Acceleration without time. Physics acceleration formulas: linear and centripetal acceleration

However, the body could start uniformly accelerated motion not from a state of rest, but already possessing some speed (or it was given an initial speed). Let's say you throw a stone vertically down from a tower with force. Such a body is subjected to an acceleration free fall, equal to 9.8 m/s2. However, your strength has given the stone even more speed. Thus, the final speed (at the moment of touching the ground) will be the sum of the speed developed as a result of acceleration and the initial speed. Thus, the final speed will be found by the formula:

at = v - v0
a = (v – v0)/t

In case of braking:

at = v0 - v
a = (v0 – v)/t

Now we derive

s = ½ * (v0 + v) * t

§ 5. Acceleration

The next step on the way to the equations of motion is the introduction of a quantity that is associated with a change in the speed of motion. It is natural to ask: how does the speed of movement change? In the previous chapters, we considered the case when the acting force led to a change in speed. There are passenger cars that pick up from a standstill for speed. Knowing this, we can determine how the speed changes, but only on average. Let's get on with the next one difficult question: how to know the rate of change of speed. In other words, how many meters per second does the speed change in . We have already established that the speed of a falling body changes with time according to the formula (see Table 8.4), and now we want to find out how much it changes in . This quantity is called acceleration.

Thus, acceleration is defined as the rate of change of speed. With all that said before, we are already sufficiently prepared to immediately write down acceleration as a derivative of velocity, just as speed is written as a derivative of distance. If we now differentiate the formula , then we get the acceleration of the falling body

(When differentiating this expression, we used the result we obtained earlier. We saw that the derivative of is equal to just (constant). If we choose this constant equal to 9.8, then we immediately find that the derivative of is equal to 9.8.) This means, that the speed of a falling body is constantly increasing by every second. The same result can be obtained from Table. 8.4. As you can see, in the case of a falling body, everything turns out quite simply, but the acceleration, generally speaking, is not constant. It turned out to be constant only because the force acting on the falling body is constant, and according to Newton's law, acceleration should be proportional to the force.

As the next example, let's find the acceleration in the problem that we have already dealt with when studying speed:

.

For speed, we got the formula

Since acceleration is the derivative of speed with respect to time, in order to find its value, you need to differentiate this formula. Let us now recall one of the rules of Table. 8.3, namely that the derivative of the sum is equal to the sum of the derivatives. To differentiate the first of these terms, we will not go through the entire long procedure that we did before, but simply recall that we encountered such a quadratic term when differentiating the function , and as a result, the coefficient doubled, and turned into . You can see for yourself that the same thing will happen now. Thus, the derivative of will be equal to . We now turn to differentiation of the second term. According to one of the rules of Table. 8.3 the derivative of the constant will be zero, therefore, this term will not give any contribution to the acceleration. Final Result: .

We derive two more useful formulas that are obtained by integration. If a body is moving from rest with a constant acceleration, then its speed at any moment of time will be equal to

and the distance traveled by him up to this point in time,

Note also that since the speed is , and the acceleration is the derivative of the speed with respect to time, we can write

. (8.10)

So now we know how the second derivative is written.

There is, of course, Feedback between acceleration and distance, which simply follows from the fact that . Since distance is an integral of velocity, it can be found by double integrating the acceleration. All the previous consideration was devoted to motion in one dimension, and now we will briefly dwell on motion in the space of three dimensions. Consider the motion of a particle in three-dimensional space. This chapter began with a discussion of one-dimensional motion passenger car, namely, from the question, at what distance from the beginning of the movement is the car at different points in time. We then discussed the relationship between speed and change in distance over time, and the relationship between acceleration and change in speed. Let's analyze the movement in three dimensions in the same sequence. It is easier, however, to start with a more illustrative two-dimensional case, and only then generalize it to the case of three dimensions. Let's draw two lines intersecting at right angles (coordinate axes) and we will set the position of the particle at any moment of time by the distances from it to each of the axes. Thus, the position of the particle is given by two numbers (coordinates) and , each of which is, respectively, the distance to the axis and to the axis (Fig. 8.3). Now we can describe the movement, for example, making a table in which these two coordinates are given as functions of time. (A generalization to the three-dimensional case requires the introduction of another axis perpendicular to the first two, and the measurement of one more coordinate. However, now the distances are taken not to the axes, but to the coordinate planes.) How to determine the speed of a particle? To do this, we first find the velocity components in each direction, or its components. The horizontal component of the velocity, or -component, will be equal to the time derivative of the coordinate , i.e.

and the vertical component, or -component, is equal to

In the case of three dimensions, you must also add

Figure 8.3. Description of the motion of a body on a plane and calculation of its velocity.

How, knowing the components of the speed, to determine the total speed in the direction of motion? Consider in the two-dimensional case two successive positions of a particle separated by a short time interval and a distance . From FIG. 8.3 shows that

(8.14)

(The symbol corresponds to the expression "approximately equal".) The average speed over the interval is obtained by simply dividing: . To find the exact speed at the moment , it is necessary, as already done at the beginning of the chapter, to tend to zero. As a result, it turns out that

. (8.15)

In the three-dimensional case, in exactly the same way, one can obtain

(8.16)

Figure 8.4. A parabola described by a falling body thrown with a horizontal initial velocity.

We define accelerations in the same way as velocities: the -component of acceleration is defined as the derivative of the -component of velocity (i.e., the second derivative with respect to time), etc.

Let's take another look interesting example mixed motion on a plane. Let the ball move in a horizontal direction with a constant speed and at the same time fall vertically downwards with a constant acceleration. What is this movement? Since and, therefore, the speed is constant, then

and since the downward acceleration is constant and equal to - , then the coordinate of the falling ball is given by the formula

What curve does our ball describe, that is, what is the relationship between the coordinates and? From equation (8.18), according to (8.17), time can be excluded, since 1 \u003d * x / u% after which we find

Uniformly accelerated motion without initial velocity

This relationship between the coordinates and can be considered as an equation for the trajectory of the ball. Ordered to depict it graphically, then we get a curve, which is called a parabola (Fig. 8.4). So any freely falling body, being thrown in some direction, moves along a parabola.

With rectilinear uniformly accelerated motion body

1. moves along a conventional straight line,
2. its speed gradually increases or decreases,
3. in equal intervals of time, the speed changes by an equal amount.

For example, a car from a state of rest begins to move along a straight road, and up to a speed of, say, 72 km / h, it moves with uniform acceleration. When the set speed is reached, the car moves without changing speed, i.e. evenly. With uniformly accelerated movement, its speed increased from 0 to 72 km/h. And let the speed increase by 3.6 km/h for every second of movement. Then the time of uniformly accelerated movement of the car will be equal to 20 seconds. Since acceleration in SI is measured in meters per second squared, the acceleration of 3.6 km / h per second must be converted to the appropriate units of measurement. It will be equal to (3.6 * 1000 m) / (3600 s * 1 s) = 1 m / s2.

Let's say that after some time of driving at a constant speed, the car began to slow down to stop. The movement during braking was also uniformly accelerated (for equal periods of time, the speed decreased by the same amount). In this case, the acceleration vector will be opposite to the velocity vector. We can say that the acceleration is negative.

So, if the initial speed of the body is zero, then its speed after a time of t seconds will be equal to the product of the acceleration by this time:

When a body falls, the acceleration of free fall "works", and the speed of the body at the very surface of the earth will be determined by the formula:

If you know the current speed of the body and the time it took to develop such a speed from rest, then you can determine the acceleration (i.e., how quickly the speed changed) by dividing the speed by the time:

However, the body could start uniformly accelerated motion not from a state of rest, but already possessing some speed (or it was given an initial speed).

Let's say you throw a stone vertically down from a tower with force. Such a body is affected by the free fall acceleration equal to 9.8 m/s2. However, your strength has given the stone even more speed. Thus, the final speed (at the moment of touching the ground) will be the sum of the speed developed as a result of acceleration and the initial speed. Thus, the final speed will be found by the formula:

However, if the stone was thrown up. Then its initial speed is directed upwards, and the acceleration of free fall is downwards. That is, the velocity vectors are directed in opposite directions. In this case (and also during braking), the product of acceleration and time must be subtracted from the initial speed:

We obtain from these formulas the acceleration formulas. In case of acceleration:

at = v - v0
a = (v – v0)/t

In case of braking:

at = v0 - v
a = (v0 – v)/t

In the case when the body stops with uniform acceleration, then at the moment of stopping its speed is 0. Then the formula is reduced to this form:

Knowing the initial speed of the body and the acceleration of deceleration, the time after which the body will stop is determined:

Now we derive formulas for the path that a body travels during rectilinear uniformly accelerated motion. Graph the dependence of speed on time for a straight line uniform motion is a segment parallel to the time axis (usually the x-axis is taken). The path is calculated as the area of ​​the rectangle under the segment.

How to find the acceleration, knowing the path and time?

That is, by multiplying the speed by the time (s = vt). With rectilinear uniformly accelerated motion, the graph is straight, but not parallel to the time axis. This straight line either increases in the case of acceleration or decreases in the case of deceleration. However, the path is also defined as the area of ​​the figure under the graph.

With rectilinear uniformly accelerated motion, this figure is a trapezoid. Its bases are a segment on the y-axis (velocity) and a segment connecting the end point of the graph with its projection on the x-axis. The sides are the velocity versus time graph itself and its projection onto the x-axis (time axis). The projection on the x-axis is not only the side, but also the height of the trapezoid, since it is perpendicular to its bases.

As you know, the area of ​​a trapezoid is half the sum of the bases times the height. The length of the first base is equal to the initial speed (v0), the length of the second base is equal to the final speed (v), the height is equal to the time. Thus we get:

s = ½ * (v0 + v) * t

Above, the formula for the dependence of the final speed on the initial and acceleration (v = v0 + at) was given. Therefore, in the path formula, we can replace v:

s = ½ * (v0 + v0 + at) * t = ½ * (2v0 + at) * t = ½ * t * 2v0 + ½ * t * at = v0t + 1/2at2

So, the distance traveled is determined by the formula:

(This formula can be arrived at by considering not the area of ​​the trapezoid, but by summing the areas of the rectangle and right triangle into which the trapezoid is divided.)

If the body began to move uniformly accelerated from rest (v0 = 0), then the path formula is simplified to s = at2/2.

If the acceleration vector was opposite to the speed, then the product at2/2 must be subtracted. It is clear that in this case the difference between v0t and at2/2 should not become negative. When will she become zero, the body will stop. The braking path will be found. Above was the formula for the time to a complete stop (t = v0/a). If we substitute the value t in the path formula, then the braking path is reduced to the following formula:

I. Mechanics

Physics->Kinematics->uniformly accelerated motion->

Online testing

Uniformly accelerated motion

In this topic, we will consider a very special kind of non-uniform motion. Based on the opposition to uniform motion, uneven movement- this is movement at an unequal speed, along any trajectory. What is the characteristic of uniformly accelerated motion? This is an uneven movement, but which "equally accelerating". Acceleration is associated with an increase in speed. Remember the word "equal", we get an equal increase in speed. And how to understand "an equal increase in speed", how to evaluate the speed is equally increasing or not? To do this, we need to detect the time, estimate the speed through the same time interval. For example, a car starts moving, in the first two seconds it develops a speed of up to 10 m/s, in the next two seconds 20 m/s, after another two seconds it is already moving at a speed of 30 m/s. Every two seconds, the speed increases and each time by 10 m/s. This is uniformly accelerated motion.

The physical quantity that characterizes how much each time the speed increases is called acceleration.

Can a cyclist's movement be considered uniformly accelerated if, after stopping, his speed is 7 km/h in the first minute, 9 km/h in the second, and 12 km/h in the third? It is forbidden! The cyclist accelerates, but not equally, first accelerating by 7 km/h (7-0), then by 2 km/h (9-7), then by 3 km/h (12-9).

Usually, the movement with increasing speed is called accelerated movement. The movement is at a decreasing speed - slow motion. But physicists call any motion with a changing speed accelerated motion. Whether the car starts off (speed increases!), or slows down (speed decreases!), in any case, it moves with acceleration.

Uniformly accelerated motion- this is such a movement of the body, in which its speed for any equal intervals of time changes(may increase or decrease) equally

body acceleration

Acceleration characterizes the rate of change of speed. This is the number by which the speed changes every second. If the modulo acceleration of the body is large, this means that the body quickly picks up speed (when it accelerates) or quickly loses it (when decelerating). Acceleration- This is a physical vector quantity, numerically equal to the ratio of the change in speed to the period of time during which this change occurred.

Let's determine the acceleration in the following problem. At the initial moment of time, the speed of the ship was 3 m/s, at the end of the first second the speed of the ship became 5 m/s, at the end of the second - 7 m/s, at the end of the third - 9 m/s, etc. Obviously, . But how do we determine? We consider the speed difference in one second. In the first second 5-3=2, in the second second 7-5=2, in the third 9-7=2. But what if the speeds are not given for every second? Such a task: the initial speed of the ship is 3 m/s, at the end of the second second - 7 m/s, at the end of the fourth 11 m/s. In this case, 11-7= 4, then 4/2=2. We divide the speed difference by the time interval.

This formula is most often used in solving problems in a modified form:

The formula is not written in vector form, so we write the "+" sign when the body accelerates, the "-" sign - when it slows down.

Direction of the acceleration vector

The direction of the acceleration vector is shown in the figures

In this figure, the car is moving in a positive direction along the Ox axis, the velocity vector always coincides with the direction of movement (directed to the right).

How to find the acceleration knowing the initial and final speed and path?

When the acceleration vector coincides with the direction of speed, this means that the car is accelerating. The acceleration is positive.

During acceleration, the direction of acceleration coincides with the direction of speed. The acceleration is positive.

In this picture, the car is moving in the positive direction on the Ox axis, the velocity vector is the same as the direction of movement (rightward), the acceleration is NOT the same as the direction of the speed, which means that the car is decelerating. The acceleration is negative.

When braking, the direction of acceleration is opposite to the direction of speed. The acceleration is negative.

Let's figure out why the acceleration is negative when braking. For example, in the first second, the ship dropped speed from 9m/s to 7m/s, in the second second to 5m/s, in the third to 3m/s. The speed changes to "-2m/s". 3-5=-2; 5-7=-2; 7-9=-2m/s. That's where it comes from negative meaning acceleration.

When solving problems, if the body slows down, the acceleration in the formulas is substituted with a minus sign!!!

Moving with uniformly accelerated motion

An additional formula called untimely

Formula in coordinates

Communication with medium speed

With uniformly accelerated motion average speed can be calculated as the arithmetic mean of the initial and final speed

From this rule follows a formula that is very convenient to use when solving many problems

Path ratio

If the body moves uniformly accelerated, the initial speed is zero, then the paths traveled in successive equal time intervals are related as a series of odd numbers.

The main thing to remember

1) What is uniformly accelerated motion;
2) What characterizes acceleration;
3) Acceleration is a vector. If the body accelerates, the acceleration is positive; if it slows down, the acceleration is negative;
3) Direction of the acceleration vector;
4) Formulas, units of measurement in SI

Exercises

Two trains go towards each other: one accelerates to the north, the other decelerates to the south. How are train accelerations directed?

Same to the north. Because the acceleration of the first train coincides in direction with the movement, and the second has the opposite movement (it slows down).

The train moves uniformly with acceleration a (a>0). It is known that by the end of the fourth second the speed of the train is 6m/s. What can be said about the distance traveled in the fourth second? Will this path be greater than, less than or equal to 6m?

Since the train is moving with acceleration, its speed increases all the time (a>0). If by the end of the fourth second the speed is 6m/s, then at the beginning of the fourth second it was less than 6m/s. Therefore, the distance traveled by the train in the fourth second is less than 6m.

Which of the following dependencies describe uniformly accelerated motion?

Equation of the speed of a moving body. What is the corresponding path equation?

* The car traveled 1m in the first second, 2m in the second second, 3m in the third second, 4m in the fourth second, and so on. Can such a movement be considered uniformly accelerated?

In uniformly accelerated motion, the paths traveled in successive equal time intervals are related as a successive series of odd numbers. Therefore, the described motion is not uniformly accelerated.

The term "acceleration" is one of the few whose meaning is clear to those who speak Russian. It denotes the value by which the velocity vector of a point is measured in its direction and numerical value. Acceleration depends on the force applied to this point, it is directly proportional to it, but inversely proportional to the mass of this very point. Here are the main criteria for how to find acceleration.

It follows from where exactly the acceleration is applied. Recall that it is denoted as "a". In the international system of units, it is customary to consider a unit of acceleration as a value that consists of an indicator of 1 m / s 2 (meter per second squared): acceleration at which for every second the speed of a body changes by 1 m per second (1 m / s). Let's say the acceleration of the body is 10m / s 2. So, for every second, its speed changes by 10 m/s. Which is 10 times faster if the acceleration were 1m/s 2 . In other words, speed means physical quantity characterizing the path traveled by the body, for certain time.

Answering the question of how to find acceleration, you need to know the path of the body, its trajectory - straight or curvilinear, and the speed - uniform or uneven. Concerning the last characteristic. those. speed, it must be remembered that it can vary vectorially or modulo, thereby imparting acceleration to the movement of the body.

Why do we need an acceleration formula

Here is an example of how to find the acceleration in terms of speed, if the body starts uniformly accelerated motion: you need to divide the change in speed by the period of time during which the change in speed occurred. It will help to solve the problem of how to find the acceleration, the acceleration formula a = (v -v0) / ?t = ?v / ?t, where the initial speed of the body is v0, the final speed is v, the time interval is ?t.

On the specific example it looks like this: let's say the car starts moving, pulling away, and in 7 seconds picks up a speed of 98 m/s. Using the above formula, the acceleration of the car is determined, i.e. taking the initial data v = 98 m/s, v0 = 0, ?t = 7s, we need to find what a is equal to. Here is the answer: a \u003d (v-v0) / ?t \u003d (98m / s - 0m / s) / 7s \u003d 14 m / s 2. We get 14 m / s 2.

Search for free fall acceleration

How to find free fall acceleration? The very principle of the search is clearly visible in this example. It is enough to take a metal body, i.e. an object made of metal, fix it at a height that can be measured in meters, and when choosing a height, air resistance must be taken into account, moreover, one that can be neglected. Optimally, this is a height of 2-4 m. A platform should be installed below, specifically for this item. Now you can detach the metal body from the bracket. Naturally, it will begin a free fall. It is necessary to fix the landing time of the body in seconds. Everything, you can find the acceleration of an object in free fall. To do this, the given height must be divided by the flight time of the body. Only this time must be taken in the second degree. The result obtained should be multiplied by 2. This will be the acceleration, more precisely, the value of the acceleration of the body in free fall, expressed in m / s 2.

It is possible to determine the acceleration due to gravity using the force of gravity. Having measured the weight of the body in kg with the scales, observing the utmost accuracy, then hang this body on a dynamometer. The resulting force of gravity will be in newtons. By dividing the value of gravity by the mass of the body that has just been hung on a dynamometer, you get the acceleration of free fall.

Acceleration determines the pendulum

It will help to establish the acceleration of free fall and the mathematical pendulum. It is a body fixed and suspended on a thread of sufficient length, which is measured in advance. Now we need to bring the pendulum into a state of oscillation. And with the help of a stopwatch, count the number of oscillations in a certain time. Then divide this fixed number of oscillations by the time (it is in seconds). Raise the number obtained after division to the second power, multiply by the length of the pendulum thread and the number 39.48. Result: the acceleration of free fall was determined.

Instruments for measuring acceleration

It is logical to complete this information block about acceleration by saying that it is measured by special devices: accelerometers. They are mechanical, electromechanical, electrical and optical. The range that they can do is from 1 cm / s 2 to 30 km / s 2, which means O, OOlg - 3000g. If you use Newton's second law, you can calculate the acceleration by finding the quotient of dividing the force F acting on a point by its mass m: a=F/m.

All tasks in which there is movement of objects, their movement or rotation, are somehow connected with speed.

This term characterizes the movement of an object in space over a certain period of time - the number of units of distance per unit of time. He is a frequent "guest" of both sections of mathematics and physics. The original body can change its location both uniformly and with acceleration. In the first case, the speed is static and does not change during the movement, in the second, on the contrary, it increases or decreases.

How to find speed - uniform motion

If the speed of the movement of the body remained unchanged from the beginning of the movement to the end of the path, then we are talking about moving with constant acceleration - uniform motion. It can be straight or curved. In the first case, the trajectory of the body is a straight line.

Then V=S/t, where:

• V is the desired speed,
• S - distance traveled (total path),
• t is the total time of movement.

How to find speed - acceleration is constant

If an object was moving with acceleration, then its speed changed as it moved. In this case, the expression will help to find the desired value:

V \u003d V (beginning) + at, where:

• V (beginning) - the initial speed of the object,
• a is the acceleration of the body,
• t is the total travel time.

How to find speed - uneven motion

In this case, there is a situation when the body passes different parts of the path in different times.
S(1) - for t(1),
S(2) - for t(2), etc.

On the first section, the movement took place at a “tempo” V(1), on the second - V(2), and so on.

To find out the speed of an object moving all the way (its average value), use the expression:

How to find speed - rotation of an object

In the case of rotation, we are talking about the angular velocity, which determines the angle through which the element rotates per unit of time. The desired value is denoted by the symbol ω (rad / s).

• ω = Δφ/Δt, where:

Δφ – passed angle (angle increment),
Δt - elapsed time (movement time - time increment).

• If the rotation is uniform, the desired value (ω) is associated with such a concept as the period of rotation - how long will it take for our object to complete 1 complete revolution. In this case:

ω = 2π/T, where:
π is a constant ≈3.14,
T is the period.

Or ω = 2πn, where:
π is a constant ≈3.14,
n is the frequency of circulation.

• With the known linear speed of the object for each point on the path of motion and the radius of the circle along which it moves, the following expression is required to find the speed ω:

ω = V/R, where:
V is the numerical value of the vector quantity (linear velocity),
R is the radius of the body's trajectory.

How to find speed - approaching and moving away points

In such tasks, it would be appropriate to use the terms approach speed and distance speed.

If the objects are heading towards each other, then the speed of approach (retreat) will be as follows:
V (approach) = V(1) + V(2), where V(1) and V(2) are the velocities of the corresponding objects.

If one of the bodies catches up with the other, then V (closer) = V(1) - V(2), V(1) is greater than V(2).

How to find speed - movement on a body of water

If events unfold on the water, then the speed of the current (i.e., the movement of water relative to a fixed shore) is added to the object’s own speed (movement of the body relative to the water). How are these concepts related?

In the case of moving downstream, V=V(own) + V(tech).
If against the current - V \u003d V (own) - V (flow).

In this lesson, we will consider an important characteristic of uneven movement - acceleration. In addition, we will consider non-uniform motion with constant acceleration. This movement is also called uniformly accelerated or uniformly slowed down. Finally, we will talk about how to graphically depict the speed of a body as a function of time in uniformly accelerated motion.

Homework

By solving the tasks for this lesson, you will be able to prepare for questions 1 of the GIA and questions A1, A2 of the Unified State Examination.

1. Tasks 48, 50, 52, 54 sb. tasks of A.P. Rymkevich, ed. 10.

2. Write down the dependences of the 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 the turning points on the graphs, if any.

3. Consider the following questions and their answers:

Question. Is gravitational acceleration an acceleration as defined above?

Answer. Of course it is. Free fall acceleration is the acceleration of a body that falls freely 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 uniformly in a circle.

Question. Is it possible to calculate the tangent of the angle of inclination using a protractor and a calculator?

Answer. Not! Because the acceleration obtained in this way will be dimensionless, and the dimension of acceleration, as we showed earlier, must have the dimension of m/s 2 .

Question. What can be said about motion if the graph of speed versus time is not a straight line?

Answer. We can say that the acceleration of this body changes with time. Such a movement will not be uniformly accelerated.

In a rectilinear uniformly accelerated motion of the body

1. moves along a conventional straight line,
2. its speed gradually increases or decreases,
3. in equal intervals of time, the speed changes by an equal amount.

For example, a car from a state of rest begins to move along a straight road, and up to a speed of, say, 72 km / h, it moves with uniform acceleration. When the set speed is reached, the car moves without changing speed, i.e. evenly. With uniformly accelerated movement, its speed increased from 0 to 72 km/h. And let the speed increase by 3.6 km/h for every second of movement. Then the time of uniformly accelerated movement of the car will be equal to 20 seconds. Since acceleration in SI is measured in meters per second squared, the acceleration of 3.6 km / h per second must be converted to the appropriate units of measurement. It will be equal to (3.6 * 1000 m) / (3600 s * 1 s) \u003d 1 m / s 2.

Let's say that after some time of driving at a constant speed, the car began to slow down to stop. The movement during braking was also uniformly accelerated (for equal periods of time, the speed decreased by the same amount). In this case, the acceleration vector will be opposite to the velocity vector. We can say that the acceleration is negative.

So, if the initial speed of the body is zero, then its speed after a time of t seconds will be equal to the product of the acceleration by this time:

When a body falls, the acceleration of free fall "works", and the speed of the body at the very surface of the earth will be determined by the formula:

If you know the current speed of the body and the time it took to develop such a speed from rest, then you can determine the acceleration (i.e., how quickly the speed changed) by dividing the speed by the time:

However, the body could start uniformly accelerated motion not from a state of rest, but already possessing some speed (or it was given an initial speed). Let's say you throw a stone vertically down from a tower with force. Such a body is affected by the acceleration of free fall, equal to 9.8 m / s 2. However, your strength has given the stone even more speed. Thus, the final speed (at the moment of touching the ground) will be the sum of the speed developed as a result of acceleration and the initial speed. Thus, the final speed will be found by the formula:

However, if the stone was thrown up. Then its initial speed is directed upwards, and the acceleration of free fall is downwards. That is, the velocity vectors are directed in opposite directions. In this case (and also during braking), the product of acceleration and time must be subtracted from the initial speed:

We obtain from these formulas the acceleration formulas. In case of acceleration:

at = v – v0
a \u003d (v - v 0) / t

In case of braking:

at = v 0 – v
a \u003d (v 0 - v) / t

In the case when the body stops with uniform acceleration, then at the moment of stopping its speed is 0. Then the formula is reduced to this form:

Knowing the initial speed of the body and the acceleration of deceleration, the time after which the body will stop is determined:

Now we derive formulas for the path that a body travels during rectilinear uniformly accelerated motion. A graph of the dependence of speed on time for rectilinear uniform motion is a segment parallel to the time axis (usually the x-axis is taken). The path is calculated as the area of ​​the rectangle under the segment. That is, by multiplying the speed by the time (s = vt). With rectilinear uniformly accelerated motion, the graph is straight, but not parallel to the time axis. This straight line either increases in the case of acceleration or decreases in the case of deceleration. However, the path is also defined as the area of ​​the figure under the graph.

With rectilinear uniformly accelerated motion, this figure is a trapezoid. Its bases are a segment on the y-axis (velocity) and a segment connecting the end point of the graph with its projection on the x-axis. The sides are the velocity versus time graph itself and its projection onto the x-axis (time axis). The projection on the x-axis is not only the side, but also the height of the trapezoid, since it is perpendicular to its bases.

As you know, the area of ​​a trapezoid is half the sum of the bases times the height. The length of the first base is equal to the initial speed (v 0), the length of the second base is equal to the final speed (v), the height is equal to the time. Thus we get:

s \u003d ½ * (v 0 + v) * t

Above, the formula for the dependence of the final speed on the initial and acceleration was given (v \u003d v 0 + at). Therefore, in the path formula, we can replace v:

s = ½ * (v 0 + v 0 + at) * t = ½ * (2v 0 + at) * t = ½ * t * 2v 0 + ½ * t * at = v 0 t + 1/2at 2

So, the distance traveled is determined by the formula:

s = v 0 t + at 2 /2

(This formula can be arrived at by considering not the area of ​​the trapezoid, but by summing the areas of the rectangle and right triangle into which the trapezoid is divided.)

If the body began to move uniformly accelerated from rest (v 0 \u003d 0), then the path formula is simplified to s \u003d at 2 /2.

If the acceleration vector was opposite to the speed, then the product at 2 /2 must be subtracted. It is clear that in this case the difference v 0 t and at 2 /2 should not become negative. When it becomes equal to zero, the body will stop. The braking path will be found. Above was the formula for the time to a complete stop (t \u003d v 0 /a). If we substitute the value t in the path formula, then the braking path is reduced to such a formula.