Projection of displacement during uniformly accelerated motion. What formula is used to calculate the projection of the displacement of a body during uniformly accelerated linear motion?

Now we must find out the most important thing - how the coordinate of a body changes during its rectilinear uniformly accelerated motion. To do this, as we know, we need to know the displacement of the body, because the projection of the displacement vector is exactly equal to the change in coordinate.

The formula for calculating displacement is easiest to obtain graphically.

When a body moves uniformly along the X axis, the speed changes with time according to the formula v x = v 0x + a x t Since time is included in this formula to the first degree, the graph for the projection of speed versus time is a straight line, as shown in Figure 39. Straight line 1 in this figure corresponds to movement with a positive projection of acceleration (speed increases), straight 2 - movement with a negative projection of acceleration (speed decreases). Both graphs refer to the case when at the moment of time t = O body has some initial speed v 0 .

Displacement is expressed by area. Let us highlight a small section on the speed graph of uniformly accelerated motion (Fig. 40) ab and drop from the points A And b perpendiculars to the axis t. Length of the segment CD on the axis t on the chosen scale is equal to the small period of time during which the speed changed from its value at the point A to its value at point b. Under the site ab the graphics turned out to be a narrow strip abсd.

If the time interval corresponding to the segment CD, is small enough, then during this short time the speed cannot change noticeably - the movement during this short period of time can be considered uniform. Strip ABCD therefore, it differs little from a rectangle, and its area is numerically equal to the projection of displacement during the time corresponding to the segment CD(see § 7).

But the entire area of ​​the figure located under the speed graph can be divided into such narrow strips. Therefore, the movement over the entire time t numerically equal to the area of ​​the trapezoid OABC. The area of ​​a trapezoid, as is known from geometry, is equal to the product of half the sum of its bases and its height. In our case, the length of one of the bases is numerically equal to v ox, the other - v x (see Fig. 40). The height of the trapezoid is numerically equal t. It follows that the projection s x displacement is expressed by the formula

3s 15.09

If the projection v ox of the initial velocity is zero (at the initial moment of time the body was at rest!), then formula (1) takes the form:

The speed graph of such movement is shown in Figure 41.

When using formulas (1) And(2) YOU NEED TO REMEMBER THAT S x , V ox And v x can be both positive and negative - after all, these are projections of vectors s, v o And v to the X axis.

Thus, we see that with uniformly accelerated motion, displacement grows with time differently than with uniform motion: now the formula includes the square of time. This means that the displacement increases over time faster than with uniform motion.

How does the coordinate of a body depend on time? Now it’s easy to get the formula for calculating the coordinates X at any moment of time for a body moving with uniform acceleration.

projection s x displacement vector is equal to the change in coordinate x-x 0. Therefore we can write

From formula (3) it is clear that, in order to calculate the x coordinate at any time t, you need to know the initial coordinate, initial velocity and acceleration.

Formula (3) describes rectilinear uniformly accelerated motion, just as formula (2) § 6 describes rectilinear uniform motion.

Another formula for moving. To calculate displacement, you can get another useful formula, which does not include time.

From the expression v x = v 0x + a x t. we get an expression for time

t= (v x - v 0x): a x and substitute it into the formula for moving s x , given above. Then we get:

These formulas allow you to find the displacement of a body if the acceleration, as well as the initial and final speeds of movement, are known. If the initial speed v o is zero, formulas (4) take the form:

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 velocity of a body that moves with an 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.

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 of 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 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 the spacecraft be 30 minutes after launch if it were constantly moving in a straight line with acceleration?

Page 8 of 12

§ 7. Movement under uniform acceleration
straight motion

1. Using a graph of speed versus time, you can obtain a formula for the displacement of a body during uniform rectilinear motion.

Figure 30 shows a graph of the projection of the speed of uniform motion onto the axis X from time. If we restore the perpendicular to the time axis at some point C, then we get a rectangle OABC. The area of ​​this rectangle is equal to the product of the sides O.A. And O.C.. But side length O.A. equal to v x, and the side length O.C. - t, from here S = v x t. Product of the projection of velocity onto an axis X and time is equal to the projection of displacement, i.e. s x = v x t.

Thus, the projection of displacement during uniform rectilinear motion is numerically equal to the area of ​​the rectangle bounded by the coordinate axes, the velocity graph and the perpendicular to the time axis.

2. We obtain in a similar way the formula for the projection of displacement in rectilinear uniformly accelerated motion. To do this, we will use the graph of the velocity projection onto the axis X from time to time (Fig. 31). Let's select a small area on the graph ab and drop the perpendiculars from the points a And b on the time axis. If time interval D t, corresponding to the site CD on the time axis is small, then we can assume that the speed does not change during this period of time and the body moves uniformly. In this case the figure cabd differs little from a rectangle and its area is numerically equal to the projection of the movement of the body over the time corresponding to the segment CD.

The whole figure can be divided into such strips OABC, and its area will be equal to the sum of the areas of all strips. Therefore, the projection of the movement of the body over time t numerically equal to the area of ​​the trapezoid OABC. From your geometry course you know that the area of ​​a trapezoid is equal to the product of half the sum of its bases and height: S= (O.A. + B.C.)O.C..

As can be seen from Figure 31, O.A. = v 0x , B.C. = v x, O.C. = t. It follows that the displacement projection is expressed by the formula: s x= (v x + v 0x)t.

With uniformly accelerated rectilinear motion, the speed of the body at any moment of time is equal to v x = v 0x + a x t, hence, s x = (2v 0x + a x t)t.

From here:

To obtain the equation of motion of a body, we substitute its expression in terms of the difference in coordinates into the displacement projection formula s x = x – x 0 .

We get: x – x 0 = v 0x t+ , or

Using the equation of motion, you can determine the coordinate of a body at any time if the initial coordinate, initial velocity and acceleration of the body are known.

3. In practice, there are often problems in which it is necessary to find the displacement of a body during uniformly accelerated rectilinear motion, but the time of motion is unknown. In these cases, a different displacement projection formula is used. Let's get it.

From the formula for the projection of the velocity of uniformly accelerated rectilinear motion v x = v 0x + a x t Let's express time:

t = .

Substituting this expression into the displacement projection formula, we get:

s x = v 0x + .

From here:

s x = , or
–= 2a x s x.

If the initial speed of the body is zero, then:

2a x s x.

4. Example of problem solution

A skier slides down a mountain slope from a state of rest with an acceleration of 0.5 m/s 2 in 20 s and then moves along a horizontal section, having traveled 40 m to a stop. With what acceleration did the skier move along a horizontal surface? What is the length of the mountain slope?

We connect the reference system with the Earth, the axis X let's direct the skier in the direction of speed at each stage of his movement (Fig. 32).

Let's write the equation for the skier's speed at the end of the descent from the mountain:

v 1 = v 01 + a 1 t 1 .

In projections onto the axis X we get: v 1x = a 1x t. Since the projections of velocity and acceleration onto the axis X are positive, the speed modulus of the skier is equal to: v 1 = a 1 t 1 .

Let us write an equation connecting the projections of speed, acceleration and displacement of the skier at the second stage of movement:

–= 2a 2x s 2x .

Considering that the initial speed of the skier at this stage of movement is equal to his final speed at the first stage

v 02 = v 1 , v 2x= 0 we get

– = –2a 2 s 2 ; (a 1 t 1) 2 = 2a 2 s 2 .

From here a 2 = ;

a 2 == 0.125 m/s 2 .

The module of movement of the skier at the first stage of movement is equal to the length of the mountain slope. Let's write the equation for displacement:

s 1x = v 01x t + .

Hence the length of the mountain slope is s 1 = ;

s 1 == 100 m.

Answer: a 2 = 0.125 m/s 2 ; s 1 = 100 m.

Self-test questions

1. As in the graph of the projection of the speed of uniform rectilinear motion onto the axis X

2. As in the graph of the projection of the speed of uniformly accelerated rectilinear motion onto the axis X determine the projection of body movement from time to time?

3. What formula is used to calculate the projection of the displacement of a body during uniformly accelerated rectilinear motion?

4. What formula is used to calculate the projection of displacement of a body moving uniformly accelerated and rectilinearly if the initial speed of the body is zero?

Task 7

1. What is the module of movement of the car in 2 minutes, if during this time its speed changed from 0 to 72 km/h? What is the coordinate of the car at the moment of time t= 2 min? The initial coordinate is considered equal to zero.

2. The train moves with an initial speed of 36 km/h and an acceleration of 0.5 m/s 2 . What is the displacement of the train in 20 s and its coordinate at the moment of time? t= 20 s if the initial coordinate of the train is 20 m?

3. What is the cyclist’s displacement in 5 s after the start of braking, if his initial speed during braking is 10 m/s and the acceleration is 1.2 m/s 2? What is the coordinate of the cyclist at the moment of time? t= 5 s, if at the initial moment of time it was at the origin?

4. A car moving at a speed of 54 km/h stops when braking for 15 s. What is the modulus of movement of a car during braking?

5. Two cars are moving towards each other from two settlements located at a distance of 2 km from each other. The initial speed of one car is 10 m/s and the acceleration is 0.2 m/s 2 , the initial speed of the other is 15 m/s and the acceleration is 0.2 m/s 2 . Determine the time and coordinates of the meeting place of the cars.

Laboratory work No. 1

Study of uniformly accelerated
rectilinear motion

Goal of the work:

learn to measure acceleration during uniformly accelerated linear motion; to experimentally establish the ratio of the paths traversed by a body during uniformly accelerated rectilinear motion in successive equal intervals of time.

Devices and materials:

trench, tripod, metal ball, stopwatch, measuring tape, metal cylinder.

Work order

1. Secure one end of the chute in the tripod leg so that it makes a small angle with the table surface. At the other end of the chute, place a metal cylinder in it.

2. Measure the paths traveled by the ball in 3 consecutive periods of time equal to 1 s each. This can be done in different ways. You can put chalk marks on the gutter that record the positions of the ball at times equal to 1 s, 2 s, 3 s, and measure the distances s_ between these marks. You can, by releasing the ball from the same height each time, measure the path s, traveled by it first in 1 s, then in 2 s and in 3 s, and then calculate the path traveled by the ball in the second and third seconds. Record the measurement results in table 1.

3. Find the ratio of the path traveled in the second second to the path traveled in the first second, and the path traveled in the third second to the path traveled in the first second. Draw a conclusion.

4. Measure the time the ball moves along the chute and the distance it travels. Calculate the acceleration of its motion using the formula s = .

5. Using the experimentally obtained acceleration value, calculate the distances that the ball must travel in the first, second and third seconds of its movement. Draw a conclusion.

Table 1

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

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

Curvilinear movement 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.

The movement can be difficult. For example, the trajectory of a body at the beginning of its journey can be rectilinear, then curved. For example, at the beginning of the journey a car moves along a straight road, and then the road begins to “wind” and the car begins to move in a curved direction.

Path

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

Move vector

Move vector(or simply moving) is a directed straight 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 movement to the ending point.

Motion vector module(that is, the length of the segment that connects the starting and ending points of the movement) can be equal to the distance traveled or less than the distance traveled. But the magnitude of the displacement vector can never be greater than the distance traveled.

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

Rice. 1.1. Displacement vector and distance traveled.

In Fig. 1.1:

Another example. If the car drives in a circle once, it turns out that the point at which the movement begins will coincide with the point at which the movement ends, and then the displacement vector will be equal to zero, and the distance traveled will be equal to the length of the circle. Thus, path and movement are two different concepts.

Vector addition rule

The displacement vectors are added geometrically according to the vector addition rule (triangle rule or 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 triangle rule
b) Addition according to the parallelogram rule

Motion vector projections

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

Let us choose the OX axis so that the vector lies in the same plane with this axis. Let's lower the perpendiculars from points A and B (from the starting and ending points of the displacement vector) until they intersect with the OX axis. Thus, we obtain the projections of points A and B onto the X axis. Let us denote the projections of points A and B, respectively, as A x and B x. The length of the segment A x B x on the OX axis is displacement vector projection on the OX axis, that is

S x = A x B x

IMPORTANT!
I remind you for those who do not know mathematics very well: do not confuse a vector with the projection of a vector onto any axis (for example, S x). A vector is always indicated by a letter or several letters, above which there is an arrow. In some electronic documents, the arrow is not placed, as this may cause difficulties when creating an electronic document. In such cases, be guided by the content of the article, where the word “vector” may be written next to the letter or in some other way they 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 = x – x 0

The projections of the displacement vector on the OY and OZ axes are determined and written similarly:

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

Rice. 1.4. Motion vector projection modules.

The difference between the subsequent and initial values ​​of some quantity is called the change in this 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 MOVE relative to the X axis. That is, the movement of the body along the X axis is zero.

Let's consider an example of body motion 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). Let's find the modulus of body displacement.

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

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

Let us determine the projections of the displacement vector on the OX and OY axes:

S x = x – x 0 S y = y – y 0

In Fig. 1.5 it is clear that 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 module of the displacement vector, since

AC = s x CB = s y

According to the Pythagorean theorem

S 2 = S x 2 + S y 2

Where can you find the module 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 some numbers in the coordinate fields and click the CALCULATE button. Your browser must support the execution of JavaScript scripts and script execution must be enabled in your browser settings, otherwise the calculation will not be performed. In real numbers, the integer and fractional parts must be separated by a dot, for example, 10.5.