Problems involving free falling bodies: examples of solving problems in kinematics. Instantaneous and average speed Application of the concept of derivative and integral

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

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

Vector instantaneous speed always directed tangentially to the trajectory of the body 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 planes, 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 devices 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

EXAMPLE 3

Instantaneous speed is the speed of the body at a given moment in time or at a given point in the trajectory. 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 first derivative of the radius vector with respect to time.

2. Average speed.

Medium speed in a certain area, a value is called equal to the ratio of the movement to the period of time during which this movement occurred.

3. Angular velocity. Formula. SI.

Angular velocity is a vector physical quantity equal to the first derivative of the angle of rotation of a body with respect to time. [rad/s]

4. Relationship between angular velocity and rotation period.

Uniform rotation is characterized by rotation period and rotation frequency.

5. Angular acceleration. Formula. SI.

This is a physical quantity equal to the first derivative of the angular velocity or the second derivative of the angle of rotation of the body with respect to time. [rad/s 2 ]

6. What is the direction of the angular velocity/angular acceleration vector.

The angular velocity vector is directed along the rotation axis so that the rotation considered from the end of the angular velocity vector occurs counterclockwise (right-hand rule).

During accelerated rotation, the angular acceleration vector is co-directed with the angular velocity vector, and during slow rotation, it is opposite to it.

7/8. Relationship between normal acceleration and angular velocity/Relation between tangential and angular acceleration.

9. What determines and how is the direction of the normal component of total acceleration? Normal SI acceleration. Normal acceleration determines the rate of change in speed in direction and is directed towards the center of curvature of the trajectory.

In SI normal acceleration[m/s 2 ]

10. What determines and how is the direction of the tangential component of total acceleration.

Tangential acceleration is equal to the first time derivative of the velocity modulus and determines the rate of change in velocity modulo, and is directed tangentially to the trajectory.

11. Tangential acceleration in SI.

12. Full body acceleration. Modulus of this acceleration.

13.Mass. Strength. Newton's laws.

Weight − is a physical quantity that is a measure of the inertial and gravitational properties of a body. SI unit of mass [ m] = kg.

Strength − is a vector physical quantity, which is a measure of the mechanical impact on a body from other bodies or fields, as a result of which the body is deformed or accelerated. The SI unit of force is Newton; kg*m/s 2

Newton's first law (or law of inertia): if no forces act on the body or their action is compensated, then this body is in a state of rest or uniform linear motion.

Newton's second law : the acceleration of a body is directly proportional to the resultant forces applied to it and inversely proportional to its mass. Newton's second law allows us to solve the basic problem of mechanics. That's why it's called basic equation of translational motion dynamics.

Newton's third law : The force with which one body acts on another is equal in magnitude and opposite in direction to the force with which the second body acts on the first.

Part 1

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 there is s (the movement of the body), and on the other side - 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 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 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

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 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

   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 value the slope of the resulting segments 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) (the coordinates of 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 estimated value 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)

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 side(uniformly accelerated motion).

3.1.6. Geometric meaning area under the graph in 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 area trapezoid: (3.9)

3.1.7. Formulas for calculating path

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 reverse side)

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 intersection of the time axis to at 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 initial velocity and acceleration depend on relative position corresponding vectors and 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 consider the movement only material point, so body shape 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 as follows:

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 concept of derivative and integral applies 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 initial conditions- values ​​and at 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 loss of 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.