Mechanical motion is called a change over time in the position in space of points and bodies relative to any main body to which the reference system is attached. Kinematics studies the mechanical movement of points and bodies, regardless of the forces causing these movements. Any movement, like rest, is relative and depends on the choice of reference system.
The trajectory of a point is a continuous line described by a moving point. If the trajectory is a straight line, then the movement of the point is called rectilinear, and if it is a curve, then it is called curvilinear. If the trajectory is flat, then the motion of the point is called flat.
The movement of a point or body is considered given or known if for each moment of time (t) it is possible to indicate the position of the point or body relative to the selected coordinate system.
The position of a point in space is determined by the task:
a) point trajectories;
b) the beginning O 1 of the distance reading along the trajectory (Figure 11): s = O 1 M - curvilinear coordinate of point M;
c) the direction of the positive count of distances s;
d) equation or law of motion of a point along a trajectory: S = s(t)
Point speed. If a point travels equal distances in equal periods of time, then its motion is called uniform. The speed of uniform motion is measured by the ratio of the path z traveled by a point in a certain period of time to the value of this period of time: v = s/1. If a point travels unequal paths in equal periods of time, then its movement is called uneven. The speed in this case is also variable and is a function of time: v = v(t). Let's consider point A, which moves along a given trajectory according to a certain law s = s(t) (Figure 12):
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Over a period of time t t. A moved to position A 1 along the arc AA. If the time period Δt is small, then the arc AA 1 can be replaced by a chord and find, as a first approximation, the average speed of the point v cp = Ds/Dt. The average speed is directed along the chord from point A to point A 1.
The true speed of a point is directed tangentially to the trajectory, and its algebraic value is determined by the first derivative of the path with respect to time:
v = limΔs/Δt = ds/dt
Dimension of point speed: (v) = length/time, for example, m/s. If the point moves in the direction of increasing curvilinear coordinate s, then ds > 0, and therefore v > 0, otherwise ds< 0 и v < 0.
Point acceleration. The change in speed per unit time is determined by acceleration. Let's consider the movement of point A along a curvilinear trajectory in time Δt from position A to position A 1 . In position A the point had a speed v, and in position A 1 - a speed v 1 (Figure 13). those. the speed of the point changed in magnitude and direction. We find the geometric difference of speeds Δv by constructing the vector v 1 from point A.
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The acceleration of a point is the vector “, which is equal to the first derivative of the point’s velocity vector with respect to time:
The found acceleration vector a can be decomposed into two mutually perpendicular components but tangent and normal to the trajectory of motion. Tangential acceleration a 1 coincides in direction with the speed during accelerated motion or is opposite to it during replaced motion. It characterizes the change in speed and is equal to the derivative of the speed with respect to time
The normal acceleration vector a is directed along the normal (perpendicular) to the curve towards the concavity of the trajectory, and its modulus is equal to the ratio of the square of the velocity of the point to the radius of curvature of the trajectory at the point in question.
Normal acceleration characterizes the change in speed along
direction.
Total acceleration value: 
, m/s 2
Types of point motion depending on acceleration.
Uniform linear movement(motion by inertia) is characterized by the fact that the speed of movement is constant, and the radius of curvature of the trajectory is equal to infinity.
That is, r = ¥, v = const, then ; and therefore . So, when a point moves by inertia, its acceleration is zero.
Rectilinear uneven movement. The radius of curvature of the trajectory is r = ¥, and n = 0, therefore a = a t and a = a t = dv/dt.
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 planes, etc.
Does the concept of "instantaneous speed" 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
| 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: |
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 polar system coordinates should be defined as functions of time, polar radius and polar angle:
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
1.2. Straight-line movement
1.2.4. Average speed
A material point (body) retains its speed unchanged only with uniform rectilinear 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 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. General path traversed by the bus, we denote S:
Stotal = 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 .
The total travel time of the bus 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: Equation of Motion 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. To determine average moving speed it is necessary to calculate the movement 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. Therefore, the modulus of the average movement speed is also zero:
| v → r | = | Δ r → | t 2 − t 1 = 0 3.0 − 0 = 0 m/s.
To determine 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.