Acceleration- a physical vector quantity that characterizes how quickly a body (material point) changes the speed of its movement. Acceleration is an important kinematic characteristic of a material point.
The simplest type of motion is uniform motion in a straight line, when the speed of the body is constant and the body covers the same path in any equal intervals of time.
But most movements are uneven. In some areas the body speed is greater, in others less. As the car begins to move, it moves faster and faster. and when stopping it slows down.
Acceleration characterizes the rate of change in speed. If, for example, the acceleration of a body is 5 m/s 2, then this means that for every second the speed of the body changes by 5 m/s, i.e. 5 times faster than with an acceleration of 1 m/s 2.
If the speed of a body during uneven motion changes equally over any equal periods of time, then the motion is called uniformly accelerated.
The SI unit of acceleration is the acceleration at which for every second the speed of the body changes by 1 m/s, i.e. meter per second per second. This unit is designated 1 m/s2 and is called “meter per second squared”.
Like speed, the acceleration of a body is characterized not only by its numerical value, but also by its direction. This means that acceleration is also a vector quantity. Therefore, in the pictures it is depicted as an arrow.
If the speed of a body during uniformly accelerated linear motion increases, then the acceleration is directed in the same direction as the speed (Fig. a); if the speed of the body decreases during a given movement, then the acceleration is directed in the opposite direction (Fig. b).
Average and instantaneous acceleration
The average acceleration of a material point over a certain period of time is the ratio of the change in its speed that occurred during this time to the duration of this interval:
\(\lt\vec a\gt = \dfrac (\Delta \vec v) (\Delta t) \)
The instantaneous acceleration of a material point at some point in time is the limit of its average acceleration at \(\Delta t \to 0\) . Keeping in mind the definition of the derivative of a function, instantaneous acceleration can be defined as the derivative of speed with respect to time:
\(\vec a = \dfrac (d\vec v) (dt) \)
Tangential and normal acceleration
If we write the speed as \(\vec v = v\hat \tau \) , where \(\hat \tau \) is the unit unit of the tangent to the trajectory of motion, then (in a two-dimensional coordinate system):
\(\vec a = \dfrac (d(v\hat \tau)) (dt) = \)
\(= \dfrac (dv) (dt) \hat \tau + \dfrac (d\hat \tau) (dt) v =\)
\(= \dfrac (dv) (dt) \hat \tau + \dfrac (d(\cos\theta\vec i + sin\theta \vec j)) (dt) v =\)
\(= \dfrac (dv) (dt) \hat \tau + (-sin\theta \dfrac (d\theta) (dt) \vec i + cos\theta \dfrac (d\theta) (dt) \vec j))v\)
\(= \dfrac (dv) (dt) \hat \tau + \dfrac (d\theta) (dt) v \hat n \),
where \(\theta \) is the angle between the velocity vector and the x-axis; \(\hat n \) - unit unit perpendicular to the speed.
Thus,
\(\vec a = \vec a_(\tau) + \vec a_n \),
Where \(\vec a_(\tau) = \dfrac (dv) (dt) \hat \tau \)- tangential acceleration, \(\vec a_n = \dfrac (d\theta) (dt) v \hat n \)- normal acceleration.
Considering that the velocity vector is directed tangent to the trajectory of motion, then \(\hat n \) is the unit unit of the normal to the trajectory of motion, which is directed to the center of curvature of the trajectory. Thus, normal acceleration is directed towards the center of curvature of the trajectory, while tangential acceleration is tangential to it. Tangential acceleration characterizes the rate of change in the magnitude of velocity, while normal acceleration characterizes the rate of change in its direction.
Movement along a curved trajectory at each moment of time can be represented as rotation around the center of curvature of the trajectory with angular velocity \(\omega = \dfrac v r\) , where r is the radius of curvature of the trajectory. In this case
\(a_(n) = \omega v = (\omega)^2 r = \dfrac (v^2) r \)
Acceleration measurement
Acceleration is measured in meters (divided) per second to the second power (m/s2). The magnitude of the acceleration determines how much the speed of a body will change per unit time if it constantly moves with such acceleration. For example, a body moving with an acceleration of 1 m/s 2 changes its speed by 1 m/s every second.
Acceleration units
- meter per second squared, m/s², SI derived unit
- centimeter per second squared, cm/s², derived unit of the GHS system
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Uniformly accelerated motion is motion with acceleration, the vector of which does not change in magnitude and direction. Examples of such movement: a bicycle rolling down a hill; a stone thrown at an angle to the horizontal.
Let's consider the last case in more detail. At any point of the trajectory, the stone is affected by the acceleration of gravity g →, which does not change in magnitude and is always directed in one direction.
The motion of a body thrown at an angle to the horizontal can be represented as the sum of motions relative to the vertical and horizontal axes.
Along the X axis the movement 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 during uniformly accelerated motion:
Here v 0 is the initial velocity 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 by the slope of the velocity graph. In the figure above, the acceleration modulus is equal to the ratio of the sides of triangle ABC.
a = v - v 0 t = B C A C
The larger the angle β, the greater the slope (steepness) of the graph relative to the time axis. Accordingly, the greater the acceleration of the body.
For the first graph: v 0 = - 2 m s; a = 0.5 m s 2.
For the second graph: v 0 = 3 m s; a = - 1 3 m s 2 .
Using this graph, you can also calculate the displacement of the body during time t. How to do it?
Let us highlight a small period of time ∆ t on the graph. We will assume that it is so small that the movement during the time ∆t can be considered a 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 us divide the entire time t into infinitesimal intervals ∆ t. The displacement s during 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 take the form:
s = v 0 t + a t 2 2
In order to find the coordinate of the body at a given moment in time, you need to add displacement to the initial coordinate of the body. The change in coordinates during uniformly accelerated motion expresses the law of uniformly accelerated motion.
Law of uniformly accelerated motion
Law of uniformly accelerated motiony = y 0 + v 0 t + a t 2 2 .
Another common problem that arises when analyzing uniformly accelerated motion is finding the displacement for given values of the initial and final velocities and acceleration.
Eliminating t from the equations written above and solving them, we obtain:
s = 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 quantities 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 under the conditions of a specific task, they can take on both positive and negative values.
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The body was constant and the body traveled the same paths for any equal periods of time.
Most movements, however, cannot be considered uniform. In some areas of the body the speed may be lower, in others it may be higher. For example, a train leaving a station begins to move faster and faster. Approaching the station, he, on the contrary, slows down.
Let's do an experiment. Let's install a dropper on the cart, from which drops of colored liquid fall at regular intervals. Let's place this cart on an inclined board and release it. We will see that the distance between the tracks left by the drops will become larger and larger as the cart moves downwards (Fig. 3). This means that the cart travels unequal distances in equal periods of time. The speed of the cart increases. Moreover, as can be proven, over the same periods of time, the speed of a cart sliding down an inclined board increases all the time by the same amount.
If the speed of a body during uneven motion changes equally over any equal periods of time, then the motion is called uniformly accelerated.
So. for example, experiments have established that the speed of any freely falling body (in the absence of air resistance) increases by approximately 9.8 m/s every second, i.e. if at first the body was at rest, then a second after the start of the fall it will have a speed of 9.8 m/s, after another second - 19.6 m/s, after another second - 29.4 m/s, etc.
A physical quantity that shows how much the speed of a body changes for each second of uniformly accelerated motion is called acceleration.
a is acceleration.
The SI unit of acceleration is the acceleration at which for every second the speed of the body changes by 1 m/s, i.e. meter per second per second. This unit is designated 1 m/s 2 and is called “meter per second squared”.
Acceleration characterizes the rate of change in speed. If, for example, the acceleration of a body is 10 m/s 2, then this means that for every second the speed of the body changes by 10 m/s, i.e. 10 times faster than with an acceleration of 1 m/s 2.
Examples of accelerations encountered in our lives can be found in Table 1. 
How do we calculate the acceleration with which bodies begin to move?
Let, for example, it is known that the speed of an electric train leaving the station increases by 1.2 m/s in 2 s. Then, in order to find out how much it increases in 1 s, we need to divide 1.2 m/s by 2 s. We get 0.6 m/s2. This is the acceleration of the train.
So, to find the acceleration of a body starting uniformly accelerated motion, it is necessary to divide the speed acquired by the body by the time during which this speed was achieved:
Let us denote all quantities included in this expression using Latin letters:
a - acceleration; V- acquired speed; t - time
Then the formula for determining acceleration can be written as follows:
This formula is valid for uniformly accelerated motion from the state peace, i.e. when the initial speed of the body is zero. The initial speed of the body is denoted by V 0 - Formula (2.1), therefore, is valid only under the condition that V 0 = 0.
If it is not the initial, but the final speed that is zero (which is simply denoted by the letter V), then the acceleration formula takes the form:
In this form, the acceleration formula is used in cases where a body having a certain speed V 0 begins to move slower and slower until it finally stops ( v= 0). It is by this formula, for example, that we will calculate the acceleration when braking cars and other vehicles. By time t we will understand the braking time.
Like speed, the acceleration of a body is characterized not only by its numerical value, but also by its direction. This means that acceleration is also vector size. Therefore, in the pictures it is depicted as an arrow.
If the speed of a body during uniformly accelerated rectilinear motion increases, then the acceleration is directed in the same direction as the speed (Fig. 4, a); if the speed of the body decreases during a given movement, then the acceleration is directed in the opposite direction (Fig. 4, b).
With uniform rectilinear motion, the speed of the body does not change. Therefore, there is no acceleration during such movement (a = 0) and cannot be depicted in the figures.
1. What kind of motion is called uniformly accelerated? 2. What is acceleration? 3. What characterizes acceleration? 4. In what cases is acceleration equal to zero? 5. What formula is used to find the acceleration of a body during uniformly accelerated motion from a state of rest? 6. What formula is used to find the acceleration of a body when the speed of motion decreases to zero? 7. What is the direction of acceleration during uniformly accelerated linear motion?
Experimental task
. Using the ruler as an inclined plane, place a coin on its top edge and release. Will the coin move? If so, how - uniformly or uniformly accelerated? How does this depend on the angle of the ruler?
S.V. Gromov, N.A. Rodina, Physics 8th grade
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at = v – v0
a = (v – v0)/t
In case of braking:
at = v0 – v
a = (v0 – v)/t
Now let's print
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 movement. It is natural to ask: how does the speed of movement change? In previous chapters, we considered the case when an acting force led to a change in speed. There are passenger cars that pick up speed from a standstill. Knowing this, we can determine how the speed changes, but only on average. Let's tackle the next more complex question: how to find out 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 everything said earlier, we are already sufficiently prepared to immediately write acceleration as a derivative of speed, just as speed is written as a derivative of distance. If we now differentiate the formula, 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 simply (a constant). If we choose this constant to be equal to 9.8, we immediately find that the derivative of is equal to 9.8.) This means, that the speed of a falling body constantly increases 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, the acceleration must be proportional to the force.
As the next example, let's find acceleration in the problem that we 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 in the table. 8.3, namely that the derivative of a sum is equal to the sum of its 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 . Let us now move on to differentiating the second term. According to one of the rules in table. 8.3, the derivative of the constant will be zero, therefore, this term will not make any contribution to the acceleration. Final result: 
.
Let us derive two more useful formulas that are obtained by integration. If a body moves from a state of rest with 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 is
Let us also note that since speed is , and acceleration is the derivative of speed with respect to time, we can write
. (8.10)
So now we know how the second derivative is written.
There is, of course, an inverse relationship between acceleration and distance, which simply follows from the fact that . Since distance is an integral of velocity, it can be found by integrating acceleration twice. The entire previous discussion was devoted to movement in one dimension, and now we will briefly dwell on movement in the space of three dimensions. Let's consider the movement of a particle in three-dimensional space. This chapter began with a discussion of the one-dimensional motion of a passenger car, namely, with the question of how far away from the origin of motion the car is at various 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 look at motion in three dimensions in the same sequence. It is easier, however, to start with the more obvious two-dimensional case, and only then generalize it to the three-dimensional case. Let's draw two lines (coordinate axes) intersecting at right angles and set the position of the particle at any moment in time by the distances from it to each of the axes. Thus, the position of the particle is specified 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 motion by creating, for example, a table in which these two coordinates are given as functions of time. (Generalization to the three-dimensional case requires introducing another axis perpendicular to the first two, and measuring another 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 components of velocity 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 speed.
How, knowing the components of speed, determine the total speed in the direction of motion? In the two-dimensional case, consider two successive positions of a particle separated by a short time interval and distance . From fig. 8.3 it is clear that
(8.14)
(The symbol corresponds to the expression “approximately equal to.”) The average speed during the interval is obtained by simple division: . To find the exact speed at moment , you need, as was already done at the beginning of the chapter, to direct 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 speeds: the acceleration component is defined as the derivative of the velocity component (i.e., the second derivative with respect to time), etc.
Let's look at another interesting example of mixed motion on a plane. Let the ball move horizontally with constant speed and at the same time fall vertically downward with constant acceleration. What kind of movement is this? 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 kind of curve does our ball describe, i.e., what is the relationship between the coordinates and ? From equation (8.18), according to (8.17), we can exclude time, since 1=*x/i% after which we find
Uniformly accelerated motion without initial speed
This relationship between the coordinates can be considered as an equation for the trajectory of the ball. If we were to depict it graphically, we would get a curve called a parabola (Fig. 8.4). So any freely falling body, being thrown in a certain direction, moves along a parabola.
In rectilinear uniformly accelerated motion the body
- moves along a conventional straight line,
- its speed gradually increases or decreases,
- over equal periods of time, the speed changes by an equal amount.
For example, a car starts moving from a state of rest along a straight road, and up to a speed of, say, 72 km/h it moves uniformly accelerated. When the set speed is reached, the car moves without changing speed, i.e. uniformly. With uniformly accelerated motion, 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, acceleration of 3.6 km/h per second must be converted into the appropriate units. 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 (over 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 a body is zero, then its speed after a time of t seconds will be equal to the product of acceleration and this time:
When a body falls, the acceleration of gravity “works”, and the speed of the body at the very surface of the earth will be determined by the formula:
If the current speed of the body and the time it took to develop such a speed from a state of rest are known, then the acceleration (i.e. how quickly the speed changed) can be determined by dividing the speed by the time:
However, the body could begin 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 using force. Such a body is subject to a gravitational acceleration equal to 9.8 m/s2. However, your strength gave 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 according to the formula:
However, if the stone was thrown upward. Then its initial speed is directed upward, and the acceleration of free fall is directed downward. That is, the velocity vectors are directed in opposite directions. In this case (as well as during braking), the product of acceleration and time must be subtracted from the initial speed:
From these formulas we obtain 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 a 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 braking acceleration, the time after which the body will stop is determined:
Now let's print formulas for the path that a body travels during rectilinear uniformly accelerated motion. The graph of speed versus 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.
How to find acceleration knowing the path and time?
That is, by multiplying speed by time (s = vt). With rectilinear uniformly accelerated motion, the graph is a straight line, but not parallel to the time axis. This straight line either increases in the case of acceleration or decreases in the case of braking. However, path is also defined as the area of the figure under the graph.
In rectilinear uniformly accelerated motion, this figure is a trapezoid. Its bases are a segment on the y-axis (speed) and a segment connecting the end point of the graph with its projection on the x-axis. The sides are the graph of speed versus time 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 equal to half the sum of the bases and 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), and the height is equal to time. Thus we get:
s = ½ * (v0 + v) * t
Above was given the formula for the dependence of the final speed on the initial and acceleration (v = v0 + at). 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 up the areas of the rectangle and right triangle into which the trapezoid is divided.)
If the body begins to move uniformly accelerated from a state of rest (v0 = 0), then the path formula simplifies 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 it becomes zero, the body will stop. A 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 into the path formula, then the braking path is reduced to the following formula:
I. Mechanics
Physics->Kinematics->uniformly accelerated motion->
Testing online
Uniformly accelerated motion
In this topic we will look at a very special type of irregular motion. Based on the contrast to uniform movement, uneven movement is movement at unequal speed along any trajectory. What is the peculiarity of uniformly accelerated motion? This is an uneven movement, but which "equally accelerated". We associate acceleration with increasing speed. Let's remember the word "equal", we get an equal increase in speed. How do we understand “equal increase in speed”, how can we evaluate whether the speed is increasing equally or not? To do this, we need to record time and estimate the speed over the same time interval. For example, a car starts to move, in the first two seconds it develops a speed of up to 10 m/s, in the next two seconds it reaches 20 m/s, and after another two seconds it already moves 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 the speed increases each time 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 he accelerated by 7 km/h (7-0), then by 2 km/h (9-7), then by 3 km/h (12-9).
Typically, movement with increasing speed is called accelerated movement. Movement with decreasing speed is called slow motion. But physicists call any movement with changing speed accelerated movement. Whether the car starts moving (the speed increases!) or brakes (the speed decreases!), in any case it moves with acceleration.
Uniformly accelerated motion- this is the movement of a body in which its speed for any equal periods of time changes(can increase or decrease) the same
Body acceleration
Acceleration characterizes the rate of change in speed. This is the number by which the speed changes every second. If the acceleration of a body is large in magnitude, this means that the body quickly gains speed (when it accelerates) or quickly loses it (when braking). Acceleration 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 next 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 did we determine? We are looking at the speed difference over 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 problem: 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, you need 11-7 = 4, then 4/2 = 2. We divide the speed difference by the time period. 
This formula is most often used in a modified form when solving problems:
The formula is not written in vector form, so we write the “+” sign when the body is accelerating, the “-” sign when it is slowing down.
Acceleration vector direction
The direction of the acceleration vector is shown in the figures
In this figure, the car moves 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 the speed, this means that the car is accelerating. Acceleration is positive.
During acceleration, the direction of acceleration coincides with the direction of speed. Acceleration is positive.
In this picture, the car is moving in the positive direction along the Ox axis, the velocity vector coincides with the direction of movement (directed to the right), the acceleration does NOT coincide with the direction of the speed, this means that the car is braking. Acceleration is negative.
When braking, the direction of acceleration is opposite to the direction of speed. Acceleration is negative.
Let's figure out why the acceleration is negative when braking. For example, in the first second the ship slowed down from 9 m/s to 7 m/s, in the second second to 5 m/s, in the third to 3 m/s. The speed changes to "-2m/s". 3-5=-2; 5-7=-2; 7-9=-2m/s. This is where the negative acceleration value comes from.
When solving problems, if the body slows down, acceleration is substituted into the formulas with a minus sign!!!
Moving during uniformly accelerated motion
An additional formula called timeless
Formula in coordinates
Medium speed communication
With uniformly accelerated motion, the average speed can be calculated as the arithmetic mean of the initial and final speeds
From this rule follows a formula that is very convenient to use when solving many problems
Path relationship
If a body moves uniformly accelerated, the initial speed is zero, then the paths traversed in successive equal intervals of time are related as a successive 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 a 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 are moving towards each other: one is heading north at an accelerated rate, the other is moving slowly to the south. How are train accelerations directed?
Equally to the north. Because the acceleration of the first train coincides in direction with the movement, while the acceleration of the second train is opposite to the movement (it is slowing 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 6 m/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 at the end of the fourth second the speed is 6 m/s, then at the beginning of the fourth second it was less than 6 m/s. Therefore, the distance covered by the train in the fourth second is less than 6 m.
Which of the given dependencies describe uniformly accelerated motion? 
Equation of the speed of a moving body. What is the corresponding path equation?
* The car covered 1m in the first second, 2m in the second, 3m in the third second, 4m in the fourth second, etc. Can such motion be considered uniformly accelerated?
In uniformly accelerated motion, paths covered in successive equal intervals of time are related as a successive series of odd numbers. Consequently, the described motion is not uniformly accelerated.
Want to do an experiment? Yes, easily. Take a long ruler, lay it horizontally and lift one end. You will end up with an inclined plane. Now take a coin and place it on the top end of the ruler. The coin will begin to slide down the ruler, watch how the coin moves at the same speed or not.
You will notice that the speed of the coin will gradually increase. And the change in speed will directly depend on the angle of inclination of the ruler. The steeper the angle of inclination, the greater the speed the coin will gain towards the end of the path.
Change the speed of a coin
You can try to find out how the speed of a coin changes over each equal period of time. In the case of a ruler and a coin, this is difficult to do at home, but in the laboratory it can be recorded that at a constant angle of inclination, the sliding coin changes its speed by the same amount every second.
Such a movement of a body, when its speed changes equally over any equal periods of time, and the body moves in a straight line, is called in physics rectilinear uniformly accelerated motion. Speed in this case refers to the speed at each specific moment in time.
This speed is called instantaneous speed. The instantaneous speed of a body can change in different ways: faster, slower, it can increase, or it can decrease. In order to characterize this change in speed, a quantity called acceleration is introduced.
Acceleration concept: formula
Acceleration is a physical quantity that shows how much the speed of a body has changed for each equal period of time. If the speed changes in the same way, then the acceleration will be constant. This happens in the case of rectilinear uniformly accelerated motion. The formula for acceleration is as follows:
a = (v - v_0)/ t,
where a is the acceleration, v is the final speed, v_0 is the initial speed, t is time.
Acceleration is measured in meters per second squared (1 m/s2). A unit that is a little strange at first glance is very easily explained: acceleration = speed/time = (m/s)/s, from where such a unit is derived.
Acceleration is a vector quantity. It can be directed either in the same direction as the speed, if the speed is increasing, or in the opposite direction, if the speed is decreasing. An example of the second option is braking. If, for example, a car slows down, its speed decreases. Then the acceleration will be a negative value, and it will be directed not in the direction of movement of the car, but in the opposite direction.
In cases where our speed changes from zero to any value, for example, when a rocket is launched, or in the case when the speed, on the contrary, decreases to zero, for example, when a train is braking to a complete stop, only one speed value can be used in calculations . The formula will then take the form: a =v /t for the first case, or: a = v_0 /t for the second.