The formula for moving with uniformly accelerated motion without time. Uniformly accelerated motion: formulas, examples

Rectilinear uniform motion is a motion in which a body travels the same distance in equal intervals of time.

Uniform movement- this is such a movement of the body in which its speed remains constant (), that is, it moves at the same speed all the time, and acceleration or deceleration does not occur ().

Rectilinear motion- this is the movement of the body in a straight line, that is, the trajectory we get is straight.

The speed of uniform rectilinear motion does not depend on time and at each point of the trajectory is directed in the same way as the movement of the body. That is, the velocity vector coincides with the displacement vector. With all this average speed in any period of time is equal to the initial and instantaneous speed:

Speed ​​of uniform rectilinear motion is a physical vector quantity equal to the ratio of the displacement of the body for any period of time to the value of this interval t:

from this formula. we can easily express body movement at uniform motion:

Consider the dependence of speed and displacement on time

Since our body moves in a straight line and uniformly accelerated (), then the graph with the dependence of speed on time will look like a parallel straight line to the time axis.

depending projections of body velocity versus time there is nothing complicated. The projection of the movement of the body is numerically equal to the area of ​​the rectangle AOBC, since the magnitude of the displacement vector is equal to the product of the velocity vector by the time during which the movement was made.

On the chart we see displacement versus time.

It can be seen from the graph that the velocity projection is equal to:

Considering this formula we can say that the larger the angle, the faster our body moves and it travels a greater distance in less time

In the previous lessons, we discussed how to determine the distance traveled with a uniform rectilinear motion. It's time to learn how to determine the coordinate of the body, the distance traveled and the displacement in a rectilinear uniformly accelerated motion. This can be done if we consider rectilinear uniformly accelerated motion as a set a large number very small uniform body movements.

The first to solve the problem of the location of the body at a certain point in time with accelerated motion was the Italian scientist Galileo Galilei (Fig. 1).

Rice. 1. Galileo Galilei (1564-1642)

He carried out his experiments with an inclined plane. Along the chute, he launched a ball, a musket bullet, 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 beating of his heart or by the pulse (Fig. 2).

Rice. 2. Experience of Galileo

Let's look at the speed graph uniformly accelerated rectilinear motion from time. You know this dependence, it is a straight line: .

Rice. 3. Definition of displacement in uniformly accelerated rectilinear motion

The speed graph is divided into small rectangular plots(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 for 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 with uniform movement is the total distance traveled.

Please note: from point to point, the speed will change, thus we will get the path traveled by the body precisely during rectilinear uniformly accelerated motion.

Note that with a rectilinear uniformly accelerated motion of the body, when the 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 area figure bounded by a graph of speed and time.

Rice. 4. The displacement modulus is equal to the distance traveled

Let's use mathematical formulas to calculate the area of ​​the specified figure.

Rice. 5 Illustration for area calculation

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, denoted by the letter . The height of the trapezoid is equal to, this is the period of time during which the movement occurred.

The final velocity discussed in the previous lesson can be written as the sum of the initial velocity and the contribution due to the constant acceleration of the body. It turns out the expression:

If you open the brackets, it becomes doubled. 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 experiments Galileo Galilei. Therefore, we can assume that it was this scientist who first made it possible to determine the location of a body in a 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 this expression is substituted into Galileo's equation, then we get the law according to which the coordinate of the body changes during rectilinear uniformly accelerated motion:

It should be remembered that the values ​​are the projections of the speed and acceleration on the selected axis. Therefore, they can be both positive and negative.

Conclusion

The next stage in the consideration of motion will be the study of motion along a curvilinear trajectory.

Bibliography

1. Kikoin I.K., Kikoin A.K. Physics: textbook for grade 9 high school. - M.: Enlightenment.
2. Peryshkin A.V., Gutnik E.M., Physics. Grade 9: textbook for general education. institutions/A. V. Peryshkin, E. M. Gutnik. - 14th ed., stereotype. - M.: Bustard, 2009. - 300.
3. Sokolovich Yu.A., Bogdanova G.S.. Physics: Handbook with examples of problem solving. - 2nd edition redistribution. - X .: Vesta: Publishing house "Ranok", 2005. - 464 p.

Additional recommended links to Internet resources

1. Internet portal "class-fizika.narod.ru" ()
2. Internet portal "videouroki.net" ()
3. Internet portal "foxford.ru" ()

Homework

1. Write down the formula by which the projection of the displacement vector of the body is determined during rectilinear uniformly accelerated motion.
2. A cyclist with an initial speed of 15 km/h has gone down a hill in 5 seconds. Determine the length of the slide if the cyclist was moving at a constant acceleration of 0.5 m/s^2 .
3. What is the difference between the dependences of displacement on time for uniform and uniformly accelerated motions?

When an accident occurs on the road, experts measure the braking distance. What for? To determine the vehicle speed at the start of braking and the acceleration during braking. All this is necessary to find out the causes of the accident: either the driver exceeded the speed, or the brakes were faulty, or everything is in order with the car, and the one who violated the rules is to blame traffic a pedestrian. How, knowing the deceleration time and braking distance, to determine the speed and acceleration of the body?

Learn about geometric sense displacement projections

In the 7th grade, you learned that for any movement, the path is numerically equal to the area of ​​\u200b\u200bthe figure under the graph of the dependence of the module of the speed of movement on the observation time. The situation is similar with the definition of the displacement projection (Fig. 29.1).

Let's get a formula for calculating the projection of the body displacement for the time interval from t: = 0 to t 2 = t. Consider a uniformly accelerated rectilinear motion, in which the initial velocity and acceleration have the same direction with the OX axis. In this case, the velocity projection graph has the form shown in Fig. 29.2, and the displacement projection is numerically equal to the area of ​​the trapezoid OABC:

On the graph, segment OA corresponds to the projection of the initial velocity v 0 x, segment BC corresponds to the projection of the final velocity v x , and segment OC corresponds to the time interval t. Replacing these segments with the corresponding ones physical quantities and given that s x = S OABC , we get a formula for determining the displacement projection:

Formula (1) is used to describe any uniformly accelerated rectilinear motion.

Determine the displacement of the body, the motion graph of which is shown in Fig. 29.1, b, 2 s and 4 s after the start of the countdown. Explain your answer.

We write the displacement projection equation

Let us exclude the variable v x from formula (1). To do this, recall that with uniformly accelerated rectilinear motion v x \u003d v 0 x + a x t. Substituting the expression for v x into formula (1), we get:

Thus, for a uniformly accelerated rectilinear motion, the displacement projection equation was obtained:

Rice. 29.3. The displacement projection graph for uniformly accelerated rectilinear motion is a parabola passing through the origin: if a x > 0, the branches of the parabola are directed upwards (a); if a x<0, ветви параболы направлены вниз (б)

Rice. 29.4. Choice of coordinate axis in case of rectilinear motion

So, the displacement projection graph for uniformly accelerated rectilinear motion is a parabola (Fig. 29.3), the top of which corresponds to the turning point:

Since the quantities v 0 x and a x do not depend on the observation time, the dependence s x (ί) is quadratic. For example, if

you can get another formula for calculating the projection of displacement for uniformly accelerated rectilinear motion:

Formula (3) is convenient to use if the condition of the problem does not refer to the time of motion of the body and it is not necessary to determine it.

Derive formula (3) yourself.

Please note: in each formula (1-3), the projections v x , v 0 x and a x can be both positive and negative - depending on how the vectors v, v 0 and a are directed relative to the OX axis.

Write down the coordinate equation

One of the main tasks of mechanics is to determine the position of the body (body coordinates) at any time. We are considering rectilinear motion, so it is enough to choose one coordinate axis (for example, the OX axis), which follows

direct along the movement of the body (Fig. 29.4). From this figure we see that, regardless of the direction of movement, the x-coordinate of the body can be determined by the formula:

Rice. 29.5. With uniformly accelerated rectilinear motion, the plot of the coordinate versus time is a parabola that intersects the x-axis at the point x 0

where x 0 is the initial coordinate (the coordinate of the body at the time of the start of observation); s x is the displacement projection.

therefore, for such a motion, the coordinate equation has the form:

For uniformly accelerated rectilinear motion

After analyzing the last equation, we conclude that the dependence x (t) is quadratic, so the coordinate graph is a parabola (Fig. 29.5).

Learning to solve problems

We will consider the main stages of solving problems for uniformly accelerated rectilinear motion using examples.

Problem solution example

Subsequence

action

1. Read the condition of the problem carefully. Determine which bodies take part in the movement, what is the nature of the movement of the bodies, what parameters of movement are known.

Problem 1. After the start of braking, the train went to a stop 225 m. What was the speed of the train before the start of braking? Consider that during deceleration the acceleration of the train is constant and equal to 0.5 m/s 2 .

In the explanatory figure, let's direct the OX axis in the direction of the train. As the train slows down,

2. Write down a brief condition of the problem. If necessary, convert the values ​​of physical quantities to SI units. 2

Problem 2. A pedestrian walks along a straight section of the road at a constant speed of 2 m/s. He is overtaken by a motorcycle, which increases its speed, moving with an acceleration of 2 m/s 3 . How long will it take for a motorcycle to overtake a pedestrian if, at the time of the start of the countdown, the distance between them was 300 m, and the motorcycle was moving at a speed of 22 m/s? How far will the bike travel in this time?

1. Read the condition of the problem carefully. Find out the nature of the movement of bodies, what parameters of movement are known.

Summing up

For uniformly accelerated rectilinear motion of a body: the projection of displacement is numerically equal to the area of ​​the figure under the graph of the projection of the velocity of motion - the graph of the dependence v x (ί):

3. Draw an explanatory drawing showing the coordinate axis, positions of bodies, directions of accelerations and velocities.

4. Write down the equation of the coordinate in general form; using the figure, specify this equation for each body.

5. Given that at the time of the meeting (overtaking) the coordinates of the bodies are the same, get a quadratic equation.

6. Solve the resulting equation and find the meeting time of the bodies.

7. Calculate the coordinate of the bodies at the time of the meeting.

8. Find the desired value and analyze the result.

9. Write down the answer.

this is the geometric meaning of displacement;

displacement projection equation has the form:

test questions

1. What formulas can be used to find the displacement projection s x for uniformly accelerated rectilinear motion? Derive these formulas. 2. Prove that the graph of body displacement versus observation time is a parabola. How are its branches directed? What moment of motion corresponds to the top of the parabola? 3. Write down the coordinate equation for uniformly accelerated rectilinear motion. What physical quantities are connected by this equation?

Exercise number 29

1. A skier moving at a speed of 1 m/s starts downhill. Determine the length of the descent if the skier rode it in 10 s. Consider that the skier's acceleration was unchanged and amounted to 0.5 m/s 2 .

2. The passenger train has changed its speed from 54 km/h to 5 m/s. Determine the distance that the train traveled during braking if the acceleration of the train was constant and amounted to 1 m / s 2.

3. The brakes of a car are in good condition if, at a speed of 8 m / s, its braking distance is 7.2 m. Determine the braking time and acceleration of the car.

4. The equations of coordinates of two bodies moving along the axis OX have the form:

1) For each body, determine: a) the nature of the movement; b) initial coordinate; c) module and direction of the initial velocity; d) acceleration.

2) Find the time and coordinate of the meeting of the bodies.

3) For each body, write down the equations v x (t) and s x (t), plot velocity and displacement projections.

5. In fig. 1 shows a graph of the projection of the speed of movement for some body.

Determine the path and displacement of the body in 4 s from the start of time. Write down the equation of the coordinate if at the time t = 0 the body was at a point with a coordinate of -20 m.

6. Two cars started moving from the same point in the same direction, with the second car leaving 20 seconds later. Both cars move uniformly with an acceleration of 0.4 m/s 2 . After what interval of time after the start of the movement of the first car, the distance between the cars will be 240 m?

7. In fig. 2 shows a graph of the dependence of the coordinate of the body on the time of its movement.

Write down the coordinate equation if it is known that the acceleration modulus is 1.6 m/s 2 .

8. An escalator in the subway rises at a speed of 2.5 m/s. Can a person on an escalator be at rest in a frame of reference linked to the Earth? If so, under what conditions? Is it possible under these conditions to consider the movement of a person as movement by inertia? Justify your answer.

This is textbook material.

How, knowing the stopping distance, determine the initial speed of the car and how, knowing the characteristics of the movement, such as the initial speed, acceleration, time, determine the movement of the car? We will get answers after we get acquainted with the topic of today's lesson: "Displacement with uniformly accelerated movement, the dependence of coordinates on time with uniformly accelerated movement"

With uniformly accelerated motion, the graph looks like a straight line going up, since its acceleration projection is greater than zero.

With uniform rectilinear motion, the area will be numerically equal to the modulus of the projection of the displacement of the body. It turns out that this fact can be generalized for the case not only of uniform motion, but also for any motion, that is, to show that the area under the graph is numerically equal to the displacement projection modulus. This is done strictly mathematically, but we will use a graphical method.

Rice. 2. Graph of the dependence of speed on time with uniformly accelerated movement ()

Let's divide the graph of the projection of speed from time for uniformly accelerated motion into small time intervals Δt. Let us assume that they are so small that during their length the speed practically did not change, that is, we will conditionally turn the linear dependence graph in the figure into a ladder. At each of its steps, we believe that the speed has not changed much. Imagine that we make the time intervals Δt infinitely small. In mathematics they say: we make a passage to the limit. In this case, the area of ​​such a ladder will indefinitely closely coincide with the area of ​​the trapezoid, which is limited by the graph V x (t). And this means that for the case of uniformly accelerated motion, we can say that the displacement projection module is numerically equal to the area bounded by the graph V x (t): the abscissa and ordinate axes and the perpendicular lowered to the abscissa axis, that is, the area of ​​the trapezoid OABS, which we see in figure 2.

The problem turns from a physical one into a mathematical one - finding the area of ​​a trapezoid. This is a standard situation when physicists make a model that describes a particular phenomenon, and then mathematics comes into play, which enriches this model with equations, laws - 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 shapes - 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 projection values, we get: 1/2t (V x - V 0x), and, remembering the law of change of speed from time with uniformly accelerated motion: V x (t) = V 0x + a x t, it is quite obvious that the difference in projections of velocities is equal to the product of the projection of acceleration a x by time t, that is, V x - V 0x = a x t.

Rice. 3. Determining the area of ​​a trapezoid ( Source)

Taking into account the fact that the area of ​​the trapezoid is numerically equal to the displacement projection module, we get:

S x (t) \u003d V 0 x t + a x t 2 / 2

We have obtained the law of the dependence of the projection of displacement on time with uniformly accelerated motion in scalar form, in vector form it will look like this:

(t) = t + t 2 / 2

Let's derive one more formula for the displacement projection, which will not include time as a variable. We solve the system of equations, excluding time from it:

S x (t) \u003d V 0 x + a x t 2 / 2

V x (t) \u003d V 0 x + a x t

Imagine that we do not know the time, then we will express the time from the second equation:

t \u003d V x - V 0x / a x

Substitute the resulting value into the first equation:

We get such a cumbersome expression, we square it and give similar ones:

We have obtained a very convenient displacement projection expression for the case when we do not know the time of motion.

Let us have the initial speed of the car, when braking began, is V 0 \u003d 72 km / h, final speed V \u003d 0, acceleration a \u003d 4 m / s 2. Find out the length of the braking distance. Converting kilometers to meters and substituting the values ​​into the formula, we get that the stopping distance will be:

S x \u003d 0 - 400 (m / s) 2 / -2 4 m / s 2 \u003d 50 m

Let's analyze the following formula:

S x \u003d (V 0 x + V x) / 2 t

The projection of movement is half the sum of the projections of the initial and final speeds, multiplied by the time of movement. Recall the displacement formula for average speed

S x \u003d V cf t

In the case of uniformly accelerated movement, the average speed will be:

V cf \u003d (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) \u003d x 0 + V 0 x t + a x t 2 / 2

In order to learn how to use this law, we will analyze a typical problem.

The car, moving from a state of rest, acquires an acceleration of 2 m / s 2. Find the distance traveled by the car in 3 seconds and in the third second.

Given: V 0 x = 0

Let us write down the law according to which the displacement changes with time at

uniformly accelerated motion: S x \u003d V 0 x t + a x t 2 /2. 2 c< Δt 2 < 3.

We can answer the first question of the problem by plugging in the data:

t 1 \u003d 3 c S 1x \u003d a x t 2 / 2 \u003d 2 3 2 / 2 \u003d 9 (m) - this is the path that went

c car in 3 seconds.

Find out how far he traveled in 2 seconds:

S x (2 s) \u003d a x t 2 / 2 \u003d 2 2 2 / 2 \u003d 4 (m)

So, you and I know that in two seconds the car drove 4 meters.

Now, knowing these two distances, we can find the path that he traveled in the third second:

S 2x \u003d S 1x + S x (2 s) \u003d 9 - 4 \u003d 5 (m)

Uniformly accelerated motion is a motion with acceleration, the vector of which does not change in magnitude and direction. Examples of such movement: a bicycle that rolls down a hill; a stone thrown at an angle to the horizon.

Let's consider the last case in more detail. At any point of the trajectory, the free fall acceleration g → acts on the stone, which does not change in magnitude and is always directed in one direction.

The motion of a body thrown at an angle to the horizon can be represented as the sum of motions about the vertical and horizontal axes.

Along the X axis the motion is uniform and rectilinear, and along the Y axis it is uniformly accelerated and rectilinear. We will consider the projections of the velocity and acceleration vectors on the axis.

Formula for speed with uniformly accelerated motion:

Here v 0 is the initial speed of the body, a = c o n s t is the acceleration.

Let us show on the graph that with uniformly accelerated motion, the dependence v (t) has the form of a straight line.

Acceleration can be determined from the slope of the velocity graph. In the figure above, the acceleration modulus is equal to the ratio of the sides of the triangle ABC.

a = v - v 0 t = B C A C

The larger the angle β, the greater the slope (steepness) of the graph with respect to the time axis. Accordingly, the greater the acceleration of the body.

For the first graph: v 0 = - 2 m s; a \u003d 0, 5 m s 2.

For the second graph: v 0 = 3 m s; a = - 1 3 m s 2 .

From this graph, you can also calculate the movement of the body in time t. How to do it?

Let's single out a small time interval ∆ t on the graph. We will assume that it is so small that the movement during the time ∆ t can be considered uniform movement with a speed equal to the speed of the body in the middle of the interval ∆ t . Then, the displacement ∆ s during the time ∆ t will be equal to ∆ s = v ∆ t .

Let's divide all time t into infinitely small intervals ∆ t . The displacement s in time t is equal to the area of ​​the trapezoid O D E F .

s = O D + E F 2 O F = v 0 + v 2 t = 2 v 0 + (v - v 0) 2 t .

We know that v - v 0 = a t , so the final formula for moving the body will be:

s = v 0 t + a t 2 2

In order to find the coordinate of the body at a given time, you need to add displacement to the initial coordinate of the body. A change in coordinates during uniformly accelerated motion expresses the law of uniformly accelerated motion.

Law of uniformly accelerated motion

Law of uniformly accelerated motion

y = y 0 + v 0 t + a t 2 2 .

Another common problem that arises in the analysis of uniformly accelerated motion is finding the displacement for given values ​​of the initial and final velocities and acceleration.

Eliminating t from the above equations and solving them, we obtain:

s \u003d v 2 - v 0 2 2 a.

From the known initial speed, acceleration and displacement, you can find the final speed of the body:

v = v 0 2 + 2 a s .

For v 0 = 0 s = v 2 2 a and v = 2 a s

Important!

The values ​​v , v 0 , a , y 0 , s included in the expressions are algebraic quantities. Depending on the nature of the movement and the direction of the coordinate axes in a particular task, they can take both positive and negative values.

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