Velocity is a vector quantity that characterizes not only the speed of movement of a particle along a trajectory, but also the direction in which the particle is moving at each moment of time.
Average speed over time from t 1 to t 2 is equal to the ratio of the movement during this time to the period of time during which this movement took place:
We will note the fact that this is the average speed by enclosing the average value in angle brackets:<...>, as done above.
The above formula for the average velocity vector is a direct consequence of the general mathematical definition of the average value<f(x)> arbitrary function f(x) on the interval [ a,b]:
Really
Average speed may be too rough a measure of movement. For example, the average speed over a period of oscillation is always zero, regardless of the nature of these oscillations, for the simple reason that over a period - by definition of a period - the oscillating body will return to its starting point and, therefore, the displacement over the period is always zero. For this and a number of other reasons, instantaneous speed is introduced - the speed at a given moment in time. In the future, meaning instantaneous speed, we will write simply: “speed”, omitting the words “instantaneous” or “at a given moment in time” whenever this cannot lead to misunderstandings. To obtain speed at a moment in time t we need to do the obvious thing: calculate the limit of the ratio as the time interval tends t 2 – t 1 to zero. Let's make some redesignations: t 1 = t And t 2 = t + and rewrite the upper relation as:
Speed at time t equal to the limit of the ratio of movement over time to the period of time during which this movement took place, as the latter tends to zero
Rice. 2.5. Towards the definition of instantaneous speed.
At the moment we do not consider the question of the existence of this limit, assuming that it exists. Note that if there is a finite displacement and a finite period of time, then and are their limiting values: an infinitesimal displacement and an infinitesimal period of time. So the right side of the definition of speed
is nothing more than a fraction - the quotient of division by, therefore the last relation can be rewritten and is very often used in the form
According to the geometric meaning of the derivative, the velocity vector at each point of the trajectory is directed tangent to the trajectory at this point in its direction of movement.
Video 2.1. The velocity vector is directed tangentially to the trajectory. Experiment with a sharpener.
Any vector can be expanded into a basis (for unit vectors of the basis, in other words, unit vectors defining the positive directions of the axes OX,OY,OZ we use the notation , , or , respectively). The coefficients of this expansion are the projections of the vector onto the corresponding axes. The following is important: in vector algebra it has been proven that the expansion with respect to the basis is unique. Let us expand the radius vector of some moving material point into a basis
Taking into account the constancy of the Cartesian unit vectors , , , we differentiate this expression with respect to time
On the other hand, the expansion in terms of the velocity vector basis has the form
juxtaposing the last two expressions, taking into account the uniqueness of the expansion of any vector with respect to the basis, gives the following result: the projections of the velocity vector onto the Cartesian axes are equal to the time derivatives of the corresponding coordinates, that is
The magnitude of the velocity vector is equal to
Let us obtain another, important, expression for the magnitude of the velocity vector.
It has already been noted that when the value || differs less and less from the corresponding path (see Fig. 2). That's why
and in the limit (>0)
In other words, the velocity module is the derivative of the distance traveled with respect to time.
Finally we have:
Average magnitude of the velocity vector, is defined as follows:
The average value of the velocity vector modulus is equal to the ratio of the distance traveled to the time during which this path was traveled:
Here s(t 1 , t 2)- path in time from t 1 to t 2 and, accordingly, s(t 0 , t 2)- path in time from t 0 to t 2 And s(t 0 , t 2)- path in time from t 0 to t 1.
The average velocity vector or simply the average velocity as stated above is
Note that, first of all, this is a vector, its module - the module of the average velocity vector should not be confused with the average value of the velocity vector module. In the general case, they are not equal: the module of the average vector is not at all equal to the average module of this vector. Two operations: calculating the modulus and calculating the average, in the general case, cannot be interchanged.
Let's look at an example. Let the point move in one direction. In Fig. 2.6. shows a graph of the path she has traveled s in from time (during the time from 0 to t). Using the physical meaning of speed, use this graph to find the moment in time at which the instantaneous speed is equal to the average ground speed for the first seconds of the point’s movement.
Rice. 2.6. Determination of instantaneous and average speed of a body
Speed module at a given time
being the derivative of the path with respect to time, it is equal to the angular coefficient of the swing to the graph of the dependence of the point corresponding to the moment of time t*. Average velocity modulus over a period of time from 0 to t* is the angular coefficient of the secant passing through the points of the same graph corresponding to the beginning t = 0 and the end t = t* time interval. We need to find such a moment in time t*, when both slopes coincide. To do this, draw a straight line through the origin, tangent to the trajectory. As can be seen from the figure, the point of tangency of this straight graph is s(t) and gives t*. In our example it turns out
Instructions
Enter the coordinate system relative to which you will determine the direction and module. If the task already contains dependencies speed from time to time, there is no need to enter a coordinate system - it is assumed that it already exists.
According to the existing dependency function speed from time you can find the value speed at any time t. Let, for example, v=2t²+5t-3. If you need to find module speed at time t=1, simply substitute this value into and calculate v: v=2+5-3=4.
Sources:
- how to find path dependence on time
Module numbers n represents the number of unit segments from the origin to point n. Moreover, it does not matter in which direction this distance will be counted - to the right or to the left of zero.
Instructions
Module numbers also called the absolute value of this numbers. It is in short vertical lines drawn to the left and right of numbers. For example, module numbers 15 is written as follows: |15|.
Remember that the modulus can only be a positive number or . Module positive numbers equal to number. Module zero. That is, for anyone numbers n, which is greater than or equal to zero, the following will be true |n| = n. For example, |15| = 15, that is, modulus numbers 15 is equal to 15.
Negative modulus numbers will be the same number, but with the opposite sign. That is, for anyone numbers n, which is less than zero, the formula |n| = -n. For example, |-28| = 28. Module numbers-28 is equal to 28.
You can find not only for integers, but also for numbers. Moreover, the same rules apply to fractional numbers. For example, |0.25| = 25, that is, module numbers 0.25 will be equal to 0.25. A |-¾| = ¾, that is, the module numbers-¾ will be equal to ¾.
When working, it is useful to know that the modules are always equal to each other, that is, |n| =|-n|. This is the main property. For example, |10| = |-10|. Module numbers 10 is equal to 10, just like modulus numbers-10. In addition, |a - b| = |b - a|, since the distance from point a to point b and the distance from b to a are equal to each other. For example, |25 - 5| = |5 - 25|, that is |20| = |- 20|.
To find the change speed decide on the type of body movement. If the body moves uniformly, change speed equals zero. If a body is moving with acceleration, then change his speed at each moment of time can be found out if we subtract from the instantaneous speed at a given moment in time its initial speed.
You will need
- stopwatch, speedometer, radar, tape measure, accelerometer.
Instructions
Definition of change speed arbitrarily moving trajectory Using a speedometer or radar, measure the speed of the body at the beginning and end of the path segment. Then subtract the initial result from the final result, this will be change speed bodies.
Definition of change speed of a body moving with acceleration Find the acceleration of the body. Use an accelerometer or dynamometer. If the mass of the body is known, then divide the force acting on the body by its mass (a=F/m). After this, measure the time during which the changes occurred speed. To find change speed, multiply the acceleration value by the time during which this happened change(Δv=a t). If acceleration is measured in meters per second, and time is measured in seconds, then the speed is measured in meters per second. If it is not possible to measure time, but that the speed changed along a certain segment of the path, use a speedometer or radar, measure the speed at the beginning of this segment, then use a tape measure or rangefinder to measure the length of this path. Using any of the methods described above, measure the acceleration that acted on the body. After this, find the final speed of the body at the end of the path. To do this, raise the initial speed to , add to it the product of the section times the acceleration and the number 2. From the result, extract . To find change speed, from the result obtained, subtract the value of the initial speed.
Definition of change speed bodies when turning If not only the magnitude, but also the direction speed, then find it change vector difference of initial and final speed. To do this, measure the angle between the vectors. Then, from the sum of the squares of the velocities, subtract their double product multiplied by the cosine of the angle between them: v1²+v2²-2v1v2 Cos(α). Take the square root of the resulting number.
Video on the topic
To determine the speed of various types movement you will need different formulas. To determine speed uniform movement, divide the distance by the time it takes to travel. Find the average speed of movement by adding all the segments that the body has passed by the total time of movement. In case of uniformly accelerated motion, find out the acceleration with which the body moved, and in case of free fall, find out the height from which it began to move.
You will need
- rangefinder, stopwatch, accelerometer.
Instructions
Speed of uniform motion and average speed Measure the distance the body has traveled using a rangefinder, and the time it took to cover it using a stopwatch. After this, divide the distance traveled by the body by the time it travels, the result will be the speed of uniform motion (v=S/t). If the body moves unevenly, make the same measurements and apply the same formula - then you will get the average speed of the body. This means that if a body along a given segment of the path moved with the obtained speed, it would be on the way for a time equal to the measured one. If the body moves along , measure it and the time it takes to complete a revolution, then multiply the radius by 6.28 and divide by the time (v=6.28 R/t). In all cases, the result will be in meters per second. To convert to an hour, multiply it by 3.6.
Velocity of uniformly accelerated motion Measure the acceleration of the body using an accelerometer or dynamometer if the mass of the body is known. Using a stopwatch, measure the time of movement of the body and its initial speed, if the body does not begin to move from a state of rest. If the body moves from a state of rest, it is equal to zero. After this, find out the speed of the body by adding the product of acceleration and time (v=v0+at) to the initial speed.
The speed of a freely falling body. Using a range finder, measure the speed of the body in meters. To find out the speed at which it will reach the Earth's surface (ignoring drag), multiply the height by 2 and the number 9.81 (gravitational acceleration). From the result, extract the square . To find the speed of a body at any height, use the same technique, only from the initial one, subtract the current one and substitute the resulting value instead of the height.
Video on the topic
A person is accustomed to perceive the concept " speed"as something simpler than it really is. Indeed, a car rushing through an intersection moves with a certain speed yu, while the person stands and watches him. But if a person is in motion, then it makes more sense to talk not about absolute speed, but about its relative value. Find relative speed very easy.
Instructions
You can continue to consider the topic of a car moving to an intersection. A person standing at a red traffic light also stands at a passing car. The person is motionless, so let’s take him as a frame of reference. A reference system is one relative to which any body or other material point moves.
Let's say a car is moving with speed 50 km/h. But let’s say that he ran after a car (you can, for example, instead of a car, imagine a minibus or a person passing by). Running speed 12 km/h. Thus, speed of this mechanical vehicle will not seem as fast as it was before when it! This is the whole point of relative speed. speed always measured relative to a moving reference frame. Thus, speed there will be no car for a pedestrian 50 km/h, but 50 - 12 = 38 km/h.
You can consider one more. It is enough to recall any of the moments when a person, sitting at the window of a bus, watches cars rushing past. Indeed, from the bus window they speed It just seems stunning. And this is not surprising, because if we take the bus as a reference system, then speed car and speed the bus will need to be folded. Let's assume that the bus is moving with speed u 50 km/h, and 60 km/h. Then 50 + 60 = 110 km/h. Exactly with this speed These same cars rush past the bus and the passengers in it.
This same speed will be fair and valid even if any of the cars passing by the buses is taken as the reference system.
Kinematics studies different types of movement body with a given speed, direction and trajectory. To determine its position relative to the starting point of the path, you need to find moving body.
Instructions
Movement body occurs along a certain trajectory. In the case of rectilinear motion of the line, therefore find moving body quite simple: it is equal to the distance traveled. Otherwise, it can be determined by the initial and final positions in space.
For general purposes, finding the speed of an object (v) is a simple task: you need to divide the displacement (s) during a certain time (s) by this time (t), that is, use the formula v = s/t. However, in this way the average speed of the body is obtained. Using some calculations, you can find the speed of the body at any point along the way. This speed is called instantaneous speed and is calculated by the formula v = (ds)/(dt), that is, it is a derivative of the formula for calculating the average speed of a body. .
Steps
Part 1
Calculation of instantaneous speed-
To calculate instantaneous speed, you need to know the equation that describes the movement of a body (its position at a certain moment in time), that is, an equation on one side of which there is s (the movement of the body), and on the other side there are terms with the variable t (time). For example:
s = -1.5t 2 + 10t + 4
- In this equation: Displacement = s. Displacement is the path traveled by an object. For example, if a body moves 10 m forward and 7 m back, then the total displacement of the body is 10 - 7 = 3 m (and 10 + 7 = 17 m). Time = t. Usually measured in seconds.
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To find the instantaneous speed of a body whose displacement is described by the above equation, you must calculate the derivative of this equation. The derivative is an equation that allows you to calculate the slope of a graph at any point (at any point in time). To find the derivative, differentiate the function as follows: if y = a*x n, then derivative = a*n*x n-1. This rule applies to each term of the polynomial.
- In other words, the derivative of each term with variable t is equal to the product of the factor (in front of the variable) and the power of the variable, multiplied by the variable to a power equal to the original power minus 1. The dummy term (the term without a variable, that is, the number) disappears because it is multiplied by 0. In our example:
s = -1.5t 2 + 10t + 4
(2)-1.5t (2-1) + (1)10t 1 - 1 + (0)4t 0
-3t 1 + 10t 0
-3t+10
- In other words, the derivative of each term with variable t is equal to the product of the factor (in front of the variable) and the power of the variable, multiplied by the variable to a power equal to the original power minus 1. The dummy term (the term without a variable, that is, the number) disappears because it is multiplied by 0. In our example:
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Replace "s" with "ds/dt" to show that the new equation is the derivative of the original equation (that is, the derivative of s with t). The derivative is the slope of the graph at a certain point (at a certain point in time). For example, to find the slope of the line described by the function s = -1.5t 2 + 10t + 4 at t = 5, simply substitute 5 into the derivative equation.
- In our example, the derivative equation should look like this:
ds/dt = -3t + 10
- In our example, the derivative equation should look like this:
-
Substitute the appropriate t value into the derivative equation to find the instantaneous speed at a certain point in time. For example, if you want to find the instantaneous speed at t = 5, simply substitute 5 (for t) into the derivative equation ds/dt = -3 + 10. Then solve the equation:
ds/dt = -3t + 10
ds/dt = -3(5) + 10
ds/dt = -15 + 10 = -5 m/s- Please note the unit of measurement for instantaneous speed: m/s. Since we are given the value of displacement in meters, and time in seconds, and the speed is equal to the ratio of displacement to time, then the unit of measurement m/s is correct.
Part 2
Graphical evaluation of instantaneous speed-
Graph the movement of the body. In the previous chapter, you calculated instantaneous velocity using a formula (a derivative equation that allows you to find the slope of a graph at a specific point). By plotting a graph of the movement of a body, you can find its inclination at any point, and therefore determine the instantaneous speed at a certain point in time.
- The Y axis is the displacement, and the X axis is the time. The coordinates of the points (x, y) are obtained by substituting various values of t into the original displacement equation and calculating the corresponding values of s.
- The graph may fall below the X-axis. If the graph of the body's movement falls below the X-axis, this means that the body is moving in the opposite direction from the point of origin of movement. Typically the graph will not extend beyond the Y axis (negative x values) - we are not measuring the speeds of objects moving backwards in time!
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Select point P and point Q close to it on the graph (curve). To find the slope of the graph at point P, we use the concept of limit. Limit – a state in which the value of the secant drawn through 2 points P and Q lying on the curve tends to zero.
- For example, consider points P(1,3) and Q(4,7) and calculate the instantaneous speed at point P.
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Find the slope of the segment PQ. The slope of the segment PQ is equal to the ratio of the difference in the values of the “y” coordinates of points P and Q to the difference in the values of the “x” coordinates of points P and Q. In other words, H = (y Q - y P)/(x Q - x P), where H – slope of segment PQ. In our example, the slope of the segment PQ is:
H = (y Q - y P)/(x Q - x P)
H = (7 - 3)/(4 - 1)
H = (4)/(3) = 1.33 -
Repeat the process several times, bringing point Q closer to point P. The smaller the distance between two points, the closer the slope of the resulting segments is to the slope of the graph at point P. In our example, we will perform calculations for point Q with coordinates (2,4.8), (1.5,3.95) and (1.25,3.49) (coordinates of the point P remain the same):
Q = (2,4.8): H = (4.8 - 3)/(2 - 1)
H = (1.8)/(1) = 1.8Q = (1.5,3.95): H = (3.95 - 3)/(1.5 - 1)
H = (.95)/(.5) = 1.9Q = (1.25,3.49): H = (3.49 - 3)/(1.25 - 1)
H = (.49)/(.25) = 1.96 -
The smaller the distance between points P and Q, the closer the value of H is to the slope of the graph at point P. If the distance between points P and Q is extremely small, the value of H will be equal to the slope of the graph at point P. Since we cannot measure or calculate the extremely small distance between two points, the graphical method gives an estimate of the slope of the graph at point P.
- In our example, as Q approached P, we obtained the following values of H: 1.8; 1.9 and 1.96. Since these numbers tend to 2, we can say that the slope of the graph at point P is 2.
- Remember that the slope of a graph at a given point is equal to the derivative of the function (from which the graph is plotted) at that point. The graph displays the movement of a body over time and, as noted in the previous section, the instantaneous speed of a body is equal to the derivative of the equation of displacement of this body. Thus, we can state that at t = 2 the instantaneous speed is 2 m/s (this is an estimate).
Part 3
Examples-
Calculate the instantaneous speed at t = 4 if the movement of the body is described by the equation s = 5t 3 - 3t 2 + 2t + 9. This example is similar to the problem from the first section, with the only difference being that here we have a third order equation (not a second one).
- First, let's calculate the derivative of this equation:
s = 5t 3 - 3t 2 + 2t + 9
s = (3)5t (3 - 1) - (2)3t (2 - 1) + (1)2t (1 - 1) + (0)9t 0 - 1
15t (2) - 6t (1) + 2t (0)
15t (2) - 6t + 2t = 1.01: s = 4(1.01) 2 - (1.01)
4(1.0201) - 1.01 = 4.0804 - 1.01 = 3.0704, so Q = (1.01,3.0704) - Now let's calculate H:
Q = (2.14): H = (14 - 3)/(2 - 1)
H = (11)/(1) = 11Q = (1.5,7.5): H = (7.5 - 3)/(1.5 - 1)
H = (4.5)/(.5) = 9Q = (1.1,3.74): H = (3.74 - 3)/(1.1 - 1)
H = (.74)/(.1) = 7.3Q = (1.01,3.0704): H = (3.0704 - 3)/(1.01 - 1)
H = (.0704)/(.01) = 7.04 - Since the obtained values of H tend to 7, we can say that the instantaneous speed of the body at point (1.3) is equal to 7 m/s (estimated value).
- First, let's calculate the derivative of this equation:
- To find acceleration (the change in speed over time), use the method in part one to obtain the derivative of the displacement function. Then take the derivative of the resulting derivative again. This will give you the equation to find the acceleration at a given time - all you have to do is plug in the value for the time.
- The equation describing y (displacement) versus x (time) can be very simple, for example: y = 6x + 3. In this case, the slope is constant and you do not need to take a derivative to find it. According to the theory of linear graphs, their slope is equal to the coefficient of the variable x, that is, in our example = 6.
- Displacement is like distance, but it has a specific direction, making it a vector quantity. Displacement can be negative, while distance will only be positive.
In the last article, we figured out a little about what mechanics are and why they are needed. We already know what a reference system, relativity of motion and a material point are. Well, it's time to move on! Here we will look at the basic concepts of kinematics, put together the most useful formulas for the basics of kinematics, and give a practical example of solving the problem.
Aristotle studied kinematics. True, then it was not called kinematics. Then a very large contribution to the development of mechanics, and kinematics in particular, was made by Galileo Galilei, who studied the free fall and inertia of bodies.
So, kinematics solves the question: how a body moves. The reasons why it came into motion are not of interest to her. Kinematics doesn’t care whether the car drove itself or was pushed by a giant dinosaur. It doesn't matter at all.
Trajectory, radius vector, law of body motion
Now we will consider the simplest kinematics - the kinematics of a point. Let's imagine that the body (material point) is moving. It doesn’t matter what kind of body it is, we still consider it as a material point. Maybe it's a UFO in the sky, or maybe it's a paper airplane that we launched out of the window. Better yet, let it be a new car in which we go on a trip. Moving from point A to point B, our point describes an imaginary line, which is called the trajectory of movement. Another definition of a trajectory is a hodograph, the radius vector, that is, the line that the end of the radius vector of a material point during movement describes.
Radius vector – a vector that specifies the position of a point in space .
In order to find out the position of a body in space at any moment in time, you need to know the law of body motion - the dependence of coordinates (or the radius vector of a point) on time.
The body has moved from point A to point B. In this case, the movement of the body - a segment connecting these points directly - is a vector quantity. The path traveled by the body is the length of its trajectory. Obviously, movement and path should not be confused. The magnitude of the displacement vector and the path length coincide only in the case of rectilinear motion.
In the SI system, displacement and path length are measured in meters.
The displacement is equal to the difference between the radius vectors at the initial and final times. In other words, it is the increment of the radius of the vector.
Speed and acceleration
Average speed is a vector physical quantity equal to the ratio of the displacement vector to the period of time during which it occurred
Now let’s imagine that the period of time decreases, decreases, and becomes very short, tends to zero. In this case, there is no need to talk about average speed; the speed becomes instantaneous. Those who remember the basics of mathematical analysis will immediately understand that in the future we cannot do without the derivative.
Instantaneous speed is a vector physical quantity equal to the time derivative of the radius vector. Instantaneous velocity is always directed tangentially to the trajectory.
In the SI system, speed is measured in meters per second.
If a body does not move uniformly and rectilinearly, then it has not only speed, but also acceleration.
Acceleration (or instantaneous acceleration) is a vector physical quantity, the second derivative of the radius vector with respect to time, and, accordingly, the first derivative of the instantaneous speed
Acceleration shows how quickly the speed of a body changes. In the case of rectilinear motion, the directions of the velocity and acceleration vectors coincide. In the case of curvilinear motion, the acceleration vector can be decomposed into two components: tangential acceleration, And acceleration is normal .
Tangential acceleration shows how quickly the speed of a body changes in magnitude and is directed tangentially to the trajectory
Normal acceleration characterizes the speed of change of speed in direction. The normal and tangential acceleration vectors are mutually perpendicular, and the normal acceleration vector is directed towards the center of the circle along which the point moves.
Here R is the radius of the circle along which the body moves
Here - x is zero - the initial coordinate. v zero - initial speed. Let's differentiate by time and get the speed
The derivative of speed with time will give the value of acceleration a, which is a constant.
Example of problem solution
Now that we have examined the physical foundations of kinematics, it’s time to consolidate our knowledge in practice and solve some problem. Moreover, the faster, the better.
For example this: a point moves in a circle with a radius of 4 meters. The law of its motion is expressed by the equation S=A+Bt^2. A=8m, B=-2m/s^2. At what point in time is the normal acceleration of a point equal to 9 m/s^2? Find the speed, tangential and total acceleration of the point for this moment in time.
Solution: we know that in order to find the speed we need to take the first time derivative of the law of motion, and the normal acceleration is equal to the quotient of the square of the speed and the radius of the circle along which the point is moving. Armed with this knowledge, we will find the required quantities.
Dear friends, congratulations! If you have read this article on the basics of kinematics, and in addition learned something new, you have already done a good deed! We sincerely hope that our “kinematics for dummies” will be useful to you. Dare and remember - we are always ready to help you with solving tricky puzzles with insidious cheap traps. . Good luck in learning mechanics!
