Page 8 of 12

§ 7. Movement with uniformly accelerated
rectilinear motion

1. Using a graph of speed versus time, you can get the formula for moving a body with uniform rectilinear motion.

Figure 30 shows a plot of the velocity projection uniform motion per axle X from time. If we set up a 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 OA And OC. But the side length OA is equal to v x, and the side length OC - t, hence S = v x t. The product of the projection of velocity on the axis X and time is equal to the displacement projection, 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 raised to the time axis.

2. We obtain in a similar way the formula for the projection of displacement in a rectilinear uniformly accelerated motion. To do this, we use the graph of the dependence of the projection of velocity on the axis X from time (Fig. 31). Select a small area on the graph ab and drop the perpendiculars from the points a And b on the time axis. If the time interval D t, corresponding to the section 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 in the time corresponding to the segment cd.

You can break the whole figure into such strips OABC, and its area will be equal to the sum of the areas of all the strips. Therefore, the projection of the movement of the body over time t numerically equal to the area of ​​the trapezoid OABC. From the 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= (OA + BC)OC.

As can be seen from figure 31, OA = v 0x , BC = v x, OC = 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 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 the body, we substitute into the displacement projection formula its expression through the difference in coordinates s x = x – x 0 .

We get: x – x 0 = v 0x t+ , or

According to the equation of motion, it is possible to determine the coordinate of the 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 speed of uniformly accelerated rectilinear motion v x = v 0x + a x t let's express the 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 velocity of the body is zero, then:

2a x s x.

4. Problem solution example

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

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

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

v 1 = v 01 + a 1 t 1 .

In projections on the axis X we get: v 1x = a 1x t. Since the projections of velocity and acceleration on the axis X are positive, the modulus of the skier's speed is: v 1 = a 1 t 1 .

Let's write an equation relating the projections of speed, acceleration and movement 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 the 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 \u003d 0.125 m / s 2; s 1 = 100 m.

Questions for self-examination

1. As according to the plot of the projection of the speed of uniform rectilinear motion on the axis X

2. As according to the graph of the projection of the speed of uniformly accelerated rectilinear motion on the axis X from time to determine the projection of the displacement of the body?

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 the displacement of a body moving uniformly accelerated and rectilinearly if the initial speed of the body is zero?

Task 7

1. What is the displacement modulus of a car in 2 minutes if during this time its speed has changed from 0 to 72 km/h? What is the coordinate of the car at the time t= 2 min? The initial coordinate is assumed to be 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 starting coordinate of the train is 20 m?

3. What is the movement of the cyclist for 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 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 seconds. What is the displacement modulus of the car when 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 coordinate of the meeting point of the cars.

Lab #1

Study of uniformly accelerated
rectilinear motion

Goal of the work:

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

Devices and materials:

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

Work order

1. Fix one end of the chute in the foot of the tripod so that it makes a small angle with the surface of the table. At the other end of the chute, put a metal cylinder into it.

2. Measure the paths traveled by the ball in 3 consecutive time intervals equal to 1 s each. This can be done in different ways. You can put marks on the chute with chalk, fixing the position of the ball at time points equal to 1 s, 2 s, 3 s, and measure the distances s_ between these marks. It is possible, releasing the ball from the same height each time, to measure the path s, passed by him 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. Make a conclusion.

4. Measure the time the ball traveled along the chute and the distance traveled by it. Calculate its acceleration using the formula s = .

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

Table 1

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.

Travel time

The time of movement (t) is equal to the ratio of the path (S) traveled by the body to the speed (v) of movement.

average speed

The average speed (vav) is equal to the ratio of the sum of the sections of the path (s 1 s 2, s 3, ...) traveled by the body to the time interval (t 1 + t 2 + t 3 + ...) for which this path was traveled .

average speed is the ratio of the length of the path traveled by the body to the time for which this path was traveled.

average speed when moving unevenly in a straight line: this is the ratio of the entire path to the total time.

Two successive stages with 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: the 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 it is against the axis. In this example

Movement module is the length of the displacement vector:

According to the Pythagorean theorem:

Projections of movement and angle of inclination

In this example:

Coordinate equation (in general):

Radius vector- vector, the beginning of which coincides with the origin of coordinates, and the end - with the position of the body in this moment time. The 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 set the position of a material point in a given reference system:

Uniform rectilinear motion - definition

Uniform rectilinear motion- a movement in which the body for any equal intervals of time, makes equal displacements.

Speed ​​in uniform rectilinear motion. Velocity is a vector physical quantity that shows how much movement a body makes per unit of 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.

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 speed projection graph is the dependence of the speed projection on time:

Graph of speed for uniform rectilinear 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).

The geometric meaning of movement.

With uniform rectilinear motion, the displacement is determined by the formula. We get the same result if we calculate the area of ​​the figure under the speed graph in the axes. So, to determine the path and the displacement module during rectilinear motion, it is necessary to calculate the area of ​​\u200b\u200bthe figure under the velocity graph in the axes:

Displacement Projection Plot- dependence of the displacement projection on time.

Displacement projection graph for uniform rectilinear motion- a straight line coming out of the origin (1, 2, 3).

If the 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 the slope (1) of the graph, the greater the speed module.

Plot coordinate- dependence of body coordinates on time:

Graph coordinates for uniform rectilinear motion - straight lines (1, 2, 3).

If over time the coordinate increases (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 slope (1), the greater the modulus of speed.

If the graphs of the coordinates of two bodies intersect, then from the intersection point one should lower the perpendiculars to the time axis and the coordinate axis.

Relativity of mechanical motion

By relativity we mean the dependence of something on the choice of reference frame. For example, peace is relative; relative movement and relative position of the body.

The rule of addition of displacements. Vector sum of displacements

where is the displacement of the body relative to the moving frame of reference (RFR); - movement of the PSO relative to the fixed frame of reference (FRS); - movement of the body relative to the fixed frame of reference (FRS).

Vector addition:

Addition of vectors directed along one straight line:

Addition of vectors, perpendicular each friend

According to the Pythagorean theorem

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. Hence,

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 to opposite sides, 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

Page 8 of 12

§ 7. Movement with uniformly accelerated
rectilinear motion

1. Using a graph of speed versus time, you can get the formula for moving a body with uniform rectilinear motion.

Figure 30 shows a graph of the projection of the speed of uniform movement on the axis X from time. If we set up a 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 OA And OC. But the side length OA is equal to v x, and the side length OC - t, hence S = v x t. The product of the projection of velocity on the axis X and time is equal to the displacement projection, 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 raised to the time axis.

2. We obtain in a similar way the formula for the projection of displacement in a rectilinear uniformly accelerated motion. To do this, we use the graph of the dependence of the projection of velocity on the axis X from time (Fig. 31). Select a small area on the graph ab and drop the perpendiculars from the points a And b on the time axis. If the time interval D t, corresponding to the section 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 in the time corresponding to the segment cd.

You can break the whole figure into such strips OABC, and its area will be equal to the sum of the areas of all the strips. Therefore, the projection of the movement of the body over time t numerically equal to the area of ​​the trapezoid OABC. From the 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= (OA + BC)OC.

As can be seen from figure 31, OA = v 0x , BC = v x, OC = 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 time is equal to v x = v 0x + a x t, hence, s x = (2v 0x + a x t)t.

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

We get: x – x 0 = v 0x t+ , or

According to the equation of motion, it is possible to determine the coordinate of the 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 speed of uniformly accelerated rectilinear motion v x = v 0x + a x t let's express the time:

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

s x = v 0x + .

s x = , or
–= 2a x s x.

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

2a x s x.

4. Problem solution example

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

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

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

v 1 = v 01 + a 1 t 1 .

In projections on the axis X we get: v 1x = a 1x t. Since the projections of velocity and acceleration on the axis X are positive, the modulus of the skier's speed is: v 1 = a 1 t 1 .

Let's write an equation relating the projections of speed, acceleration and movement 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 the 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 \u003d 0.125 m / s 2; s 1 = 100 m.

Questions for self-examination

1. As according to the plot of the projection of the speed of uniform rectilinear motion on the axis X

2. As according to the graph of the projection of the speed of uniformly accelerated rectilinear motion on the axis X from time to determine the projection of the displacement of the body?

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 the displacement of a body moving uniformly accelerated and rectilinearly if the initial speed of the body is zero?

Task 7

1. What is the displacement modulus of a car in 2 minutes if during this time its speed has changed from 0 to 72 km/h? What is the coordinate of the car at the time t= 2 min? The initial coordinate is assumed to be 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 starting coordinate of the train is 20 m?

3. What is the movement of the cyclist for 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 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 seconds. What is the displacement modulus of the car when 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 coordinate of the meeting point of the cars.

Lab #1

Study of uniformly accelerated
rectilinear motion

Goal of the work:

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

Devices and materials:

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

Work order

1. Fix one end of the chute in the foot of the tripod so that it makes a small angle with the surface of the table. At the other end of the chute, put a metal cylinder into it.

2. Measure the paths traveled by the ball in 3 consecutive time intervals equal to 1 s each. This can be done in different ways. You can put marks on the chute with chalk, fixing the position of the ball at time points equal to 1 s, 2 s, 3 s, and measure the distances s_ between these marks. It is possible, releasing the ball from the same height each time, to measure the path s, passed by him 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. Make a conclusion.

4. Measure the time the ball traveled along the chute and the distance traveled by it. Calculate its acceleration using the formula s = .

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

Table 1

How, knowing the stopping distance, determine the initial speed of the car and how, knowing the characteristics of the movement, such as the initial speed, acceleration, time, determine the movement of the car? We will get answers after we get acquainted with the topic of today's lesson: "Displacement with uniformly accelerated movement, the dependence of coordinates on time with uniformly accelerated movement"

With uniformly accelerated motion, the graph looks like a straight line going up, since its acceleration projection is greater than zero.

With uniform rectilinear motion, the area will be numerically equal to the modulus of the projection of the displacement of the body. It turns out that this fact can be generalized for the case not only of uniform motion, but also for any motion, that is, to show that the area under the graph is numerically equal to the displacement projection modulus. This is done strictly mathematically, but we will use a graphical method.

Rice. 2. Graph of the dependence of speed on time with uniformly accelerated movement ()

Let's divide the graph of the projection of speed from time for uniformly accelerated motion into small time intervals Δt. Let us assume that they are so small that during their length the speed practically did not change, that is, we will conditionally turn the linear dependence graph in the figure into a ladder. At each of its steps, we believe that the speed has not changed much. Imagine that we make the time intervals Δt infinitely small. In mathematics they say: we make a passage to the limit. In this case, the area of ​​such a ladder will indefinitely closely coincide with the area of ​​the trapezoid, which is limited by the graph V x (t). And this means that for the case of uniformly accelerated motion, we can say that the displacement projection module is numerically equal to area, bounded by the graph V x (t): the abscissa and ordinate axes and the perpendicular lowered to the abscissa axis, that is, the area of ​​the trapezoid OABS, which we see in Figure 2.

The task turns from a physical one into a math problem- Finding the area of ​​a trapezoid. This is the standard situation when physicists make up a model that describes a particular phenomenon, and then mathematics comes into play, which enriches this model with equations, laws - that turns the model into a theory.

We find the area of ​​the trapezoid: the trapezoid is rectangular, since the angle between the axes is 90 0, we divide the trapezoid into two shapes - a rectangle and a triangle. Obviously, the total area will be equal to the sum of the areas of these figures (Fig. 3). Let's find their areas: the area of ​​the rectangle is equal to the product of the sides, that is, V 0x t, the area right triangle will be equal to half the product of the legs - 1/2AD BD, substituting the projection values, we get: 1/2t (V x - V 0x), and, remembering the law of change in speed with time during uniformly accelerated motion: V x (t) = V 0x + a x t, it is quite obvious that the difference in the projections of the speeds is equal to the product of the projection of the acceleration a x by the time t, that is, V x - V 0x = a x t.

Rice. 3. Determining the area of ​​a trapezoid ( Source)

Taking into account the fact that the area of ​​the trapezoid is numerically equal to the displacement projection module, we get:

S x (t) \u003d V 0 x t + a x t 2 / 2

We have obtained the law of the dependence of the projection of displacement on time with uniformly accelerated motion in scalar form, in vector form it will look like this:

(t) = t + t 2 / 2

Let's derive one more formula for the displacement projection, which will not include time as a variable. We solve the system of equations, excluding time from it:

S x (t) \u003d V 0 x + a x t 2 / 2

V x (t) \u003d V 0 x + a x t

Imagine that we do not know the time, then we will express the time from the second equation:

t \u003d V x - V 0x / a x

Substitute the resulting value into the first equation:

We get such a cumbersome expression, we square it and give similar ones:

We have obtained a very convenient displacement projection expression for the case when we do not know the time of motion.

Let us have the initial speed of the car, when braking began, is V 0 \u003d 72 km / h, final speed V \u003d 0, acceleration a \u003d 4 m / s 2. Find out the length of the braking distance. Converting kilometers to meters and substituting the values ​​into the formula, we get that the stopping distance will be:

S x \u003d 0 - 400 (m / s) 2 / -2 4 m / s 2 \u003d 50 m

Let's analyze the following formula:

S x \u003d (V 0 x + V x) / 2 t

The projection of movement is half the sum of the projections of the initial and final speeds, multiplied by the time of movement. Recall the displacement formula for average speed

S x \u003d V cf t

In the case of uniformly accelerated movement, the average speed will be:

V cf \u003d (V 0 + V k) / 2

We have come close to solving the main problem of the mechanics of uniformly accelerated motion, that is, obtaining the law according to which the coordinate changes with time:

x(t) \u003d x 0 + V 0 x t + a x t 2 / 2

In order to learn how to use this law, we will analyze a typical problem.

The car, moving from a state of rest, acquires an acceleration of 2 m / s 2. Find the distance traveled by the car in 3 seconds and in the third second.

Given: V 0 x = 0

Let us write down the law according to which the displacement changes with time at

uniformly accelerated motion: S x \u003d V 0 x t + a x t 2 /2. 2 c

We can answer the first question of the problem by plugging in the data:

t 1 \u003d 3 c S 1x \u003d a x t 2 / 2 \u003d 2 3 2 / 2 \u003d 9 (m) - this is the path that went

c car in 3 seconds.

Find out how far he traveled in 2 seconds:

S x (2 s) \u003d a x t 2 / 2 \u003d 2 2 2 / 2 \u003d 4 (m)

So, you and I know that in two seconds the car drove 4 meters.

Now, knowing these two distances, we can find the path that he traveled in the third second:

S 2x \u003d S 1x + S x (2 s) \u003d 9 - 4 \u003d 5 (m)

Uniformly accelerated motion called such a movement in which the acceleration vector remains unchanged in magnitude and direction. An example of such a movement is the movement of a stone thrown at a certain angle to the horizon (ignoring air resistance). At any point of the trajectory, the acceleration of the stone is equal to the acceleration free fall. Thus, the study of uniformly accelerated motion is reduced to the study of rectilinear uniformly accelerated motion. In the case of rectilinear motion, the velocity and acceleration vectors are directed along the straight line of motion. Therefore, the speed and acceleration in projections on the direction of motion can be considered as algebraic quantities. With uniformly accelerated rectilinear motion, the speed of the body is determined by the formula (1)

In this formula, the speed of the body at t = 0 (starting speed ), = const – acceleration. In the projection onto the selected x-axis, equation (1) will be written in the form: (2). On the velocity projection graph υ x ( t), this dependence has the form of a straight line.

The slope of the velocity graph can be used to determine the acceleration a body. The corresponding constructions are made in Figs. for graph I Acceleration is numerically equal to the ratio of the sides of the triangle ABC: .

The greater the angle β that forms the velocity graph with the time axis, i.e. the greater the slope of the graph ( steepness), the greater the acceleration of the body.

For graph I: υ 0 \u003d -2 m / s, a\u003d 1/2 m / s 2. For graph II: υ 0 \u003d 3 m / s, a\u003d -1/3 m / s 2.

The speed graph also allows you to determine the projection of the displacement s of the body for some time t. Let us allocate some small time interval Δt on the time axis. If this time interval is small enough, then the change in speed over this interval is small, that is, the movement during this time interval can be considered uniform with some average speed, which is equal to instantaneous speedυ of the body in the middle of the interval Δt. Therefore, the displacement Δs during the time Δt will be equal to Δs = υΔt. This displacement is equal to the area shaded in Fig. stripes. By dividing the time interval from 0 to a certain moment t into small intervals Δt, we can obtain that the displacement s for a given time t during uniformly accelerated rectilinear motion is equal to the area of ​​the trapezoid ODEF. The corresponding constructions are made in Figs. for schedule II. The time t is taken equal to 5.5 s.

(3) - the resulting formula allows you to determine the displacement with uniformly accelerated motion if the acceleration is not known.

If we substitute the expression for velocity (2) into equation (3), then we obtain (4) - this formula is used to write the equation of body motion: (5).

If we express from equation (2) the time of motion (6) and substitute into equality (3), then

This formula allows you to determine the movement at an unknown time of movement.

Let's consider how the projection of the displacement vector of a body moving uniformly accelerated is calculated if its initial speed v 0 is equal to zero. In this case, the equation

will look like this:

Let us rewrite this equation by substituting into it, instead of the projections s x and a x, the modules s and a of the vectors

displacement and acceleration. Since in this case the vectors sua are directed in the same direction, their projections have the same signs. Therefore, the equation for the modules of vectors can be written:

It follows from this formula that with a rectilinear uniformly accelerated movement without an initial speed, the module of the displacement vector is directly proportional to the square of the time interval during which this movement was made. This means that with an increase in the time of movement by n times (counted from the moment the movement began), the movement increases by n 2 times.

For example, if for an arbitrary period of time t 1 from the beginning of the movement, the body moved

then for a period of time t 2 \u003d 2t 1 (counted from the same moment as t 1) it will move

for a period of time t n \u003d nt l - displacement s n \u003d n 2 s l (where n is a natural number).

This dependence of the module of the displacement vector on time for rectilinear uniformly accelerated motion without initial speed is clearly reflected in Figure 15, where the segments OA, OB, OS, OD and OE are the modules of the displacement vectors (s 1, s 2, s 3, s 4 and s 5), committed by the body, respectively, for time intervals t 1 , t 2 = 2t 1 , t 3 = 3t 1 , t 4 = 4t 1 and t 5 = 5t 1 .

Rice. 15. Patterns of uniformly accelerated motion: OA:OB:OS:OD:0E = 1:4:9:16:25; OA:AB:BC:CD:DE = 1:3:5:7:9

From this figure, it is clear that

OA:OB:OS:OD:OE = 1:4:9:16:25, (1)

i.e., with an increase in the time intervals counted from the beginning of the movement, by an integer number of times compared to t 1, the modules of the corresponding displacement vectors increase as a series of squares of successive natural numbers.

Figure 15 shows another pattern:

OA:AB:BC:CD:DE = 1:3:5:7:9, (2)

i.e., the modules of the vectors of displacements performed by the body in successive equal periods of time (each of which is equal to t 1) are related as a series of consecutive odd numbers.

Regularities (1) and (2) are inherent only to uniformly accelerated motion. Therefore, they can be used if it is necessary to determine whether the movement is uniformly accelerated or not.

Let us determine, for example, whether the movement of the cochlea was uniformly accelerated, which moved 0.5 cm in the first 20 seconds of movement, 1.5 cm in the second 20 seconds, and 2.5 cm in the third 20 seconds.

To do this, let's find how many times the movements made in the second and third time intervals are greater than in the first:

This means that 0.5 cm: 1.5 cm: 2.5 cm = 1: 3: 5. Since these ratios are a series of consecutive odd numbers, the movement of the body was uniformly accelerated.

In this case, the uniformly accelerated nature of the movement was revealed on the basis of regularity (2).

Questions

1. What formulas are used to calculate the projection and module of the displacement vector of a body during its uniformly accelerated movement from a state of rest?
2. How many times will the modulus of the displacement vector of the body increase with an increase in the time of its movement from rest by n times?
3. Write down how the modules of the displacement vectors of a body moving uniformly accelerated from a state of rest relate to each other with an increase in the time of its movement by an integer number of times compared to t 1.
4. Write down how the modules of the vectors of displacements performed by the body in successive equal time intervals relate to each other if this body moves uniformly accelerated from a state of rest.
5. What is the purpose of using regularities (1) and (2)?

Exercise 8

1. The train departing from the station during the first 20 s moves in a straight line and uniformly accelerated. It is known that in the third second from the start of the movement the train traveled 2 m. Determine the module of the displacement vector made by the train in the first second and the module of the acceleration vector with which it moved.
2. A car, moving uniformly accelerated from a state of rest, travels 6.3 m in the fifth second of acceleration. What speed has the car developed by the end of the fifth second from the start of movement?
3. Some body in the first 0.03 s of movement without an initial velocity moved 2 mm, in the first 0.06 s - 8 mm, in the first 0.09 s - 18 mm. Based on regularity (1), prove that during all 0.09 s the body moved uniformly accelerated.

Questions.

1. What formulas are used to calculate the projection and modulus of the displacement vector of a body during its uniformly accelerated movement from a state of rest?

2. How many times will the modulus of the displacement vector of the body increase with an increase in the time of its movement from rest by n times?

3. Write down how the moduli of the displacement vectors of a body moving uniformly accelerated from a state of rest relate to each other with an increase in the time of its movement by an integer number of times compared to t 1.

4. Write down how the modules of the vectors of displacements performed by the body in successive equal time intervals relate to each other if this body moves uniformly accelerated from a state of rest.

5. For what purpose can regularities (3) and (4) be used?

Regularities (3) and (4) are used to determine whether the movement is uniformly accelerated or not (see p.33).

Exercises.

1. The train departing from the station during the first 20 s moves in a straight line and uniformly accelerated. It is known that in the third second from the start of the movement the train traveled 2 m. Determine the module of the displacement vector made by the train in the first second and the module of the acceleration vector with which it moved.