Mechanics
Kinematic formulas:
Kinematics
mechanical movement
Mechanical movement is called a change in the position of a body (in space) relative to other bodies (over time).
Relativity of motion. Reference system
To describe the mechanical motion of a body (point), you need to know its coordinates at any time. To determine the coordinates, select reference body and connect with him coordinate system. Often the reference body is the Earth, which is associated with a rectangular Cartesian coordinate system. To determine the position of a point at any point in time, it is also necessary to set the origin of the time reference.
The coordinate system, the body of reference with which it is associated, and the device for measuring time form reference system, relative to which the motion of the body is considered.
Material point
A body whose dimensions can be neglected under given conditions of motion is called material point.
A body can be considered as a material point if its dimensions are small compared to the distance it travels, or compared to the distances from it to other bodies.
Trajectory, path, movement
Trajectory of movement called the line along which the body moves. The length of the trajectory is called the way we have traveled. Path- scalar physical quantity, can only be positive.
moving is called a vector connecting the start and end points of the trajectory.
The movement of a body, in which all its points at a given moment in time move in the same way, is called progressive movement. For description forward movement body, it is enough to choose one point and describe its movement.
A movement in which the trajectories of all points of the body are circles with centers on one straight line and all the planes of the circles are perpendicular to this straight line is called rotational movement.
Meter and second
To determine the coordinates of a body, it is necessary to be able to measure the distance on a straight line between two points. Any process of measuring a physical quantity consists in comparing the measured quantity with the unit of measurement of this quantity.
The unit of length in the International System of Units (SI) is meter. A meter is approximately 1/40,000,000 of the earth's meridian. According to the modern idea, a meter is the distance that light travels in the void in 1/299,792,458 of a second.
To measure time, some periodically repeating process is selected. The unit of time in SI is accepted second. A second is equal to 9,192,631,770 periods of radiation of a cesium atom during the transition between two levels of the hyperfine structure of the ground state.
In SI, length and time are taken to be independent of other quantities. Such quantities are called main.
Instant Speed
For quantitative characteristics the process of body movement, the concept of speed of movement is introduced.
instantaneous speed translational motion of the body at time t is the ratio of a very small displacement Ds to a small period of time Dt during which this displacement occurred:
Instantaneous speed is a vector quantity. The instantaneous velocity of movement is always directed tangentially to the trajectory in the direction of body motion.
The unit of speed is 1 m/s. A meter per second is equal to the speed of a point moving in a straight line and uniformly, at which the point moves a distance of 1 m in a time of 1 s.
Acceleration
acceleration is called a vector physical quantity equal to the ratio of a very small change in the velocity vector to a small period of time during which this change occurred, i.e. is a measure of the rate of change of speed:
A meter per second per second is such an acceleration at which the speed of a body moving in a straight line and uniformly accelerated changes by 1 m / s in a time of 1 s.
The direction of the acceleration vector coincides with the direction of the velocity change vector () for very small values of the time interval during which the velocity changes.
If the body moves in a straight line and its speed increases, then the direction of the acceleration vector coincides with the direction of the velocity vector; when the speed decreases, it is opposite to the direction of the speed vector.
When moving along a curvilinear trajectory, the direction of the velocity vector changes in the process of movement, and the acceleration vector can be directed at any angle to the velocity vector.
Uniform, uniformly accelerated rectilinear motion
Moving at a constant speed is called uniform rectilinear motion. In uniform rectilinear motion, the body moves in a straight line and for any equal intervals of time covers the same path.
A movement in which a body makes unequal movements in equal intervals of time is called uneven movement. With such a movement, the speed of the body changes with time.
equivariable is called such a movement in which the speed of the body for any equal time intervals changes by the same amount, i.e. movement with constant acceleration.
uniformly accelerated called uniformly variable motion, in which the magnitude of the speed increases. equally slow- uniformly variable motion, in which the magnitude of the speed decreases.
dependency graph V(t) for this case is shown in Fig.1.2.1. Time interval Δt in formula (1.4) one can take any. Attitude ∆V/∆t does not depend on it. Then ΔV=аΔt. Applying this formula to the interval from t about= 0 up to some point t, you can write an expression for the speed:
V(t)=V0 + at. (1.5)
Here V0– speed value at t about= 0. If the directions of velocity and acceleration are opposite, then they speak of uniformly slow motion (Fig. 1.2.2).
For uniformly slow motion, we similarly obtain
V(t) = V0 – at.
Let us analyze the derivation of the formula for the displacement of a body during uniformly accelerated motion. Note that in this case the displacement and the distance traveled are the same number.
Consider a short period of time Δt. From the definition of average speed Vcp = ∆S/∆t you can find the path ∆S = V cp ∆t. The figure shows that the path ∆S numerically equal to area rectangle with width Δt and height Vcp. If the time interval Δt choose small enough average speed on the interval Δt coincides with the instantaneous speed at the midpoint. ∆S ≈ V∆t. This ratio is more accurate, the less Δt. Dividing the total travel time into such small intervals and taking into account that the full path S is the sum of the paths traveled during these intervals, you can make sure that on the velocity graph it is numerically equal to the area of the trapezoid:
S= ½ (V 0 + V)t,
substituting (1.5), we obtain for uniformly accelerated motion:
S \u003d V 0 t + (at 2 / 2)(1.6)
For uniformly slow motion L calculated like this:
L= V 0 t–(at 2 /2).
Let's analyze task 1.3.
Let the speed graph have the form shown in Fig. 1.2.4. Draw qualitatively synchronous graphs of the path and acceleration versus time.
Student:– I have never come across the concept of “synchronous graphics”, I also have no idea what it means to “draw with high quality”.
– Synchronous graphs have the same scales along the abscissa axis, on which time is plotted. The graphs are arranged one below the other. Synchronous graphs are convenient for comparing several parameters at once at one point in time. In this problem, we will depict the movement qualitatively, that is, without taking into account specific numerical values. For us, it is quite enough to establish whether the function decreases or increases, what form it has, whether it has breaks or kinks, etc. I think that for a start we should reason together.
Divide the entire time of movement into three intervals OV, BD, DE. Tell me, what is the nature of the movement on each of them and by what formula will we calculate the distance traveled?
Student:- Location on OV the body was moving uniformly with zero initial speed, so the formula for the path is:
S 1 (t) = at2/2.
The acceleration can be found by dividing the change in speed, i.e. length AB, for a period of time OV.
Student:- Location on BD the body moves uniformly with a speed V 0 acquired by the end of the section OV. Path Formula - S=Vt. There is no acceleration.
S 2 (t) = at 1 2 /2 + V 0 (t–t1).
Given this explanation, write a formula for the path on the site DE.
Student:- In the last section, the movement is uniformly slow. I will argue like this. Until the point in time t 2 the body has already traveled a distance S 2 \u003d at 1 2 / 2 + V (t 2 - t 1).
To it must be added an expression for the equally slow case, given that the time is counted from the value t2 we get the distance traveled, in time t - t 2:
S 3 \u003d V 0 (t–t 2)–/2.
I foresee the question of how to find the acceleration a 1 . It equals CD/DE. As a result, we get the path traveled in time t>t 2
S (t)= at 1 2 /2+V 0 (t–t 1)– /2.
Student:- In the first section we have a parabola with branches pointing upwards. On the second - a straight line, on the last - also a parabola, but with branches down.
Your drawing is inaccurate. The path graph has no kinks, i.e., parabolas should be smoothly mated with a straight line. We have already said that the speed is determined by the tangent of the slope of the tangent. According to your drawing, it turns out that at the moment t 1 the speed has two values at once. If you build a tangent on the left, then the speed will be numerically equal to tgα, and if you approach the point on the right, then the speed is equal to tgβ. But in our case, the speed is continuous function. The contradiction is removed if the graph is constructed in this way.
There is another useful relationship between S, a, V And V 0 . We will assume that the movement occurs in one direction. In this case, the movement of the body from the starting point coincides with the path traveled. Using (1.5), express the time t and exclude it from equality (1.6). This is how you get this formula.
Student:– V(t) = V0 + at, Means,
t = (V–V 0)/a,
S = V 0 t + at 2 /2 = V 0 (V– V 0)/a + a[(V– V 0)/a] 2 = .
Finally we have:
S= . (1.6a)
Story.
Once, while studying in Göttingen, Niels Bohr was poorly prepared for a colloquium, and his performance turned out to be weak. Bor, however, did not lose heart and concluded with a smile:
“I have heard so many bad speeches here that I ask you to consider mine as revenge.
In general uniformly accelerated motion called such a movement in which the acceleration vector remains unchanged in magnitude and direction. An example of such a movement is the movement of a stone thrown at a certain angle to the horizon (ignoring air resistance). At any point of the trajectory, the acceleration of the stone is equal to the acceleration free fall. For a kinematic description of the movement of a stone, it is convenient to choose a coordinate system so that one of the axes, for example, the axis OY, was directed parallel to the acceleration vector. Then curvilinear motion stone can be represented as the sum of two movements - rectilinear uniformly accelerated motion along the axis OY And uniform rectilinear motion in the perpendicular direction, i.e. along the axis OX(Fig. 1.4.1).
Thus, the study of uniformly accelerated motion is reduced to the study of rectilinear uniformly accelerated motion. In the case of rectilinear motion, the velocity and acceleration vectors are directed along the straight line of motion. Therefore, the speed v and acceleration a in projections on the direction of motion can be considered as algebraic quantities.
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Figure 1.4.1. Projections of the velocity and acceleration vectors on the coordinate axes. ax = 0, ay = –g |
With uniformly accelerated rectilinear motion, the speed of the body is determined by the formula
(*)
In this formula, υ 0 is the speed of the body at t = 0 (starting speed ), a= const - acceleration. On the velocity graph υ ( t), this dependence looks like a straight line (Fig. 1.4.2).
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Figure 1.4.2. Graphs of the speed of uniformly accelerated motion |
The slope of the velocity graph can be used to determine the acceleration a body. The corresponding constructions are made in Figs. 1.4.2 for graph I. The acceleration is numerically equal to the ratio of the sides of the triangle ABC:
The greater the angle β that forms the velocity graph with the time axis, i.e. the greater the slope of the graph ( steepness), the greater the acceleration of the body.
For graph I: υ 0 \u003d -2 m / s, a\u003d 1/2 m / s 2.
For graph II: υ 0 \u003d 3 m / s, a\u003d -1/3 m / s 2
The velocity graph also allows you to determine the displacement projection s body for a while t. Let us allocate on the time axis some small time interval Δ t. If this time interval is small enough, then the change in speed over this interval is small, i.e., the movement during this time interval can be considered uniform with a certain average speed, which is equal to the instantaneous speed υ of the body in the middle of the interval Δ t. Therefore, displacement Δ s in time Δ t will be equal to Δ s = υΔ t. This displacement is equal to the area of the shaded strip (Fig. 1.4.2). Breaking down the time span from 0 to some point t for small intervals Δ t, we get that the displacement s for a given time t with uniformly accelerated rectilinear motion is equal to the area of the trapezoid ODEF. Corresponding constructions are made for graph II in fig. 1.4.2. Time t taken equal to 5.5 s.
Since υ - υ 0 = at, the final formula for moving s bodies with uniformly accelerated motion over a time interval from 0 to t will be written in the form:
(**)
To find the coordinate y body at any given time. t to the starting coordinate y 0 add displacement over time t:
(***)
This expression is called law of uniformly accelerated motion .
When analyzing uniformly accelerated motion, sometimes the problem arises of determining the displacement of the body according to the given values of the initial υ 0 and final υ velocities and acceleration a. This problem can be solved using the equations written above by eliminating time from them. t. The result is written as
From this formula, you can get an expression for determining the final speed υ of the body, if the initial speed υ 0 is known, acceleration a and moving s:
If the initial speed υ 0 is equal to zero, these formulas take the form
It should again be noted that the quantities υ 0, υ, included in the formulas of uniformly accelerated rectilinear motion, s, a, y 0 are algebraic quantities. Depending on the specific type of movement, each of these quantities can take both positive and negative values.
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, the graph linear dependence in the figure, we will conditionally turn it 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 movement: V x (t) = V 0x + a x t, it is quite obvious that the difference in the projections of speeds 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 are close to a solution main task mechanics of uniformly accelerated motion, that is, obtaining a law according to which the coordinate changes over 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)
1) Analytical method.We consider the highway to be straight. Let's write down the equation of motion of a cyclist. Since the cyclist was moving uniformly, his equation of motion is:
(the origin of coordinates is placed at the starting point, so the initial coordinate of the cyclist is zero).
The motorcyclist was moving at a uniform speed. He also started moving from the starting point, so his initial coordinate is zero, the initial speed of the motorcyclist is also equal to zero (the motorcyclist began to move from a state of rest).
Considering that the motorcyclist started moving a little later, the motorcyclist's equation of motion is:
In this case, the speed of the motorcyclist changed according to the law:
At the moment when the motorcyclist caught up with the cyclist, their coordinates are equal, i.e. or:
Solving this equation with respect to , we find the meeting time:
This quadratic equation. We define the discriminant:
Define roots:
Substitute the numerical values into the formulas and calculate:
The second root is discarded as inappropriate physical conditions tasks: the motorcyclist could not catch up with the cyclist 0.37 s after the cyclist started moving, since he himself left the starting point only 2 s after the cyclist started.
Thus, the time when the motorcyclist caught up with the cyclist:
Substitute this value of time into the formula for the law of change in the speed of a motorcyclist and find the value of his speed at this moment:
2) Graphical way.
On the same coordinate plane, we build graphs of changes in the coordinates of the cyclist and motorcyclist over time (the graph for the coordinates of the cyclist is in red, for the motorcyclist - in green). It can be seen that the dependence of the coordinate on time for a cyclist is a linear function, and the graph of this function is a straight line (the case of uniform rectilinear motion). The motorcyclist was moving with uniform acceleration, so the dependence of the motorcyclist’s coordinates on time is quadratic function, whose graph is a parabola.
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