The speed of a point is the vector that determines at each this moment time, speed and direction of movement of the point.

The speed of uniform motion is determined by the ratio of the path traveled by a point in a certain period of time to the value of this period of time.

Speed; S-path; t- time.

Speed is measured in units of length divided by unit of time: m/s; cm/s; km/h, etc.

In the case of rectilinear motion, the velocity vector is directed along the trajectory in the direction of its movement.

If a point travels unequal paths in equal periods of time, then this movement is called uneven. Speed is a variable quantity and is a function of time.

The average speed of a point over a given period of time is the speed of such uniform rectilinear motion at which the point during this period of time would receive the same displacement as in its movement under consideration.

Let's consider point M, which moves along a curvilinear trajectory specified by the law

Over a period of time?t, point M will move to position M1 along the arc MM 1. If the time period?t is small, then arc MM 1 can be replaced by a chord and, to a first approximation, find the average speed of the point

This speed is directed along the chord from point M to point M 1. We find the true speed by going to the limit at?t> 0

When?t> 0, the direction of the chord in the limit coincides with the direction of the tangent to the trajectory at point M.

Thus, the value of the speed of a point is defined as the limit of the ratio of the increment of the path to the corresponding period of time as the latter tends to zero. The direction of the velocity coincides with the tangent to the trajectory at a given point.

Point acceleration

Note that in the general case, when moving along a curved path, the speed of a point changes both in direction and in magnitude. The change in speed per unit time is determined by acceleration. In other words, the acceleration of a point is a quantity that characterizes the rate of change in speed over time. If during the time interval?t the speed changes by an amount, then the average acceleration

The true acceleration of a point at a given time t is the value to which the average acceleration tends at?t> 0, that is

As the time interval tends to zero, the acceleration vector will change both in magnitude and direction, tending to its limit.

Acceleration dimension

Acceleration can be expressed in m/s 2 ; cm/s 2, etc.

In the general case, when the motion of a point is given in a natural way, the acceleration vector is usually decomposed into two components, directed tangentially and normal to the trajectory of the point.

Then the acceleration of the point at time t can be represented as follows

Let us denote the component limits by and.

The direction of the vector does not depend on the value of the time interval?t.

This acceleration always coincides with the direction of the velocity, that is, it is directed tangentially to the trajectory of the point and is therefore called tangential or tangential acceleration.

The second component of the acceleration of a point is directed perpendicular to the tangent to the trajectory at a given point towards the concavity of the curve and affects the change in the direction of the velocity vector. This component of acceleration is called normal acceleration.

Since the numerical value of the vector is equal to the increment in the speed of the point over the considered period?t of time, then the numerical value of the tangential acceleration

The numerical value of the tangential acceleration of a point is equal to the time derivative of the numerical value of the velocity. The numerical value of the normal acceleration of a point is equal to the square of the point’s speed divided by the radius of curvature of the trajectory at the corresponding point on the curve

The total acceleration during uneven curvilinear motion of a point is composed geometrically of the tangential and normal accelerations.

Methods for specifying the movement of a point.

Set point movement - this means indicating a rule by which at any moment in time one can determine its position in a given frame of reference.

The mathematical expression for this rule is called law of motion , or equation of motion points.

There are three ways to specify the movement of a point:

vector;

coordinate;

natural.

To set the movement in a vector way, need to:

à select a fixed center;

à determine the position of the point using the radius vector, starting at the stationary center and ending at the moving point M;

à define this radius vector as a function of time t: .

Expression

called vector law of motion dots, or vector equation of motion.

!! Radius vector – this is the distance (vector modulus) + direction from the center O to the point M, which can be determined in different ways, for example, by angles with given directions.

To set movement coordinate method , need to:

à select and fix a coordinate system (any: Cartesian, polar, spherical, cylindrical, etc.);

à determine the position of a point using the appropriate coordinates;

à set these coordinates as a function of time t.

In the Cartesian coordinate system, therefore, it is necessary to indicate the functions

In the polar coordinate system, the polar radius and polar angle should be defined as functions of time:

In general, with the coordinate method of specifying, those coordinates with which the current position of the point is determined should be specified as a function of time.

To be able to set the movement of a point in a natural way, you need to know it trajectory . Let us write down the definition of the trajectory of a point.

Trajectory points are called the set of its positions over any period of time(usually from 0 to +¥).

In the example with a wheel rolling along the road, the trajectory of point 1 is cycloid, and points 2 – roulette; in the reference system associated with the center of the wheel, the trajectories of both points are circle.

To set the movement of a point in a natural way, you need:

à know the trajectory of the point;

à on the trajectory, select the origin and positive direction;

à determine the current position of a point by the length of the trajectory arc from the origin to this current position;

à indicate this length as a function of time.

The expression defining the above function is

called law of motion of a point along a trajectory, or natural equation of motion points.

Depending on the type of function (4), a point along a trajectory can move in different ways.

3. Trajectory of a point and its definition.

The definition of the concept “trajectory of a point” was given earlier in question 2. Let us consider the question of determining the trajectory of a point when in different ways movement tasks.

The natural way: The trajectory must be given, so there is no need to find it.

Vector method : you need to go to the coordinate method according to the equalities

Coordinate method: it is necessary to exclude time t from the equations of motion (2), or (3).

Coordinate equations of motion define the trajectory parametrically, through the parameter t (time). To obtain an explicit equation for the curve, the parameter must be excluded from the equations.

After eliminating time from equations (2), two equations of cylindrical surfaces are obtained, for example, in the form

The intersection of these surfaces will be the trajectory of the point.

When a point moves along a plane, the problem becomes simpler: after eliminating time from the two equations

The trajectory equation will be obtained in one of the following forms:

When will be , therefore the trajectory of the point will be the right branch of the parabola:

From the equations of motion it follows that

therefore, the trajectory of the point will be the part of the parabola located in the right half-plane:

Then we get

Since the entire ellipse will be the trajectory of the point.

At the center of the ellipse will be at the origin O; at we get a circle; the parameter k does not affect the shape of the ellipse; the speed of movement of the point along the ellipse depends on it. If you swap cos and sin in the equations, then the trajectory will not change (the same ellipse), but the initial position of the point and the direction of movement will change.

The speed of a point characterizes the “speed” of change in its position. Formally: speed – movement of a point per unit of time.

Precise definition.

Then Attitude

And why is it 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.

Let's solve this problem: 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.

Need help solving problems? Professional student service is ready to provide it.

1.2. Straight-line movement

1.2.4. average speed

A material point (body) retains its speed unchanged only with uniform straight motion. If the movement is uneven (including uniformly variable), then the speed of the body changes. This movement is characterized by average speed. A distinction is made between average travel speed and average ground speed.

Average moving speed is a vector physical quantity, which is determined by the formula

v → r = Δ r → Δ t,

where Δ r → is the displacement vector; ∆t is the time interval during which this movement occurred.

Average ground speed is a scalar physical quantity and is calculated by the formula

v s = S total t total,

where S total = S 1 + S 1 + ... + S n; ttot = t 1 + t 2 + ... + t N .

Here S 1 = v 1 t 1 - the first section of the path; v 1 - speed of passage of the first section of the path (Fig. 1.18); t 1 - time of movement on the first section of the route, etc.

Rice. 1.18

Example 7. One quarter of the way the bus moves at a speed of 36 km/h, the second quarter of the way - 54 km/h, the remaining way - at a speed of 72 km/h. Calculate the average ground speed of the bus.

Solution. We denote the total path traveled by the bus as S:

Stot = S.

S 1 = S /4 - the path traveled by the bus on the first section,

S 2 = S /4 - the path traveled by the bus on the second section,

S 3 = S /2 - the path traveled by the bus in the third section.

The bus travel time is determined by the formulas:

in the first section (S 1 = S /4) -
t 1 = S 1 v 1 = S 4 v 1 ;
in the second section (S 2 = S /4) -
t 2 = S 2 v 2 = S 4 v 2 ;
in the third section (S 3 = S /2) -
t 3 = S 3 v 3 = S 2 v 3 .

Total time bus movement is:

t total = t 1 + t 2 + t 3 = S 4 v 1 + S 4 v 2 + S 2 v 3 = S (1 4 v 1 + 1 4 v 2 + 1 2 v 3) .

v s = S total t total = S S (1 4 v 1 + 1 4 v 2 + 1 2 v 3) =

1 (1 4 v 1 + 1 4 v 2 + 1 2 v 3) = 4 v 1 v 2 v 3 v 2 v 3 + v 1 v 3 + 2 v 1 v 2 .

v s = 4 ⋅ 36 ⋅ 54 ⋅ 72 54 ⋅ 72 + 36 ⋅ 72 + 2 ⋅ 36 ⋅ 54 = 54 km/h.

Example 8. A city bus spends a fifth of its time stopping, the rest of the time it moves at a speed of 36 km/h. Determine the average ground speed of the bus.

Solution. Let us denote the total travel time of the bus on the route by t:

ttot = t.

t 1 = t /5 - time spent stopping,

t 2 = 4t /5 - bus travel time.

Distance covered by the bus:

during time t 1 = t /5 -
S 1 = v 1 t 1 = 0,

since the speed of the bus v 1 at a given time interval is zero (v 1 = 0);

during time t 2 = 4t /5 -
S 2 = v 2 t 2 = v 2 4 t 5 = 4 5 v 2 t ,
where v 2 is the speed of the bus at a given time interval (v 2 = 36 km/h).

The general route of the bus is:

S total = S 1 + S 2 = 0 + 4 5 v 2 t = 4 5 v 2 t.

We will calculate the average ground speed of the bus using the formula

v s = S total t total = 4 5 v 2 t t = 4 5 v 2 .

The calculation gives the value of the average ground speed:

v s = 4 5 ⋅ 36 = 30 km/h.

Example 9. The equation of motion of a material point has the form x (t) = (9.0 − 6.0t + 2.0t 2) m, where the coordinate is given in meters, time in seconds. Determine the average ground speed and the average speed of movement of a material point in the first three seconds of movement.

Solution. For determining average moving speed it is necessary to calculate the displacement of a material point. The module of movement of a material point in the time interval from t 1 = 0 s to t 2 = 3.0 s will be calculated as the difference in coordinates:

| Δ r → | = | x (t 2) − x (t 1) | ,

Substituting the values into the formula to calculate the displacement modulus gives:

| Δ r → | = | x (t 2) − x (t 1) | = 9.0 − 9.0 = 0 m.

Thus, the displacement of the material point is zero. Consequently, the module of the average movement speed is also equal to zero:

| v → r | = | Δ r → | t 2 − t 1 = 0 3.0 − 0 = 0 m/s.

For determining average ground speed you need to calculate the path traveled by a material point during the time interval from t 1 = 0 s to t 2 = 3.0 s. The movement of the point is uniformly slow, so it is necessary to find out whether the stopping point falls within the specified interval.

To do this, we write the law of change in the speed of a material point over time in the form:

v x = v 0 x + a x t = − 6.0 + 4.0 t ,

where v 0 x = −6.0 m/s is the projection of the initial velocity onto the Ox axis; a x = = 4.0 m/s 2 - projection of acceleration onto the indicated axis.

Let's find the stopping point from the condition

v (τ rest) = 0,

those.

τ rest = v 0 a = 6.0 4.0 = 1.5 s.

The stopping point falls within the time interval from t 1 = 0 s to t 2 = 3.0 s. Thus, we calculate the distance traveled using the formula

S = S 1 + S 2,

where S 1 = | x (τ rest) − x (t 1) | - the path traveled by the material point to the stop, i.e. during the time from t 1 = 0 s to τ rest = 1.5 s; S 2 = | x (t 2) − x (τ rest) | - the path traveled by the material point after stopping, i.e. during the time from τ rest = 1.5 s to t 1 = 3.0 s.

Let's calculate the coordinate values at the specified times:

x (t 1) = 9.0 − 6.0 t 1 + 2.0 t 1 2 = 9.0 − 6.0 ⋅ 0 + 2.0 ⋅ 0 2 = 9.0 m;

x (τ rest) = 9.0 − 6.0 τ rest + 2.0 τ rest 2 = 9.0 − 6.0 ⋅ 1.5 + 2.0 ⋅ (1.5) 2 = 4.5 m ;

x (t 2) = 9.0 − 6.0 t 2 + 2.0 t 2 2 = 9.0 − 6.0 ⋅ 3.0 + 2.0 ⋅ (3.0) 2 = 9.0 m .

The coordinate values allow you to calculate the paths S 1 and S 2:

S 1 = | x (τ rest) − x (t 1) | = | 4.5 − 9.0 | = 4.5 m;

S 2 = | x (t 2) − x (τ rest) | = | 9.0 − 4.5 | = 4.5 m,

as well as the total distance traveled:

S = S 1 + S 2 = 4.5 + 4.5 = 9.0 m.

Consequently, the desired value of the average ground speed of the material point is equal to

v s = S t 2 − t 1 = 9.0 3.0 − 0 = 3.0 m/s.

Example 10. The graph of the projection of the velocity of a material point versus time is a straight line and passes through the points (0; 8.0) and (12; 0), where the velocity is given in meters per second, time in seconds. How many times does the average ground speed for 16 seconds of movement exceed the average speed of movement for the same time?

Solution. A graph of the projection of body velocity versus time is shown in the figure.

To graphically calculate the path traveled by a material point and the modulus of its movement, it is necessary to determine the value of the velocity projection at a time equal to 16 s.

There are two ways to determine the value of v x at a specified point in time: analytical (through the equation of a straight line) and graphical (through the similarity of triangles). To find v x, we use the first method and draw up an equation of a straight line using two points:

t − t 1 t 2 − t 1 = v x − v x 1 v x 2 − v x 1 ,

where (t 1 ; v x 1) - coordinates of the first point; (t 2 ; v x 2) - coordinates of the second point. According to the conditions of the problem: t 1 = 0, v x 1 = 8.0, t 2 = 12, v x 2 = 0. Taking into account specific coordinate values, this equation takes the form:

t − 0 12 − 0 = v x − 8.0 0 − 8.0 ,

v x = 8.0 − 2 3 t .

At t = 16 s the velocity projection value is

| v x | = 8 3 m/s.

This value can also be obtained from the similarity of triangles.

Let us calculate the path traveled by the material point as the sum of the values S 1 and S 2:
S = S 1 + S 2,
where S 1 = 1 2 ⋅ 8.0 ⋅ 12 = 48 m - the path traveled by the material point during the time interval from 0 s to 12 s; S 2 = 1 2 ⋅ (16 − 12) ⋅ | v x | = 1 2 ⋅ 4.0 ⋅ 8 3 = = 16 3 m - the path traveled by the material point during the time interval from 12 s to 16 s.

The total distance traveled is

S = S 1 + S 2 = 48 + 16 3 = 160 3 m.

The average ground speed of a material point is equal to

v s = S t 2 − t 1 = 160 3 ⋅ 16 = 10 3 m/s.

Let us calculate the value of the movement of a material point as the modulus of the difference between the values S 1 and S 2:
S = | S 1 − S 2 | = | 48 − 16 3 | = 128 3 m.

The average speed of movement is

| v → r | = | Δ r → | t 2 − t 1 = 128 3 ⋅ 16 = 8 3 m/s.

The required speed ratio is

v s | v → r | = 10 3 ⋅ 3 8 = 10 8 = 1.25.

The average ground speed of a material point is 1.25 times higher than the module of the average speed of movement.