Speed ​​average and instantaneous value. Problems involving free falling bodies: examples of solving problems in kinematics

3.1. Uniform motion in a straight line.

3.1.1. Uniform motion in a straight line- movement in a straight line with acceleration constant in magnitude and direction:

3.1.2. Acceleration()- a physical vector quantity showing how much the speed will change in 1 s.

In vector form:

where is the initial speed of the body, is the speed of the body at the moment of time t.

In projection onto the axis Ox:

where is the projection of the initial velocity onto the axis Ox, - projection of the body velocity onto the axis Ox at a point in time t.

The signs of the projections depend on the direction of the vectors and the axis Ox.

3.1.3. Projection graph of acceleration versus time.

With uniformly alternating motion, the acceleration is constant, therefore it will appear as straight lines parallel to the time axis (see figure):

3.1.4. Speed ​​during uniform motion.

In vector form:

In projection onto the axis Ox:

For uniformly accelerated motion:

For uniform slow motion:

3.1.5. Projection graph of speed versus time.

The graph of the projection of speed versus time is a straight line.

Direction of movement: if the graph (or part of it) is above the time axis, then the body is moving in the positive direction of the axis Ox.

Acceleration value: the greater the tangent of the angle of inclination (the steeper it goes up or down), the greater the acceleration module; where is the change in speed over time

Intersection with the time axis: if the graph intersects the time axis, then before the intersection point the body slowed down (uniformly slow motion), and after the intersection point it began to accelerate in the opposite direction (uniformly accelerated motion).

3.1.6. Geometric meaning of the area under the graph in the axes

Area under the graph when on the axis Oy the speed is delayed, and on the axis Ox- time is the path traveled by the body.

In Fig. 3.5 shows the case of uniformly accelerated motion. The path in this case will be equal to the area of ​​the trapezoid: (3.9)

3.1.7. Formulas for calculating path

Uniformly accelerated motion	Equal slow motion
(3.10)	(3.12)
(3.11)	(3.13)
(3.14)

All formulas presented in the table work only when the direction of movement is maintained, that is, until the straight line intersects with the time axis on the graph of the velocity projection versus time.

If the intersection has occurred, then the movement is easier to divide into two stages:

before crossing (braking):

After the intersection (acceleration, movement in the opposite direction)

In the formulas above - the time from the beginning of movement to the intersection with the time axis (time before stopping), - the path that the body has traveled from the beginning of movement to the intersection with the time axis, - the time elapsed from the moment of crossing the time axis to this moment t, - the path that the body has traveled in the opposite direction during the time elapsed from the moment of crossing the time axis to this moment t, - the module of the displacement vector for the entire time of movement, L- the path traveled by the body during the entire movement.

3.1.8. Movement in the th second.

During this time the body will travel the following distance:

During this time the body will travel the following distance:

Then during the th interval the body will travel the following distance:

Any period of time can be taken as an interval. Most often with.

Then in 1 second the body travels the following distance:

In 2 seconds:

In 3 seconds:

If we look carefully, we will see that, etc.

Thus, we arrive at the formula:

In words: the paths traversed by a body over successive periods of time are related to each other as a series of odd numbers, and this does not depend on the acceleration with which the body moves. We emphasize that this relation is valid for

3.1.9. Equation of body coordinates for uniform motion

Coordinate equation

The signs of the projections of the initial velocity and acceleration depend on the relative position of the corresponding vectors and the axis Ox.

To solve problems, it is necessary to add to the equation the equation for changing the velocity projection onto the axis:

3.2. Graphs of kinematic quantities for rectilinear motion

3.3. Free fall body

By free fall we mean the following physical model:

1) The fall occurs under the influence of gravity:

2) There is no air resistance (in problems they sometimes write “neglect air resistance”);

3) All bodies, regardless of mass, fall with the same acceleration (sometimes they add “regardless of the shape of the body,” but we are considering the movement of only a material point, so the shape of the body is no longer taken into account);

4) The acceleration of gravity is directed strictly downwards and is equal on the surface of the Earth (in problems we often assume for convenience of calculations);

3.3.1. Equations of motion in projection onto the axis Oy

Unlike movement along a horizontal straight line, when not all tasks involve a change in direction of movement, in free fall it is best to immediately use the equations written in projections onto the axis Oy.

Body coordinate equation:

Velocity projection equation:

As a rule, in problems it is convenient to select the axis Oy in the following way:

Axis Oy directed vertically upward;

The origin coincides with the level of the Earth or the lowest point of the trajectory.

With this choice, the equations and will be rewritten in the following form:

3.4. Movement in a plane Oxy.

We considered the motion of a body with acceleration along a straight line. However, the uniformly variable motion is not limited to this. For example, a body thrown at an angle to the horizontal. In such problems, it is necessary to take into account movement along two axes at once:

Or in vector form:

And changing the projection of speed on both axes:

3.5. Application of the concept of derivative and integral

We will not provide a detailed definition of the derivative and integral here. To solve problems we need only a small set of formulas.

Derivative:

Where A, B and that is, constant values.

Integral:

Now let's see how the concepts of derivative and integral apply to physical quantities. In mathematics, the derivative is denoted by """, in physics, the derivative with respect to time is denoted by "∙" above the function.

Speed:

that is, the speed is a derivative of the radius vector.

For velocity projection:

Acceleration:

that is, acceleration is a derivative of speed.

For acceleration projection:

Thus, if the law of motion is known, then we can easily find both the speed and acceleration of the body.

Now let's use the concept of integral.

Speed:

that is, the speed can be found as the time integral of the acceleration.

Radius vector:

that is, the radius vector can be found by taking the integral of the velocity function.

Thus, if the function is known, we can easily find both the speed and the law of motion of the body.

The constants in the formulas are determined from the initial conditions - values ​​and at the moment of time

3.6. Velocity triangle and displacement triangle

3.6.1. Speed ​​triangle

In vector form with constant acceleration, the law of speed change has the form (3.5):

This formula means that a vector is equal to the vector sum of vectors and the vector sum can always be depicted in a figure (see figure).

In each problem, depending on the conditions, the velocity triangle will have its own form. This representation allows the use of geometric considerations in the solution, which often simplifies the solution of the problem.

3.6.2. Triangle of movements

In vector form, the law of motion with constant acceleration has the form:

When solving a problem, you can choose the reference system in the most convenient way, therefore, without losing generality, we can choose the reference system in such a way that, that is, we place the origin of the coordinate system at the point where the body is located at the initial moment. Then

that is, the vector is equal to the vector sum of the vectors and Let us depict it in the figure (see figure).

As in the previous case, depending on the conditions, the displacement triangle will have its own shape. This representation allows the use of geometric considerations in the solution, which often simplifies the solution of the problem.

Part 1

Calculation of instantaneous speed

1. Start with an equation. 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 is s (the movement of the body), and on the other side 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 = 3m(and at 10 + 7 = 17 m). Time = t. Usually measured in seconds.

2. Calculate the derivative of the equation. To find the instantaneous speed of a body whose movements are described by the above equation, you need to 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 free 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

3. 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

4. Substitute the appropriate t value into the derivative equation to find the instantaneous speed at a particular 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

   1. Construct a graph of the body's displacement. 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 displacement, and the X axis is 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, then this means that the body is moving in the opposite direction from the point where the movement began. Typically the graph does not extend beyond the Y axis (negative x values) - we are not measuring the speeds of objects moving backwards in time!

   2. 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 the points P(1,3) And Q(4,7) and calculate the instantaneous speed at point P.

   3. Find the slope of segment PQ. The slope of the segment PQ is equal to the ratio of the difference in the y-coordinate values ​​of points P and Q to the difference in the x-coordinate values ​​of points P and Q. In other words, H = (y Q - y P)/(x Q - x P), where H is the slope of the 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

   4. 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) = 1.8

      Q = (1.5,3.95): H = (3.95 - 3)/(1.5 - 1)
      H = (.95)/(.5) = 1.9

      Q = (1.25,3.49): H = (3.49 - 3)/(1.25 - 1)
      H = (.49)/(.25) = 1.96

   5. 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 equal to 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

   1. 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 (rather than 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 + 2

      • Now let’s substitute the value t = 4 into the derivative equation:
        

        s = 15t (2) - 6t + 2
        15(4) (2) - 6(4) + 2
        15(16) - 6(4) + 2
        240 - 24 + 2 = 22 m/s

   2. Let us estimate the value of the instantaneous speed at the point with coordinates (1.3) on the graph of the function s = 4t 2 - t. In this case, point P has coordinates (1,3) and it is necessary to find several coordinates of point Q, which lies close to point P. Then we calculate H and find the estimated values ​​of the instantaneous speed.

      • First, let's find the coordinates of Q at t = 2, 1.5, 1.1 and 1.01.
        

        s = 4t 2 - t

        t = 2: s = 4(2) 2 - (2)
        4(4) - 2 = 16 - 2 = 14, so Q = (2.14)

        t = 1.5: s = 4(1.5) 2 - (1.5)
        4(2.25) - 1.5 = 9 - 1.5 = 7.5, so Q = (1.5,7.5)

        t = 1.1: s = 4(1.1) 2 - (1.1)
        4(1.21) - 1.1 = 4.84 - 1.1 = 3.74, so Q = (1.1,3.74)

        t = 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)

This is a vector physical quantity, numerically equal to the limit to which the average speed tends over an infinitesimal period of time:

In other words, instantaneous speed is the radius vector over time.

The instantaneous velocity vector is always directed tangentially to the body's trajectory in the direction of the body's movement.

Instantaneous speed provides precise information about movement at a specific point in time. For example, when driving a car at some point in time, the driver looks at the speedometer and sees that the device shows 100 km/h. After some time, the speedometer needle points to 90 km/h, and a few minutes later – to 110 km/h. All of the listed speedometer readings are the values ​​of the instantaneous speed of the car at certain points in time. The speed at each moment of time and at each point of the trajectory must be known when docking space stations, when landing aircraft, etc.

Does the concept of “instantaneous speed” have a physical meaning? Velocity is a characteristic of change in space. However, in order to determine how the movement has changed, it is necessary to observe the movement for some time. Even the most advanced instruments for measuring speed, such as radar installations, measure speed over a period of time - albeit quite small, but this is still a finite time interval, and not a moment in time. The expression “velocity of a body at a given moment in time” is not correct from the point of view of physics. However, the concept of instantaneous speed is very convenient in mathematical calculations, and is constantly used.

Examples of solving problems on the topic “Instantaneous speed”

EXAMPLE 1

EXAMPLE 2

Exercise	The law of motion of a point in a straight line is given by the equation. Find the instantaneous speed of the point 10 seconds after the start of movement.
Solution	The instantaneous speed of a point is the radius vector in time. Therefore, for the instantaneous speed we can write: 10 seconds after the start of movement, the instantaneous speed will have the value:
Answer	10 seconds after the start of movement, the instantaneous speed of the point is m/s.

EXAMPLE 3

Exercise	A body moves in a straight line so that its coordinate (in meters) changes according to the law. How many seconds after the movement starts will the body stop?
Solution	Let's find the instantaneous speed of the body:

Rolling the body down an inclined plane (Fig. 2);

Rice. 2. Rolling the body down an inclined plane ()

Free fall (Fig. 3).

All these three types of movement are not uniform, that is, their speed changes. In this lesson we will look at uneven motion.

Uniform movement - mechanical movement in which a body travels the same distance in any equal periods of time (Fig. 4).

Rice. 4. Uniform movement

Movement is called uneven, in which the body travels unequal paths in equal periods of time.

Rice. 5. Uneven movement

The main task of mechanics is to determine the position of the body at any moment in time. When the body moves unevenly, the speed of the body changes, therefore, it is necessary to learn to describe the change in the speed of the body. To do this, two concepts are introduced: average speed and instantaneous speed.

The fact of a change in the speed of a body during uneven movement does not always need to be taken into account; when considering the movement of a body over a large section of the path as a whole (the speed at each moment of time is not important to us), it is convenient to introduce the concept of average speed.

For example, a delegation of schoolchildren travels from Novosibirsk to Sochi by train. The distance between these cities by rail is approximately 3,300 km. The speed of the train when it just left Novosibirsk was , does this mean that in the middle of the journey the speed was like this same, but at the entrance to Sochi [M1]? Is it possible, having only these data, to say that the travel time will be (Fig. 6). Of course not, since residents of Novosibirsk know that it takes approximately 84 hours to get to Sochi.

Rice. 6. Illustration for example

When considering the movement of a body over a large section of the path as a whole, it is more convenient to introduce the concept of average speed.

Medium speed they call the ratio of the total movement that the body has made to the time during which this movement was made (Fig. 7).

Rice. 7. Average speed

This definition is not always convenient. For example, an athlete runs 400 m - exactly one lap. The athlete’s displacement is 0 (Fig. 8), but we understand that his average speed cannot be zero.

Rice. 8. Displacement is 0

In practice, the concept of average ground speed is most often used.

Average ground speed is the ratio of the total path traveled by the body to the time during which the path was traveled (Fig. 9).

Rice. 9. Average ground speed

There is another definition of average speed.

average speed- this is the speed with which a body must move uniformly in order to cover a given distance in the same time in which it passed it, moving unevenly.

From the mathematics course we know what the arithmetic mean is. For numbers 10 and 36 it will be equal to:

In order to find out the possibility of using this formula to find the average speed, let's solve the following problem.

Task

A cyclist climbs a slope at a speed of 10 km/h, spending 0.5 hours. Then it goes down at a speed of 36 km/h in 10 minutes. Find the average speed of the cyclist (Fig. 10).

Rice. 10. Illustration for the problem

Given:; ; ;

Find:

Solution:

Since the unit of measurement for these speeds is km/h, we will find the average speed in km/h. Therefore, we will not convert these problems into SI. Let's convert to hours.

The average speed is:

The full path () consists of the path up the slope () and down the slope ():

The path to climb the slope is:

The path down the slope is:

The time it takes to travel the full path is:

Answer:.

Based on the answer to the problem, we see that it is impossible to use the arithmetic mean formula to calculate the average speed.

The concept of average speed is not always useful for solving the main problem of mechanics. Returning to the problem about the train, it cannot be said that if the average speed along the entire journey of the train is equal to , then after 5 hours it will be at a distance from Novosibirsk.

The average speed measured over an infinitesimal period of time is called instantaneous speed of the body(for example: a car’s speedometer (Fig. 11) shows instantaneous speed).

Rice. 11. Car speedometer shows instantaneous speed

There is another definition of instantaneous speed.

Instantaneous speed– the speed of movement of the body at a given moment in time, the speed of the body at a given point of the trajectory (Fig. 12).

Rice. 12. Instant speed

To better understand this definition, let's look at an example.

Let the car move straight along a section of highway. We have a graph of the projection of displacement versus time for a given movement (Fig. 13), let’s analyze this graph.

Rice. 13. Graph of displacement projection versus time

The graph shows that the speed of the car is not constant. Let's say you need to find the instantaneous speed of a car 30 seconds after the start of observation (at the point A). Using the definition of instantaneous speed, we find the magnitude of the average speed over the time interval from to . To do this, consider a fragment of this graph (Fig. 14).

Rice. 14. Graph of displacement projection versus time

In order to check the correctness of finding the instantaneous speed, let’s find the average speed module for the time interval from to , for this we consider a fragment of the graph (Fig. 15).

Rice. 15. Graph of displacement projection versus time

We calculate the average speed over a given period of time:

We obtained two values ​​of the instantaneous speed of the car 30 seconds after the start of observation. More accurate will be the value where the time interval is smaller, that is. If we decrease the time interval under consideration more strongly, then the instantaneous speed of the car at the point A will be determined more accurately.

Instantaneous speed is a vector quantity. Therefore, in addition to finding it (finding its module), it is necessary to know how it is directed.

(at ) – instantaneous speed

The direction of instantaneous velocity coincides with the direction of movement of the body.

If a body moves curvilinearly, then the instantaneous speed is directed tangentially to the trajectory at a given point (Fig. 16).

Exercise 1

Can instantaneous speed () change only in direction, without changing in magnitude?

Solution

To solve this, consider the following example. The body moves along a curved path (Fig. 17). Let's mark a point on the trajectory of movement A and period B. Let us note the direction of the instantaneous velocity at these points (the instantaneous velocity is directed tangentially to the trajectory point). Let the velocities and be equal in magnitude and equal to 5 m/s.

Answer: Maybe.

Task 2

Can instantaneous speed change only in magnitude, without changing in direction?

Solution

Rice. 18. Illustration for the problem

Figure 10 shows that at the point A and at the point B instantaneous speed is in the same direction. If a body moves uniformly accelerated, then .

Answer: Maybe.

In this lesson, we began to study uneven movement, that is, movement with varying speed. The characteristics of uneven motion are average and instantaneous speeds. The concept of average speed is based on the mental replacement of uneven motion with uniform motion. Sometimes the concept of average speed (as we have seen) is very convenient, but it is not suitable for solving the main problem of mechanics. Therefore, the concept of instantaneous speed is introduced.

Bibliography

1. G.Ya. Myakishev, B.B. Bukhovtsev, N.N. Sotsky. Physics 10. - M.: Education, 2008.
2. A.P. Rymkevich. Physics. Problem book 10-11. - M.: Bustard, 2006.
3. O.Ya. Savchenko. Physics problems. - M.: Nauka, 1988.
4. A.V. Peryshkin, V.V. Krauklis. Physics course. T. 1. - M.: State. teacher ed. min. education of the RSFSR, 1957.

1. Internet portal “School-collection.edu.ru” ().
2. Internet portal “Virtulab.net” ().

Homework

1. Questions (1-3, 5) at the end of paragraph 9 (page 24); G.Ya. Myakishev, B.B. Bukhovtsev, N.N. Sotsky. Physics 10 (see list of recommended readings)
2. Is it possible, knowing the average speed over a certain period of time, to find the displacement made by a body during any part of this interval?
3. What is the difference between instantaneous speed during uniform linear motion and instantaneous speed during uneven motion?
4. While driving a car, speedometer readings were taken every minute. Is it possible to determine the average speed of a car from these data?
5. The cyclist rode the first third of the route at a speed of 12 km per hour, the second third at a speed of 16 km per hour, and the last third at a speed of 24 km per hour. Find the average speed of the bike over the entire journey. Give your answer in km/hour