Analytical description of uniformly accelerated motion. Derivation of the formula for moving with uniformly accelerated motion. Trajectory

The most important thing for us is to be able to calculate the displacement of the body, because, knowing the displacement, we can also find the coordinates of the body, and this is the main task of mechanics. How to calculate displacement uniformly accelerated motion?

The formula for determining the displacement is easiest to obtain if you use the graphical method.

In § 9, we saw that with a rectilinear uniform motion, the displacement of the body is numerically equal to the area of ​​\u200b\u200bthe figure (rectangle) located under the velocity graph. Is this true for uniformly accelerated motion?

With uniformly accelerated motion of the body 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 graph of the speed of uniformly accelerated movement (Fig. 41) and lower from points a and perpendiculars to the axis. the graphics turned out to be a narrow strip

If the time interval 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 a rectangle. The area of ​​the strip is therefore numerically equal to the displacement of the body in the time corresponding to the segment

But it is possible to divide the entire area of ​​the figure located under the velocity graph into such narrow strips. Consequently, the displacement for all 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:

We substitute expression (1a) into this formula instead, then

Dividing term by term the numerator by the denominator, we get:

Substituting expression (16) into formula (2), we obtain (see Fig. 42):

Formula (2a) is used when the acceleration vector is directed in the same direction 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).

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 received the formulas for calculating the displacement, it is easy for us to obtain the formula for calculating the coordinates of the body. We have seen (see § 8) that in order to find the coordinate of the body at some point in time, it is necessary to add to the initial coordinate the projection of the displacement vector of the body onto the coordinate axis:

(For) if the acceleration vector is directed in the same direction 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 the body at any time in a rectilinear uniformly accelerated motion. To do this, you need to know the initial coordinate of the body, its initial velocity and acceleration a.

Task 1. The driver of a car moving at a speed of 72 km/h saw a red traffic light and applied the brakes. After that, the car began to slow down, moving with acceleration

What is the distance traveled by the car in the time sec after the start of braking? How far will the car travel before it comes to a complete stop?

Solution. For the origin of coordinates, we choose the point of the road at which the car began to slow down. Let's direct the coordinate axis in the direction of the car's movement (Fig. 44), and refer the time reference to the moment at which the driver pressed the brake. The speed of the car is directed in the same direction as the X axis, and the acceleration of the car is opposite to the direction of this axis. Therefore, the velocity projection on the X axis is positive, and the acceleration projection is negative, and the vehicle coordinate must be found using the formula (36):

Substituting in this formula the values

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 time of movement. It can be found using the formula

Since at the moment when the car stops, its speed is zero, then

The distance that the car will travel to a complete stop is equal to the coordinate of the car at the 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 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 can be seen that the time of movement is therefore:

The answer obtained shows that the graph shown in Figure 45 corresponds to the movement of the body first in one direction, and then the same distance in the opposite direction, as a result of which the body is at the starting point. Such a graph may, for example, refer to the motion of a body thrown vertically upwards.

Problem 3. A body moves along a straight line with uniform 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 movement in time is equal to:

At point B, the body had a speed and its displacement over the next period of time is:

2. Figure 47 shows the 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 velocities of bodies at the 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, perform the following tasks: a) Determine the accelerations of these bodies; b) compose for

of each body the formula for the dependence of speed on time: c) how are the movements corresponding to graphs 2 and 3 similar and how do they differ?

4. Figure 49 shows graphs of the speed of movement of three bodies. According to 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. During takeoff, the aircraft passes the runway in 15 seconds and at the moment of takeoff from the landing has a speed of 100 m/s. How fast was the plane moving and how long was the runway?

6. The car stopped at a traffic light. After the green signal lights up, it starts to move with acceleration and moves like this until its speed becomes equal to 16 m / s, after which it continues to move at a constant speed. How far from the traffic light will the car be 15 seconds after the green signal appears?

7. A projectile with a speed of 1,000 m/s breaks through the wall of the dugout in 10 minutes and then has a speed of 200 m/s. Considering the motion of the projectile in the thickness of the wall to be uniformly accelerated, find the thickness of the wall.

8. The rocket moves with acceleration and by some point in time reaches a speed of 900 m/sec. Which path will she take in the next

9. How far from the Earth would spaceship 30 minutes after the start, if he moved straight ahead with acceleration all the time

Uniform movement- this is movement at a constant speed, that is, when the speed does not change (v \u003d const) and there is no acceleration or deceleration (a \u003d 0).

Rectilinear motion is a movement in a straight line, that is, a trajectory rectilinear motion is a straight line.

is a movement in which the body makes the same movements for any equal intervals of time. For example, if we divide some time interval into segments of one second, then with uniform motion the body will move the same distance for each of these segments of time.

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. Wherein 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 displacement of the body for any period of time to the value of this interval t:

V(vector) = s(vector) / t

Thus, the speed of uniform rectilinear motion shows what movement a material point makes per unit of time.

moving with uniform rectilinear motion is determined by the formula:

s(vector) = V(vector) t

Distance traveled in rectilinear motion is equal to the displacement modulus. If the positive direction of the OX axis coincides with the direction of movement, then the projection of the velocity on the OX axis is equal to the velocity and is positive:

v x = v, i.e. v > 0

The projection of displacement onto the OX axis is equal to:

s \u003d vt \u003d 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)

Motion equation, that is, the dependence of the body coordinate 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 velocity on the OX axis is negative, the velocity is less than zero (v< 0), и тогда уравнение движения принимает вид:

4. Equal-variable movement.

Uniform rectilinear motion This is a special case of non-uniform motion.

Uneven movement- this is a movement in which a body (material point) makes unequal movements in equal intervals of time. For example, a city bus moves unevenly, since its movement consists mainly of acceleration and deceleration.

Equal-variable motion- this is a movement in which the speed of a body (material point) changes in the same way for any equal time intervals.

Acceleration of a body in uniform motion remains constant in magnitude and direction (a = const).

Uniform motion can be uniformly accelerated or uniformly slowed down.

Uniformly accelerated motion- this is the movement of a body (material point) with a positive acceleration, that is, with such a movement, the body accelerates with a constant acceleration. In the case of uniformly accelerated motion, the modulus of the body's velocity increases with time, the direction of acceleration coincides with the direction of the velocity of motion.

Uniformly slow motion- this is the movement of a body (material point) with negative acceleration, that is, with such a movement, the body slows down uniformly. With uniformly slow motion, the velocity and acceleration vectors are opposite, and the modulus of velocity decreases with time.

In mechanics, any rectilinear motion is accelerated, so slow motion differs from accelerated motion only by the sign of the projection of the acceleration vector onto the selected axis of the coordinate system.

Average speed of variable motion 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.

Instant Speed is the speed of the body (material point) in this moment 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 uniform 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- this is a value that determines the rate of change in the speed of the body, that is, the limit to which the change in speed tends with an infinite decrease in the time interval Δt:

a(vector) = lim(t-0) ^v(vector)/^t

Acceleration vector of uniform 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)"

Given 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 time interval during which the change in speed occurred, acceleration formula will be as follows:

a(vector) = v(vector)-v0(vector)/t

From here uniform velocity formula at any given time:

v(vector) = v 0 (vector) + a(vector)t

If the 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. Equations of projections of the velocity vector onto other coordinate axes are written similarly.

Since the acceleration is constant (a \u003d const) with uniformly variable motion, 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.

Speed ​​versus 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.

Graph of speed versus time(Fig. 1.16) shows that

In this case, the displacement is numerically equal to the area of ​​\u200b\u200bthe figure 0abc (Fig. 1.16).

The area of ​​a trapezoid is half the sum of the lengths of its bases times the height. The bases of the trapezoid 0abc are numerically equal:

The height of the trapezoid is t. Thus, the area of ​​the trapezoid, and hence the projection of displacement onto the OX axis, is equal to:

In the case of uniformly slow motion, the projection of acceleration is negative, and in the formula for the projection of displacement, the sign “–” (minus) is placed in front of the acceleration.

The general formula for determining the displacement projection is:

The graph of the dependence of the speed of the body on time at various accelerations is shown in Fig. 1.17. The graph of the dependence of displacement on time at 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 displacement 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 \u003d tg α, and the movement is determined by the formula:

If the time of motion of the body is unknown, you can use another displacement formula by solving a system of two equations:

The formula for the abbreviated multiplication of the difference of squares will help us to derive the formula for the displacement projection:

Since the coordinate of the body at any moment of time is determined by the sum of the initial coordinate and the displacement projection, then body motion equation will look like this:

The graph of the x(t) coordinate is also a parabola (as is the displacement graph), but the vertex of the parabola generally does not coincide with the origin. For a x< 0 и х 0 = 0 ветви параболы направлены вниз (рис. 1.18).

Let's derive a formula that can be used to calculate the projection of the displacement vector of a body moving in a straight line and uniformly accelerated for any period of time. To do this, let's turn to Figure 14. Both in Figure 14, a, and in Figure 14, b, the segment AC is a graph of the projection of the velocity vector of a body moving with constant acceleration a (at the initial speed v 0).

Rice. 14. The projection of the displacement vector of a body moving in a straight line and uniformly accelerated is numerically equal to the area S under the graph

Recall that with a rectilinear uniform motion of a body, the projection of the displacement vector made by this body is determined by the same formula as the area of ​​the rectangle enclosed under the velocity vector projection graph (see Fig. 6). Therefore, the projection of the displacement vector is numerically equal to the area of ​​this rectangle.

Let us prove that in the case of a rectilinear uniformly accelerated motion, the projection of the displacement vector s x can be determined by the same formula as the area of ​​the figure enclosed between the AC graph, the Ot axis and the segments OA and BC, i.e. that in this case the projection of the displacement vector numerically equal to the area of ​​the figure under the speed graph. To do this, on the Ot axis (see Fig. 14, a) we select a small time interval db. From points d and b we draw perpendiculars to the Ot axis until they intersect with the velocity vector projection graph at points a and c.

Thus, for a period of time corresponding to the segment db, the speed of the body changes from v ax to v cx.

For a sufficiently short period of time, the projection of the velocity vector changes very slightly. Therefore, the movement of the body during this period of time differs little from uniform, that is, from movement at a constant speed.

It is possible to divide the entire area of ​​the OASV figure, which is a trapezoid, into such strips. Therefore, the projection of the displacement vector sx for the time interval corresponding to the segment OB is numerically equal to the area S of the trapezoid OASV and is determined by the same formula as this area.

According to the rule in school courses geometry, the area of ​​a trapezoid is equal to the product of half the sum of its bases and its height. Figure 14, b shows that the bases of the trapezoid OASV are the segments OA = v 0x and BC = v x, and the height is the segment OB = t. Consequently,

Since v x \u003d v 0x + a x t, a S \u003d s x, then we can write:

Thus, we have obtained a formula for calculating the projection of the displacement vector during uniformly accelerated motion.

Using the same formula, the projection of the displacement vector is also calculated when the body moves with a decreasing modulus of speed, only in this case the velocity and acceleration vectors will be directed in opposite directions, so their projections will have different signs.

Questions

1. Using Figure 14, a, prove that the projection of the displacement vector during uniformly accelerated motion is numerically equal to the area of ​​the OASV figure.
2. Write down an equation to determine the projection of the displacement vector of a body during its rectilinear uniformly accelerated motion.

Exercise 7

Let's try to derive a formula for finding the projection of the displacement vector of a body that moves in a straight line and uniformly accelerated for any period of time.

To do this, let us turn to the graph of the dependence of the projection of the speed of rectilinear uniformly accelerated motion on time.

Graph of the projection of the speed of rectilinear uniformly accelerated motion on 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 a 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 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.

Find the area of ​​the figure bounded by the ot axis, the segments AO and BC, as well as the segment AC.

Let us allocate a small time interval db on the ot axis. Let's draw perpendiculars to the time axis through these points until they intersect with the velocity projection graph. Note 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 deal with uniform rectilinear motion on this interval.

Then we can consider the segment ac as horizontal, and abcd as 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 time intervals.

That is, we have obtained that the projection of the displacement vector Sx for the time interval corresponding to the segment OB will be numerically equal to the area S of the OACB trapezoid, and will be determined by the same formula as this area.

Consequently,

• 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.

Trajectory(from late Latin trajectories - referring to movement) - this is the line along which the body moves (material point). The trajectory of movement can be straight (the body moves in one direction) and curvilinear, that is mechanical movement can be straight or curved.

Rectilinear trajectory in this coordinate system is a straight line. For example, we can assume that the trajectory of a car on a flat road without turns is a straight line.

Curvilinear motion- this is the movement of bodies in a circle, ellipse, parabola or hyperbola. An example of curvilinear motion is the movement of a point on the wheel of a moving car, or the movement of a car in a turn.

Movement can be tricky. For example, the trajectory of the movement of the body at the beginning of the path can be rectilinear, then curvilinear. For example, a car at the beginning of the journey moves along a straight road, and then the road begins to "wind" and the car begins to curve.

Path

Path is the length of the path. The path is a scalar and in international system SI units are measured in meters (m). Path calculation is performed in many problems in physics. Some examples will be discussed later in this tutorial.

Displacement vector

Displacement vector(or simply moving) is a directed line segment connecting the initial position of the body with its subsequent position (Fig. 1.1). Displacement is a vector quantity. The displacement vector is directed from the starting point of the movement to the end point.

Displacement vector modulus(that is, the length of the segment that connects the start and end points of the movement) can be equal to the distance traveled or less than the distance traveled. But never the module of the displacement vector can be greater than the distance traveled.

The module of the displacement vector is equal to the distance traveled when the path coincides with the trajectory (see sections and), for example, if the car moves from point A to point B along a straight road. The module of the displacement vector is less than the distance traveled when the material point moves along a curved path (Fig. 1.1).

Rice. 1.1. The displacement vector and the distance traveled.

On fig. 1.1:

Another example. If the car passes in a circle once, then it turns out that the start point of the movement will coincide with the end point of the movement, and then the displacement vector will be zero, and the distance traveled will be equal to the circumference of the circle. Thus, the path and movement are two different concepts.

Vector addition rule

The displacement vectors are added geometrically according to the vector addition rule (the triangle rule or the parallelogram rule, see Fig. 1.2).

Rice. 1.2. Addition of displacement vectors.

Figure 1.2 shows the rules for adding vectors S1 and S2:

a) Addition according to the rule of a triangle
b) Addition according to the parallelogram rule

Displacement vector projections

When solving problems in physics, projections of the displacement vector onto coordinate axes are often used. The projections of the displacement vector onto the coordinate axes can be expressed in terms of the difference between the coordinates of its end and beginning. For example, if a material point has moved from point A to point B, then the displacement vector (see Fig. 1.3).

We choose the OX axis so that the vector lies with this axis in the same plane. Let's lower the perpendiculars from points A and B (from the start and end points of the displacement vector) to the intersection with the OX axis. Thus, we get the projections of points A and B on the X axis. Let us denote the projections of points A and B, respectively, A x and B x. The length of the segment A x B x on the OX axis - this is displacement vector projection on the x-axis, that is

S x = A x B x

IMPORTANT!
A reminder for those who do not know mathematics very well: do not confuse a vector with the projection of a vector on any axis (for example, S x). A vector is always denoted by a letter or several letters with an arrow above it. In some electronic documents, the arrow is not put, as this can cause difficulties when creating electronic document. In such cases, be guided by the content of the article, where the word “vector” can be written next to the letter or in some other way indicate to you that this is a vector, and not just a segment.

Rice. 1.3. Projection of the displacement vector.

The projection of the displacement vector onto the OX axis is equal to the difference between the coordinates of the end and beginning of the vector, that is

S x \u003d x - x 0

The projections of the displacement vector on the OY and OZ axes are defined and written in the same way:

S y = y – y 0 S z = z – z 0

Here x 0 , y 0 , z 0 are the initial coordinates, or the coordinates of the initial position of the body (material point); x, y, z - final coordinates, or coordinates of the subsequent position of the body (material point).

The projection of the displacement vector is considered positive if the direction of the vector and the direction of the coordinate axis coincide (as in Figure 1.3). If the direction of the vector and the direction of the coordinate axis do not coincide (opposite), then the projection of the vector is negative (Fig. 1.4).

If the displacement vector is parallel to the axis, then the module of its projection is equal to the module of the Vector itself. If the displacement vector is perpendicular to the axis, then the module of its projection is zero (Fig. 1.4).

Rice. 1.4. Modules of displacement vector projection.

The difference between the subsequent and initial values ​​of a quantity is called the change in that quantity. That is, the projection of the displacement vector onto the coordinate axis is equal to the change in the corresponding coordinate. For example, for the case when the body moves perpendicular to the X axis (Fig. 1.4), it turns out that the body DOES NOT MOVEMENT relative to the X axis. That is, the displacement of the body along the X axis is zero.

Consider an example of the motion of a body on a plane. The initial position of the body is point A with coordinates x 0 and y 0, that is, A (x 0, y 0). The final position of the body is point B with coordinates x and y, that is, B (x, y). Find the modulus of displacement of the body.

From points A and B we lower the perpendiculars on the coordinate axes OX and OY (Fig. 1.5).

Rice. 1.5. Movement of a body on a plane.

Let's define the projections of the displacement vector on the axes OX and OY:

S x = x – x 0 S y = y – y 0

On fig. 1.5 it can be seen that the triangle ABC is a right triangle. It follows from this that when solving the problem, one can use Pythagorean theorem, with which you can find the modulus of the displacement vector, since

AC = s x CB = s y

According to the Pythagorean theorem

S 2 \u003d S x 2 + S y 2

Where can you find the modulus of the displacement vector, that is, the length of the body's path from point A to point B:

And finally, I suggest you consolidate your knowledge and calculate a few examples at your discretion. To do this, enter any numbers in the coordinate fields and click the CALCULATE button. Your browser must support the execution of scripts (scripts) JavaScript and the execution of scripts must be allowed in your browser settings, otherwise the calculation will not be performed. In real numbers, the integer and fractional parts must be separated by a period, for example, 10.5.