acceleration is the change in speed per unit time(see Fig. 4) : Rice. 4. Acceleration Speed ​​is a vector quantity, therefore the change in speed, i.e. the difference between the vectors of the final and initial speed, is a vector. Acceleration is also a vector, directed in the same direction as the vector of the speed difference (see Fig. 5). We are considering linear motion, so we can select a coordinate axis along the straight line along which the motion occurs, and consider the projections of the velocity and acceleration vectors onto this axis:

Then its speed changes uniformly: (if its initial speed was zero). How to find the displacement now? It is impossible to multiply speed by time: the speed was constantly changing; which one to take? How to determine where the body will be at any moment during such a movement - today we will solve this problem.

Let’s immediately define the model: we are considering the rectilinear translational motion of a body. In this case, we can use the material point model. Acceleration is directed along the same straight line along which the material point moves (see Fig. 6).

Forward movement

Translational motion is a movement in which all points of the body move the same way: at the same speed, making the same movement (see Fig. 7). Rice. 7. Forward movement How else could it be? Wave your hand and observe: it is clear that the palm and shoulder moved differently. Look at the Ferris wheel: the points near the axis hardly move, but the cabins move at different speeds and along different trajectories (see Fig. 8). Rice. 8. Movement of selected points on the Ferris wheel Look at a moving car: if you do not take into account the rotation of the wheels and the movement of engine parts, all points of the car move equally, we consider the movement of the car to be translational (see Fig. 9). Rice. 9. Car movement Then there is no point in describing the movement of each point; you can describe the movement of one. We consider a car to be a material point. Please note that during translational movement, the line connecting any two points of the body during movement remains parallel to itself (see Fig. 10). Rice. 10. Position of the line connecting two points

The car drove straight for an hour. At the beginning of the hour his speed was 10 km/h, and at the end - 100 km/h (see Fig. 11).

Rice. 11. Drawing for the problem

The speed changed uniformly. How many kilometers did the car travel?

Let us analyze the condition of the problem.

The speed of the car changed uniformly, that is, its acceleration was constant throughout the journey. Acceleration by definition is equal to:

The car was driving straight, so we can consider its movement in projection onto one coordinate axis:

Let's find the displacement.

Increasing speed example

Nuts are placed on the table, one nut per minute. It’s clear: no matter how many minutes pass, so many nuts will appear on the table. Now let’s imagine that the rate of placing nuts increases uniformly from zero: the first minute no nuts are placed, the second minute they put one nut, then two, three, and so on. How many nuts will be on the table after some time? It is clear that it is less than if the maximum speed was always maintained. Moreover, it is clearly visible that it is 2 times less (see Fig. 12). Rice. 12. Number of nuts at different laying speeds It’s the same with uniformly accelerated motion: let’s say that at first the speed was zero, but at the end it became equal (see Fig. 13). Rice. 13. Change speed If the body were constantly moving at such a speed, its displacement would be equal to , but since the speed increased uniformly, it would be 2 times less.

We know how to find displacement during UNIFORM movement: . How to work around this problem? If the speed does not change much, then the movement can be approximately considered uniform. The change in speed will be small over a short period of time (see Fig. 14).

Rice. 14. Change speed

Therefore, we divide the travel time T into N small segments of duration (see Fig. 15).

Rice. 15. Splitting a period of time

Let's calculate the displacement at each time interval. The speed increases at each interval by:

On each segment we will consider the movement to be uniform and the speed approximately equal to the initial speed for a given period of time. Let's see if our approximation will lead to an error if we assume the motion to be uniform over a short interval. The maximum error will be:

and the total error for the entire journey -> . For large N we assume the error is close to zero. We will see this on the graph (see Fig. 16): there will be an error at each interval, but the total error with a sufficiently large number of intervals will be negligible.

Rice. 16. Interval error

So, each subsequent speed value is the same amount greater than the previous one. From algebra we know that this is an arithmetic progression with a progression difference:

The path in the sections (with uniform rectilinear motion (see Fig. 17) is equal to:

Rice. 17. Consideration of areas of body movement

On the second section:

On the nth section the path is:

Arithmetic progression

Arithmetic progression is a number sequence in which each subsequent number differs from the previous one by the same amount. An arithmetic progression is specified by two parameters: the initial term of the progression and the difference of the progression. Then the sequence is written like this: The sum of the first terms of an arithmetic progression is calculated using the formula:

Let's sum up all the paths. This will be the sum of the first N terms of the arithmetic progression:

Since we have divided the movement into many intervals, we can assume that then:

We had many formulas, and in order not to get confused, we did not write the x indices each time, but considered everything in projection onto the coordinate axis.

So, we have obtained the main formula for uniformly accelerated motion: displacement during uniformly accelerated motion in time T, which, along with the definition of acceleration (change in speed per unit time), we will use to solve problems:

We were working on solving a problem about a car. Let's substitute numbers into the solution and get the answer: the car traveled 55.4 km.

Mathematical part of solving the problem

We figured out the movement. How to determine the coordinate of a body at any moment in time?

By definition, the movement of a body over time is a vector, the beginning of which is at the initial point of movement, and the end is at the final point at which the body will be after time. We need to find the coordinate of the body, so we write an expression for the projection of displacement onto the coordinate axis (see Fig. 18):

Rice. 18. Motion projection

Let's express the coordinate:

That is, the coordinate of the body at the moment of time is equal to the initial coordinate plus the projection of the movement that the body made during the time. We have already found the projection of displacement during uniformly accelerated motion, all that remains is to substitute and write:

This is the equation of motion with constant acceleration. It allows you to find out the coordinates of a moving material point at any time. It is clear that we choose the moment of time within the interval when the model works: the acceleration is constant, the movement is rectilinear.

Why the equation of motion cannot be used to find a path

In what cases can we consider movement modulo equal to path? When a body moves along a straight line and does not change direction. For example, with uniform rectilinear motion, we do not always clearly define whether we are finding a path or a displacement; they still coincide. With uniformly accelerated motion, the speed changes. If the speed and acceleration are directed in opposite directions (see Fig. 19), then the velocity modulus decreases, and at some point it will become equal to zero and the speed will change direction, that is, the body will begin to move in the opposite direction. Rice. 19. Velocity modulus decreases And then, if at a given moment in time the body is at a distance of 3 m from the beginning of observation, then its displacement is equal to 3 m, but if the body first traveled 5 m, then turned around and traveled another 2 m, then the path will be equal to 7 m. And how How can you find it if you don’t know these numbers? You just need to find the moment when the speed is zero, that is, when the body turns around, and find the path to and from this point (see Fig. 20). Rice. 20. The moment when the speed is 0

Bibliography

1. 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.
2. Landsberg G.S. Elementary physics textbook; v.1. Mechanics. Heat. Molecular physics - M.: Publishing house "Science", 1985.

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Homework

1. What is an arithmetic progression?
2. What kind of movement is called translational?
3. What is a vector quantity characterized by?
4. Write down the formula for acceleration through a change in speed.
5. What is the form of the equation of motion with constant acceleration?
6. The acceleration vector is directed towards the movement of the body. How will the body change its speed?

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From DA

12.12.2018 17:31

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An example of accelerated motion would be a flower pot falling from the balcony of a low building. At the beginning of the fall, the speed of the pot is zero, but in a few seconds it manages to increase to tens of m/s. An example of slow motion is the movement of a stone thrown vertically upward, the speed of which is initially high, but then gradually decreases to zero at the top point of the trajectory. If we neglect the force of air resistance, then the acceleration in both of these cases will be the same and equal to the acceleration of gravity, which is always directed vertically downward, denoted by the letter g and equal to approximately 9.8 m/s 2 .

Acceleration of gravity, g caused by the force of gravity of the Earth. This force accelerates all bodies moving towards the earth and slows down those moving away from it.

To find the equation for speed during rectilinear motion with constant acceleration, we will assume that at time t=0 the body had an initial speed v 0 . Since acceleration a is constant, then for any moment t the following equation is valid:

Where v– speed of the body at the moment of time t, from where, after simple transformations, we obtain the equation for speed when moving with constant acceleration:

v = v 0 + a t (5.1)

To derive an equation for the path traveled during rectilinear motion with constant acceleration, we first construct a graph of speed versus time (5.1). For a>0 the graph of this dependence is shown on the left in Fig. 5 (blue straight line). As we established in §3, the movement made during time t can be determined by calculating the area under the speed versus time curve between moments t=0 and t. In our case, the figure under the curve, bounded by two vertical lines t = 0 and t, is a trapezoid OABC, the area of ​​which S, as is known, is equal to the product of half the sum of the lengths of the bases OA and CB and the height OC:

As can be seen in Fig. 5, OA = v0, CB = v0 + a t, and OC = t. Substituting these values ​​into (5.2), we obtain the following equation for the displacement S made in time t during rectilinear motion with constant acceleration a at an initial speed v 0:

It is easy to show that formula (5.3) is valid not only for motion with acceleration a>0, for which it was derived, but also in those cases when a<0. На рис.5 справа красными линиями показаны графики зависимости S при положительных (верх) и отрицательных (низ) значениях a, constructed according to formula (5.3) for various values ​​of v0. It can be seen that, in contrast to uniform motion (see Fig. 3), the graph of displacement versus time is a parabola, and not a straight line, shown for comparison with a dotted line.

Review questions:

· Is motion with constant acceleration uniform?

· Define uniformly accelerated and uniformly decelerated motion.

· What is the acceleration due to gravity, and what causes it?

· By what law does speed change during uniformly accelerated or uniformly decelerated motion?

· How does displacement during uniformly accelerated motion depend on time, acceleration and initial speed?

Rice. 5. On the left - the dependence of speed on time (blue straight line) for uniformly accelerated motion; on the right - the dependence of displacement on time (red curves) for uniformly accelerated (top) and uniformly decelerated motion (bottom).

§ 6. UNIFORM CIRCULAR MOTION: CENTRIPETAL ACCELERATION.