Rectilinear uniform motion.
Speed (v) - physical quantity, is numerically equal to the path (s) traveled by the body per unit time (t).
Path
Path (S) - the length of the trajectory along which the body moved, is numerically equal to the product of the speed (v) of the body and the time (t) of movement.
Driving time
The time of movement (t) is equal to the ratio of the distance (S) traveled by the body to the speed (v) of movement.
average speed
The average speed (vср) is equal to the ratio of the sum of the path sections (s 1 s 2, s 3, ...) traveled by the body to the time period (t 1 + t 2 + t 3 + ...) during which this path was traveled .
average speed- this is the ratio of the length of the path traveled by the body to the time during which this path was covered.
average speed for uneven movement in a straight line: this is the ratio of the entire path to the entire time.
Two successive stages at different speeds: 
Where
When solving problems - how many stages of movement there will be so many components: 
Projections of the displacement vector on the coordinate axes
Projection of the displacement vector onto the OX axis:
Projection of the displacement vector onto the OY axis:
The projection of a vector onto an axis is zero if the vector is perpendicular to the axis.
Signs of displacement projections: a projection is considered positive if the movement from the projection of the beginning of the vector to the projection of the end occurs in the direction of the axis, and negative if against the axis. IN in this example
Motion module is the length of the displacement vector:
According to the Pythagorean theorem:
Motion projections and tilt angle
In this example:
Coordinate equation (in general form):
Radius vector- a vector, the beginning of which coincides with the origin of coordinates, and the end - with the position of the body in this moment time. Projections of the radius vector on the coordinate axes determine the coordinates of the body at a given time.
The radius vector allows you to specify the position of a material point in a given reference system:
Uniform linear motion - definition
Uniform linear movement- a movement in which a body makes equal movements over any equal periods of time.
Speed during uniform linear motion. Speed is a vector physical quantity that shows how much movement a body makes per unit time.
In vector form:
In projections onto the OX axis:
Additional speed units:
1 km/h = 1000 m/3600 s,
1 km/s = 1000 m/s,
1 cm/s = 0.01 m/s,
1 m/min =1 m/60 s.
The measuring device - speedometer - shows the speed module.
The sign of the velocity projection depends on the direction of the velocity vector and the coordinate axis:
The velocity projection graph represents the dependence of the velocity projection on time:
Velocity graph for uniform linear motion- straight line parallel to the time axis (1, 2, 3).
If the graph lies above the time axis (.1), then the body moves in the direction of the OX axis. If the graph is located under the time axis, then the body moves against the OX axis (2, 3).
Geometric meaning of movement.
With uniform linear motion, the displacement is determined by the formula. We get the same result if we calculate the area of the figure under the velocity graph in the axes. This means that to determine the path and modulus of displacement during linear motion, it is necessary to calculate the area of the figure under the velocity graph in the axes:
Displacement Projection Graph- dependence of the displacement projection on time.
Displacement projection graph at uniform rectilinear motion- a straight line coming from the origin of coordinates (1, 2, 3).
If straight line (1) lies above the time axis, then the body moves in the direction of the OX axis, and if under the axis (2, 3), then against the OX axis.
The greater the tangent of slope (1) of the graph, the greater the velocity module.
Graph coordinates- dependence of the body coordinates on time:
Graph of coordinates for uniform rectilinear motion - straight lines (1, 2, 3).
If the coordinate increases over time (1, 2), then the body moves in the direction of the OX axis; if the coordinate decreases (3), then the body moves against the direction of the OX axis.
The greater the tangent of the angle of inclination (1), the greater the speed module.
If the coordinate graphs of two bodies intersect, then from the intersection point perpendiculars should be lowered onto the time axis and coordinate axis.
Relativity of mechanical motion
By relativity we understand the dependence of something on the choice of frame of reference. For example, peace is relative; movement is relative and the position of the body is relative.
The rule for adding displacements. Vector sum of displacements
where is the movement of the body relative to the moving frame of reference (MSF); - movement of the PSO relative to the fixed reference system (FRS); - movement of the body relative to a fixed frame of reference (FFR).
Vector addition:
Addition of vectors directed along one straight line:
Addition of vectors perpendicular to each other
According to the Pythagorean theorem 
The most important thing for us is to be able to calculate the displacement of a body, because, knowing the displacement, we can also find the coordinates of the body, and this is the main task mechanics. How to calculate displacement during uniformly accelerated motion?
The easiest way to obtain the formula for determining displacement is to use the graphical method.
In § 9 we saw that in case of rectilinear uniform motion, the displacement of the body is numerically equal to the area of the figure (rectangle) located under the velocity graph. Is this true for uniformly accelerated motion?
With uniformly accelerated motion of a body occurring along the coordinate axis X, the speed does not remain constant over time, but changes with time according to the formulas:
Therefore, the speed graphs have the form shown in Figure 40. Line 1 in this figure corresponds to movement with “positive” acceleration (speed increases), line 2 corresponds to movement with “negative” acceleration (speed decreases). Both graphs refer to the case when at the moment of time the body had a speed
Let us select a small section on the speed graph of uniformly accelerated motion (Fig. 41) and drop from points a and perpendiculars to the axis. The length of the segment on the axis is numerically equal to the small period of time during which the speed changed from its value at point a to its value at point Below the section the graphics turned out to be a narrow strip
If the period of time numerically equal to the segment is small enough, then during this time the change in speed is also small. The movement during this period of time can be considered uniform, and the strip will then differ little from the rectangle. The area of the strip is therefore numerically equal to the displacement of the body during the time corresponding to the segment
But the entire area of the figure located under the speed graph can be divided into such narrow strips. Consequently, the displacement over the entire time is numerically equal to the area of the trapezoid. The area of the trapezoid, as is known from geometry, is equal to the product of half the sum of its bases and the height. In our case, the length of one of the bases of the trapezoid is numerically equal to the length of the other - V. Its height is numerically equal. It follows that the displacement is equal to:
Let us substitute expression (1a) into this formula, then
Dividing the numerator by the denominator term by term, we get:
Substituting expression (16) into formula (2), we obtain (see Fig. 42):
Formula (2a) is used in the case when the acceleration vector is directed in the same way as the coordinate axis, and formula (26) when the direction of the acceleration vector is opposite to the direction of this axis.
If the initial speed is zero (Fig. 43) and the acceleration vector is directed along the coordinate axis, then from formula (2a) it follows that
If the direction of the acceleration vector is opposite to the direction of the coordinate axis, then from formula (26) it follows that
(the “-” sign here means that the displacement vector, as well as the acceleration vector, is directed opposite to the selected coordinate axis).
Let us recall that in formulas (2a) and (26) the quantities and can be both positive and negative - these are projections of the vectors and
Now that we have obtained formulas for calculating displacement, it is easy for us to obtain a formula for calculating the coordinates of the body. We saw (see § 8) that in order to find the coordinate of a body at some point in time, we need to add to the initial coordinate the projection of the body’s displacement vector onto the coordinate axis:
(For) if the acceleration vector is directed in the same way as the coordinate axis, and
if the direction of the acceleration vector is opposite to the direction of the coordinate axis.
These are the formulas that allow you to find the position of a body at any moment in time during rectilinear uniformly accelerated motion. To do this, you need to know the initial coordinate of the body, its initial speed and acceleration a.
Problem 1. The driver of a car moving at a speed of 72 km/h saw a red traffic light and pressed the brake. After this, the car began to slow down, moving with acceleration
How far will the car travel in seconds after the start of braking? How far will the car travel before coming to a complete stop?
Solution. For the origin of coordinates, we choose the point on the road at which the car began to slow down. We will direct the coordinate axis in the direction of movement of the car (Fig. 44), and we will refer the beginning of the time count to the moment at which the driver pressed the brake. The speed of the car is in the same direction as the X-axis, and the acceleration of the car is opposite to the direction of that axis. Therefore, the projection of velocity onto the X axis is positive, and the projection of acceleration is negative, and the coordinate of the car must be found using formula (36):
Substituting the values into this formula
Now let’s find how far the car will travel before it comes to a complete stop. To do this we need to know the travel time. It can be found out using the formula
Since at the moment when the car stops, its speed is zero, then
The distance that the car will travel before coming to a complete stop is equal to the coordinates of the car at the moment of time
Task 2. Determine the displacement of the body, the velocity graph of which is shown in Figure 45. The acceleration of the body is equal to a.
Solution. Since at first the modulus of the body’s velocity decreases with time, the acceleration vector is directed opposite to the direction . To calculate the displacement we can use the formula
From the graph it is clear that the movement time is therefore:
The answer obtained shows that the graph shown in Figure 45 corresponds to the movement of a body first in one direction, and then by the same distance in the opposite direction, as a result of which the body ends up at the starting point. Such a graph could, for example, relate to the motion of a body thrown vertically upward.
Problem 3. A body moves along a straight line uniformly accelerated with acceleration a. Find the difference in the distances traveled by the body in two successive equal periods of time i.e.
Solution. Let us take the straight line along which the body moves as the X axis. If at point A (Fig. 46) the speed of the body was equal, then its displacement over time is equal to:
At point B the body had a speed and its displacement over the next period of time is equal to:
2. Figure 47 shows graphs of the speed of movement of three bodies? What is the nature of the movement of these bodies? What can be said about the speeds of movement of bodies at moments of time corresponding to points A and B? Determine the accelerations and write the equations of motion (formulas for speed and displacement) of these bodies.
3. Using the graphs of the velocities of three bodies shown in Figure 48, complete the following tasks: a) Determine the accelerations of these bodies; b) make up for
of each body, the formula for the dependence of speed on time: c) in what ways are the movements corresponding to graphs 2 and 3 similar and different?
4. Figure 49 shows graphs of the speed of movement of three bodies. Using these graphs: a) determine what the segments OA, OB and OS correspond to on the coordinate axes; 6) find the accelerations with which the bodies move: c) write the equations of motion for each body.
5. When taking off, an airplane passes the runway in 15 seconds and at the moment it takes off from the ground it has a speed of 100 m/sec. How fast was the plane moving and what was the length of the runway?
6. The car stopped at a traffic light. After the green signal lights up, it begins to move with acceleration and moves until its speed becomes equal to 16 m/sec, after which it continues to move at a constant speed. At what distance from the traffic light will the car be 15 seconds after the green signal appears?
7. A projectile whose speed is 1,000 m/sec penetrates the wall of the dugout for and after that has a speed of 200 m/sec. Assuming that the motion of a projectile in the thickness of a wall is uniformly accelerated, find the thickness of the wall.
8. The rocket moves with acceleration and at some point in time reaches a speed of 900 m/sec. Which path will she take next?
9. At what distance from the Earth would you be? spaceship 30 minutes after the start, if it were moving in a straight line with acceleration all the time
Rectilinear uniform motion - this is a movement in which, in equal periods of time, the body travels the same distance.
Uniform movement- this is the movement of a body in which its speed remains constant (), that is, it moves at the same speed all the time, and acceleration or deceleration does not occur ().
Straight-line movement- this is the movement of a body in a straight line, that is, the trajectory we get is straight.
The speed of uniform rectilinear motion does not depend on time and at each point of the trajectory is directed in the same way as the movement of the body. That is, the velocity vector coincides with the displacement vector. With all this average speed at any time interval is equal to the initial and instantaneous speed:
Speed of uniform rectilinear motion is a physical vector quantity equal to the ratio of the movement of a body over any period of time to the value of this interval t:
From this formula. we can easily express body movement with uniform motion:
Let's consider the dependence of speed and displacement on time
Since our body moves rectilinearly and uniformly accelerated (), the graph with the dependence of speed on time will look like a parallel straight line to the time axis.
Depending projections of body velocity versus time there is nothing complicated. The projection of the body's movement is numerically equal to the area of the rectangle AOBC, since the magnitude of the movement vector is equal to the product of the velocity vector and the time during which the movement was made.
On the graph we see dependence of movement on time.
The graph shows that the projection of the velocity is equal to:
Having considered this formula. we can say that the larger the angle, the faster our body moves and it covers a greater distance in less time
Let's try to derive a formula for finding the projection of the displacement vector of a body that moves rectilinearly and uniformly accelerated for any period of time.
To do this, let us turn to the graph of the projection of the velocity of rectilinear uniformly accelerated motion versus time.
Graph of the projection of the velocity of rectilinear uniformly accelerated motion versus time
The figure below shows a graph for the projection of the speed of some body that moves with initial speed V0 and constant acceleration a.
If we had uniform rectilinear motion, then to calculate the projection of the displacement vector, it would be necessary to calculate the area of the figure under the graph of the projection of the velocity vector.
Now we will prove that in the case of uniformly accelerated rectilinear motion, the projection of the displacement vector Sx will be determined in the same way. That is, the projection of the displacement vector will be equal to the area of the figure under the graph of the projection of the velocity vector.
Let us find the area of the figure limited by the ot-axis, segments AO and BC, as well as segment AC.
Let us select a small time interval db on the ot axis. Let us draw perpendiculars to the time axis through these points until they intersect with the graph of the velocity projection. Let us mark the intersection points a and c. During this period of time, the speed of the body will change from Vax to Vbx.
If we take this interval small enough, then we can assume that the speed remains practically unchanged, and therefore we will be dealing with uniform rectilinear motion in this interval.
Then we can consider the segment ac to be horizontal, and abcd to be a rectangle. The area abcd will be numerically equal to the projection of the displacement vector over the time interval db. We can divide the entire area of the OACB figure into such small periods of time.
That is, we found that the projection of the displacement vector Sx for the period of time corresponding to the segment OB will be numerically equal to the area S of the trapezoid OACB, and will be determined by the same formula as this area.
Hence,
- S=((V0x+Vx)/2)*t.
Since Vx=V0x+ax*t and S=Sx, the resulting formula will take the following form:
- Sx=V0x*t+(ax*t^2)/2.
We have obtained a formula with which we can calculate the projection of the displacement vector during uniformly accelerated motion.
In the case of uniformly slow motion, the formula will take the following form.
Uniform movement– this is movement at a constant speed, that is, when the speed does not change (v = const) and acceleration or deceleration does not occur (a = 0).
Straight-line movement- this is movement in a straight line, that is, the trajectory of rectilinear movement is a straight line.
- this is a movement in which a body makes equal movements at any equal intervals of time. For example, if we divide a certain time interval into one-second intervals, then with uniform motion the body will move the same distance for each of these time intervals.
The speed of uniform rectilinear motion does not depend on time and at each point of the trajectory is directed in the same way as the movement of the body. That is, the displacement vector coincides in direction with the velocity vector. In this case, the average speed for any period of time is equal to the instantaneous speed:
Speed of uniform rectilinear motion is a physical vector quantity equal to the ratio of the movement of a body over any period of time to the value of this interval t:
V(vector) = s(vector) / t
Thus, the speed of uniform rectilinear motion shows how much movement a material point makes per unit time.
Moving with uniform linear motion is determined by the formula:
s(vector) = V(vector) t
Distance traveled in linear motion is equal to the displacement module. If the positive direction of the OX axis coincides with the direction of movement, then the projection of the velocity onto the OX axis is equal to the magnitude of the velocity and is positive:
v x = v, that is v > 0
The projection of displacement onto the OX axis is equal to:
s = vt = x – x 0
where x 0 is the initial coordinate of the body, x is the final coordinate of the body (or the coordinate of the body at any time)
Equation of motion, that is, the dependence of the body coordinates on time x = x(t), takes the form:
If the positive direction of the OX axis is opposite to the direction of motion of the body, then the projection of the body’s velocity onto the OX axis is negative, the speed is less than zero (v< 0), и тогда уравнение движения принимает вид:
4. Equally alternating motion.
Uniform linear movement- This is a special case of uneven motion.
Uneven movement- this is a movement in which a body (material point) makes unequal movements over equal periods of time. For example, a city bus moves unevenly, since its movement consists mainly of acceleration and deceleration.
Equally alternating motion- this is a movement in which the speed of a body (material point) changes equally over any equal periods of time.
Acceleration of a body during uniform motion remains constant in magnitude and direction (a = const).
Uniform motion can be uniformly accelerated or uniformly decelerated.
Uniformly accelerated motion- this is the movement of a body (material point) with positive acceleration, that is, with such movement the body accelerates with constant acceleration. In the case of uniformly accelerated motion, the modulus of the body’s velocity increases over time, and the direction of acceleration coincides with the direction of the speed of movement.
Equal slow motion- this is the movement of a body (material point) with negative acceleration, that is, with such movement the body uniformly slows down. In uniformly slow motion, the velocity and acceleration vectors are opposite, and the velocity modulus decreases over time.
In mechanics, any rectilinear motion is accelerated, therefore slow motion differs from accelerated motion only in the sign of the projection of the acceleration vector onto the selected axis of the coordinate system.
Average variable speed is determined by dividing the movement of the body by the time during which this movement was made. The unit of average speed is m/s.
Instantaneous speed is the speed of a body (material point) at a given moment of time or at a given point of the trajectory, that is, the limit to which the average speed tends with an infinite decrease in the time interval Δt:
V=lim(^t-0) ^s/^t
Instantaneous velocity vector uniformly alternating motion can be found as the first derivative of the displacement vector with respect to time:
V(vector) = s’(vector)
Velocity vector projection on the OX axis:
this is the derivative of the coordinate with respect to time (the projections of the velocity vector onto other coordinate axes are similarly obtained).
Acceleration is a quantity that determines the rate of change in the speed of a body, that is, the limit to which the change in speed tends with an infinite decrease in the time period Δt:
a(vector) = lim(t-0) ^v(vector)/^t
Acceleration vector of uniformly alternating motion can be found as the first derivative of the velocity vector with respect to time or as the second derivative of the displacement vector with respect to time:
a(vector) = v(vector)" = s(vector)"
Considering that 0 is the speed of the body at the initial moment of time (initial speed), is the speed of the body at a given moment of time (final speed), t is the period of time during which the change in speed occurred, acceleration formula will be as follows:
a(vector) = v(vector)-v0(vector)/t
From here uniform speed formula at any time:
v(vector) = v 0 (vector) + a(vector)t
If a body moves rectilinearly along the OX axis of a rectilinear Cartesian coordinate system, coinciding in direction with the body’s trajectory, then the projection of the velocity vector onto this axis is determined by the formula:
v x = v 0x ± a x t
The “-” (minus) sign in front of the projection of the acceleration vector refers to uniformly slow motion. The equations for projections of the velocity vector onto other coordinate axes are written similarly.
Since in uniform motion the acceleration is constant (a = const), the acceleration graph is a straight line parallel to the 0t axis (time axis, Fig. 1.15).
Rice. 1.15. Dependence of body acceleration on time.
Dependence of speed on time is a linear function, the graph of which is a straight line (Fig. 1.16).
Rice. 1.16. Dependence of body speed on time.
Speed versus time graph(Fig. 1.16) shows that
In this case, the displacement is numerically equal to the area of the figure 0abc (Fig. 1.16).
The area of a trapezoid is equal to the product of half the sum of the lengths of its bases and its height. The bases of the trapezoid 0abc are numerically equal:
The height of the trapezoid is t. Thus, the area of the trapezoid, and therefore the projection of displacement onto the OX axis is equal to:
In the case of uniformly slow motion, the acceleration projection is negative and in the formula for the displacement projection a “–” (minus) sign is placed before the acceleration.
General formula for determining the displacement projection:
A graph of the velocity of a body versus time at various accelerations is shown in Fig. 1.17. The graph of displacement versus time for v0 = 0 is shown in Fig. 1.18.
Rice. 1.17. Dependence of body speed on time for different meanings acceleration.
Rice. 1.18. Dependence of body movement on time.
The speed of the body at a given time t 1 is equal to the tangent of the angle of inclination between the tangent to the graph and the time axis v = tg α, and the displacement is determined by the formula:
If the time of movement of the body is unknown, you can use another displacement formula by solving a system of two equations:
Formula for abbreviated multiplication of square difference will help us derive the formula for displacement projection:
Since the coordinate of the body at any moment in time is determined by the sum of the initial coordinate and the projection of displacement, then equation of body motion will look like this:
The graph of the coordinate x(t) is also a parabola (like the graph of displacement), but the vertex of the parabola in the general case does not coincide with the origin. When a x< 0 и х 0 = 0 ветви параболы направлены вниз (рис. 1.18).
