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Consider the motion of a body thrown horizontally and moving under the action of gravity alone (neglecting air resistance). For example, imagine that a ball lying on a table is given a push, and it rolls to the edge of the table and begins to fall freely, having an initial velocity directed horizontally (Fig. 174).

Let's project the movement of the ball on the vertical axis and on the horizontal axis. The movement of the projection of the ball onto the axis is a movement without acceleration with a speed of ; the motion of the projection of the ball on the axis is a free fall with acceleration beyond the initial velocity under the action of gravity. We know the laws of both motions. The velocity component remains constant and equal to . The component grows in proportion to time: . The resulting speed is easily found using the parallelogram rule, as shown in Fig. 175. It will lean downward and its slope will increase with time.

Rice. 174. Movement of a ball rolling off a table

Rice. 175. A ball thrown horizontally with a speed has a speed at the moment

Find the trajectory of a body thrown horizontally. The coordinates of the body at the moment of time matter

To find the trajectory equation, we express from (112.1) the time through and substitute this expression in (112.2). As a result, we get

The graph of this function is shown in Fig. 176. The ordinates of the trajectory points turn out to be proportional to the squares of the abscissas. We know that such curves are called parabolas. A parabola depicted a graph of the path of uniformly accelerated motion (§ 22). Thus, a freely falling body whose initial velocity is horizontal moves along a parabola.

The path traveled in the vertical direction does not depend on the initial speed. But the path traveled in the horizontal direction is proportional to the initial speed. Therefore, with a large horizontal initial velocity, the parabola along which the body falls is more elongated in the horizontal direction. If a jet of water is fired from a horizontally located tube (Fig. 177), then individual particles of water will, like the ball, move along a parabola. The more open the tap through which water enters the tube, the greater the initial velocity of the water and the farther from the tap the jet gets to the bottom of the cuvette. By placing a screen with parabolas pre-drawn on it behind the jet, one can verify that the water jet really has the shape of a parabola.

In this lesson, we will consider an important characteristic of uneven movement - acceleration. In addition, we will consider uneven movement with constant acceleration. This movement is also called uniformly accelerated or uniformly slowed down. Finally, we will talk about how to graphically depict the speed of a body as a function of time in uniformly accelerated motion.

Homework

By solving the tasks for this lesson, you will be able to prepare for questions 1 of the GIA and questions A1, A2 of the Unified State Examination.

1. Tasks 48, 50, 52, 54 sb. tasks of A.P. Rymkevich, ed. ten.

2. Write down the dependences of the speed on time and draw graphs of the dependence of the speed of the body on time for the cases shown in fig. 1, cases b) and d). Mark the turning points on the graphs, if any.

3. Consider next questions and their answers:

Question. Is gravitational acceleration an acceleration as defined above?

Answer. Of course it is. Free fall acceleration is the acceleration of a body that falls freely from a certain height (air resistance must be neglected).

Question. What happens if the acceleration of the body is directed perpendicular to the speed of the body?

Answer. The body will move uniformly in a circle.

Question. Is it possible to calculate the tangent of the angle of inclination using a protractor and a calculator?

Answer. Not! Because the acceleration obtained in this way will be dimensionless, and the dimension of acceleration, as we showed earlier, must have the dimension of m/s 2 .

Question. What can be said about motion if the graph of speed versus time is not a straight line?

Answer. We can say that the acceleration of this body changes with time. Such a movement will not be uniformly accelerated.

3.2.1. How to correctly understand the conditions of the problem?

The speed of the body has increased n once:

The speed has decreased in n once:

Speed increased by 2 m/s:

By how much did the speed increase?

By how much did the speed decrease?

How has the speed changed?

How much has the speed increased?

How much has the speed decreased?

The body has reached its greatest height:

The body has traveled half the distance:

The body is thrown from the ground: (the last condition is often overlooked - if the body's speed is zero, for example, the handle lying on the table, can it fly up by itself?), The initial speed is directed upwards.

The body is thrown down: the initial velocity is directed downwards.

The body is thrown upwards: the initial velocity is directed upwards.

At the moment of falling to the ground:

The body falls out of the balloon (balloon): the initial speed is equal to the speed of the balloon (balloon) and is directed in the same direction.

3.2.2. How to determine acceleration from a graph of speed?

The law of change of speed has the form:

The graph of this equation is a straight line. Since - coefficient before t, then is the slope of the straight line.

For chart 1:

The fact that graph 1 “rises up” means that the acceleration projection is positive, i.e. the vector is directed in the positive direction of the axis Ox

For chart 2:

The fact that graph 2 “goes down” means that the acceleration projection is negative, i.e. the vector is directed in the negative direction of the axis Ox. The intersection of the graph with the axis - a change in the direction of movement to the opposite.

To determine and, we select such points on the graph at which it is possible to accurately determine the values, as a rule, these are points located at the vertices of the cells.

3.2.3. How to determine the distance traveled and displacement from the speed graph?

As stated in paragraph 3.1.6, the path can be as the area under the graph of speed versus acceleration. A simple case is shown in Section 3.1.6. Let's consider a more complicated option, when the speed graph crosses the time axis.

Recall that the path can only increase, so the path that the body traveled in the example in Figure 9 is:

where and are the areas of the figures shaded in the figure.

To determine the displacement, it should be noted that at the points and the body changes the direction of motion. While passing the way the body moves in the positive direction of the axis Ox, since the graph lies above the time axis. Traveling the way the body moves in the opposite direction, in the negative direction of the axis Ox since the graph lies under the time axis. Passing the path, the body moves in the positive direction of the axis Ox, since the graph lies above the time axis. So the displacement is:

Let's pay attention again:

1) intersection with the time axis means turning in the opposite direction;

2) the area of the graph lying under the time axis is positive and is included with the sign "+" in the definition of the distance traveled, but with the sign "-" in the definition of displacement.

3.2.4. How to determine the dependence of speed on time and coordinates on time from the graph of acceleration versus time?

In order to determine the required dependencies, initial conditions are necessary - the values of velocity and coordinates at the moment of time Without initial conditions, it is impossible to solve this problem unambiguously, therefore, as a rule, they are given in the condition of the problem.

In this example, we will try to give all the reasoning in letters, so that a particular example (when substituting numbers) does not lose the essence of the actions.

Let at the moment of time the speed of the body be equal to zero and the initial coordinate

The initial values of speed and coordinates are determined from the initial conditions, and the acceleration from the graph:

therefore, the motion is uniformly accelerated and the law of change of speed has the form:

By the end of this time interval (), the speed () and coordinate () will be equal (instead of time in the formulas and you need to substitute ):

The initial value of the speed on this interval must be equal to the final value on the previous interval, the initial value of the coordinate is equal to the final value of the coordinate on the previous interval, and the acceleration is determined from the graph:

therefore, the motion is uniformly accelerated and the law of change of speed has the form:

By the end of this time interval (), the speed () and coordinate () will be equal (instead of time in the formulas and you need to substitute ):

For a better understanding, we plot the results obtained on a graph (see Fig.)

On the speed chart:

1) From 0 to a straight line, “rising up” (because);

2) From to a horizontal straight line (because );

3) From to: a straight line, "falling down" (because).

Coordinates on the chart:

1) From 0 to : a parabola whose branches are directed upwards (because );

2) From to: a straight line rising up (since);

3) From to: a parabola whose branches are directed downwards (because).

3.2.5. How to write down the analytical formula of the law of motion from the graph of the law of motion?

Let the graph of uniform motion be given.

There are three unknowns in this formula: and

To determine, it is enough to look at the value of the function at. To determine the other two unknowns, we select two points on the graph, the values of which we can accurately determine - the tops of the cells. We get the system:

We assume that we already know. Multiply the 1st equation of the system by and the 2nd equation by:

We subtract the 2nd equation from the 1st equation, after which we get:

We substitute the value obtained from this expression into any of the equations of the system (3.67) and solve the resulting equation with respect to:

3.2.6. How to determine the law of change of speed according to the known law of motion?

The law of uniform motion has the form:

This is its standard appearance for this type of movement and it cannot look otherwise, so it is worth remembering.

In this law, the coefficient before t is the value of the initial speed, the coefficient pre is the acceleration divided in half.

For example, given the law:

And the velocity equation is:

Thus, to solve such problems, it is necessary to remember exactly the form of the law of uniform motion and the meaning of the coefficients included in this equation.

However, you can go the other way. Let's remember the formula:

In our example:

3.2.7. How to determine the place and time of the meeting?

Let the laws of motion of two bodies be given:

At the moment of the meeting, the bodies are in the same coordinate, that is, it is necessary to solve the equation:

Let's rewrite it in the form:

This is quadratic equation, common decision which we will not present due to its cumbersomeness. The quadratic equation either has no solutions, which means that the bodies have not met; either has one solution - one single meeting; or has two solutions - two meetings of bodies.

The resulting solutions must be checked for physical feasibility. The most important condition: and that is, the time of the meeting must be positive.

3.2.8. How to determine the path in -th second?

Let the body start moving from a state of rest and cover the path in the -th second. It is required to find which path the body travels in n th second.

To solve this problem, it is necessary to use formula (3.25):

Denote Then

We divide the equation by and we get:

3.2.9. How does a body thrown up from a height move? h?

Body thrown up from a height h with speed

Coordinate equation y

The rise time to the highest point of the flight is determined from the condition:

H it is necessary in it is necessary to substitute :

Falling speed:

3.2.10. How does a body thrown down from a height move? h?

Body thrown up from a height h with speed

Coordinate equation y at an arbitrary point in time:

The equation :

The time of the entire flight is determined from the equation:

This is a quadratic equation that has two solutions, but in this problem the body can appear in the coordinate only once. Therefore, among the solutions obtained, one must be “removed”. The main dropout criterion is that the flight time cannot be negative:

Falling speed:

3.2.11. How does a body thrown up from the surface of the earth move?

A body is thrown upwards from the earth's surface with a speed

Coordinate equation y at an arbitrary point in time:

Velocity projection equation at an arbitrary moment of time:

The rise time to the highest point of the flight is determined from the condition

To find the maximum height H it is necessary in (3.89) it is necessary to substitute

The time of the entire flight is determined from the condition We obtain the equation:

Falling speed:

Note that it means that the time to rise is equal to the time to fall to the same height.

Also received: that is - with what speed they threw, with the same speed the body fell. The “−” sign in the formula indicates that the speed at the time of the fall is directed downward, that is, against the axis Oy.

3.2.12. The body has been at the same height twice...

When throwing a body, it can twice be at the same height - the first time when moving up, the second - when falling down.

1) When the body is on top h?

For a body thrown up from the surface of the earth, the law of motion is valid:

When the body is up h its coordinate will be equal to We get the equation:

whose solution looks like:

2) Times are known and when the body was at its best h. When will the body reach its maximum height?

Flight time from altitude h back to height h equals As already shown, the time of ascent is equal to the time of falling to the same height, so the time of flight from altitude h up to the maximum height is equal to:

Then the flight time from the start of movement to the maximum altitude:

3) Times are known and when the body was at its best h. What is the flight time of the body?

The total flight time is:

4) Times are known and when the body was at its best h. What is the maximum lifting height?

3.2.13. How does a body thrown horizontally from a height move? h?

Body thrown horizontally from a height h with speed

Acceleration projections:

Velocity projections at an arbitrary point in time t:

The flight time is determined from the condition

To determine the flight range, it is necessary in the equation for the coordinate x instead of t substitute

To determine the speed of a body at the moment of fall, it is necessary to put into the equation instead of t substitute

The angle at which the body falls to the ground:

3.2.14. How does a body thrown at an angle α to the horizon move from a height h?

A body thrown at an angle α to the horizon from a height h with speed

Projections of the initial speed on the axis:

Acceleration projections:

Velocity projections at an arbitrary point in time t:

Velocity modulus at an arbitrary point in time t:

Body coordinates at an arbitrary point in time t:

Max Height H

x L:

Speed at the time of fall

Angle of incidence:

3.2.15. How does a body thrown at an angle α to the earth's horizon move?

A body thrown at an angle α to the horizon from the earth's surface with a speed

Projections of the initial speed on the axis:

Acceleration projections:

Velocity projections at an arbitrary point in time t:

Velocity modulus at an arbitrary point in time t:

Body coordinates at an arbitrary point in time t:

The flight time to the highest point is determined from the condition

Speed at the highest point of flight

Max Height H is determined by substituting into the law of change of the coordinate y of time

The entire flight time is found from the condition we obtain the equation:

We get

Again we got that, that is, we showed once again that the rise time is equal to the fall time.

If we substitute into the law of coordinate change x time we get the flight range L:

Speed at the time of fall

The angle that the velocity vector forms with the horizontal at an arbitrary point in time:

Angle of incidence:

3.2.16. What are flat and mounted trajectories?

Let's solve the following problem: at what angle should a body be thrown from the surface of the earth so that the body falls at a distance L from the drop point?

Flight range is determined by the formula:

From physical considerations, it is clear that the angle α cannot be greater than 90°, therefore, two roots are suitable from a series of solutions to the equation:

A motion trajectory for which is called a flat trajectory. The trajectory of movement, for which it is called the hinged trajectory.

3.2.17. How to use the triangle of speeds?

As it was said in 3.6.1, the speed triangle in each task will have its own form. Let's look at a specific example.

A body is thrown from the top of a tower at a speed such that the flight range is maximum. By the time it hits the ground, the speed of the body is How long did the flight last?

Let's construct a triangle of speeds (see Fig.). We draw a height in it, which, obviously, is equal to Then the area of the triangle of speeds is equal to:

Here we have used formula (3.121).

Find the area of the same triangle using a different formula:

Since these are the areas of the same triangle, we equate the formulas and:

Where do we get

As can be seen from the formulas for the final velocity obtained in the previous paragraphs, the final velocity does not depend on the angle at which the body was thrown, but only the values of the initial velocity and initial height depend. Therefore, the flight range according to the formula depends only on the angle between the initial and final speed β. Then the flight range L will be maximum if it takes the maximum possible value, that is,

Thus, if the flight range is maximum, then the velocity triangle will be rectangular, therefore, the Pythagorean theorem is fulfilled:

Where do we get

The property of the velocity triangle, which has just been proved, can be used in solving other problems: the velocity triangle is rectangular in the maximum range problem.

3.2.18. How to use the displacement triangle?

As mentioned in 3.6.2, the displacement triangle in each task will have its own form. Let's look at a specific example.

A body is thrown at an angle β to the surface of a mountain with an angle of inclination α. With what speed must the body be thrown so that it falls exactly at a distance L from the drop point?

Let's build a displacement triangle - this is a triangle ABC(see fig. 19). Let's draw a height in it BD. Obviously the angle DBC is equal to α.

Let's express the side BD from a triangle BCD:

Let's express the side BD from a triangle ABD:

Equate and :

Where do we find the flight time:

Express AD from a triangle ABD:

Let's express the side DC from a triangle BCD:

But We Get

Substitute in this equation, the resulting expression for the flight time:

Finally we get

3.2.19. How to solve problems using the law of motion? (horizontally)

As a rule, at school, when solving problems for uniformly variable motion, formulas are used

However, this approach to the solution is difficult to apply to solving many problems. Let's consider a specific example.

The late passenger approached the last car of the train at the moment when the train started moving, starting to move with constant acceleration The only open door in one of the cars turned out to be at a distance from the passenger What is the least constant speed he must develop in order to have time to get on the train?

Let's introduce the axis Ox, directed along the movement of a person and a train. For the zero position, we take the initial position of the person (“2”). Then the initial coordinate of the open door ("1") L:

The door (“1”), like the whole train, has an initial speed equal to zero. The person ("2") starts moving at a speed

The door (“1”), like the whole train, moves with an acceleration a. The person ("2") moves at a constant speed:

The law of motion of both the door and the person has the form:

We substitute the conditions and into the equation for each of the moving bodies:

We have compiled an equation of motion for each of the bodies. Now let's use the already known algorithm to find the place and time of the meeting of two bodies - we need to equate and :

Where do we get the quadratic equation for determining the meeting time:

This is a quadratic equation. Both solutions have physical meaning - smallest root, this is the first meeting of a person and a door (a person can run quickly from a place, and the train will not immediately pick up high speed, so a person can overtake the door), the second root is the second meeting (when the train has already accelerated and caught up with a person). But the presence of both roots means that a person can run more slowly. The speed will be minimal when the equation has one single root, that is

Where do we find the minimum speed:

In such problems, it is important to analyze in the conditions of the problem: what are the initial coordinate, initial velocity and acceleration. After that, we compose the equation of motion and think about how to solve the problem further.

3.2.20. How to solve problems using the law of motion? (vertically)

Consider an example.

A freely falling body traveled the last 10 m in 0.5 s. Find the time of fall and the height from which the body fell. Ignore air resistance.

For the free fall of a body, the law of motion is valid:

In our case:

start coordinate:

starting speed:

Substitute the conditions in the law of motion:

Substituting the required values of time into the equation of motion, we will obtain the coordinates of the body at these moments.

At the time of the fall, the coordinate of the body

From before the moment of fall, that is, at the coordinate of the body

Equations and constitute a system of equations in which the unknowns H and Solving this system, we get:

So, knowing the form of the law of motion (3.30), and using the conditions of the problem to find and we obtain the law of motion for a given specific task. After that, substituting the required time values, we obtain the corresponding coordinate values. And we solve the problem!

Uniformly accelerated motion- this is a movement in which the acceleration vector does not change in magnitude and direction. Examples of such movement: a bicycle that rolls down a hill; a stone thrown at an angle to the horizon. Uniform motion is a special case of uniformly accelerated motion with an acceleration equal to zero.

Let us consider the case of free fall (a body is thrown at an angle to the horizon) in more detail. Such movement can be represented as the sum of movements about the vertical and horizontal axes.

At any point of the trajectory, the free fall acceleration g → acts on the body, which does not change in magnitude and is always directed in one direction.

Along the X axis the motion is uniform and rectilinear, and along the Y axis it is uniformly accelerated and rectilinear. We will consider the projections of the velocity and acceleration vectors on the axis.

Formula for speed with uniformly accelerated motion:

Here v 0 is the initial speed of the body, a = c o n s t is the acceleration.

Let us show on the graph that with uniformly accelerated motion, the dependence v (t) has the form of a straight line.

Acceleration can be determined from the slope of the velocity graph. In the figure above, the acceleration modulus is equal to the ratio of the sides of the triangle ABC.

a = v - v 0 t = B C A C

The larger the angle β, the greater the slope (steepness) of the graph with respect to the time axis. Accordingly, the greater the acceleration of the body.

For the first graph: v 0 = - 2 m s; a \u003d 0, 5 m s 2.

For the second graph: v 0 = 3 m s; a = - 1 3 m s 2 .

From this graph, you can also calculate the movement of the body in time t. How to do it?

Let's single out a small time interval ∆ t on the graph. We will assume that it is so small that the movement during the time ∆ t can be considered uniform movement with a speed equal to the speed of the body in the middle of the interval ∆ t . Then, the displacement ∆ s during the time ∆ t will be equal to ∆ s = v ∆ t .

Let's divide all time t into infinitely small intervals ∆ t . The displacement s in time t is equal to the area of the trapezoid O D E F .

s = O D + E F 2 O F = v 0 + v 2 t = 2 v 0 + (v - v 0) 2 t .

We know that v - v 0 = a t , so the final formula for moving the body will be:

s = v 0 t + a t 2 2

To find the coordinate of the body in this moment time, you need to add displacement to the initial coordinate of the body. The change in coordinates depending on time expresses the law of uniformly accelerated motion.

Law of uniformly accelerated motion

y = y 0 + v 0 t + a t 2 2 .

Another common task of kinematics that arises in the analysis of uniformly accelerated motion is finding the coordinate for given values of the initial and final velocities and acceleration.

Eliminating t from the above equations and solving them, we obtain:

s \u003d v 2 - v 0 2 2 a.

From the known initial speed, acceleration and displacement, you can find the final speed of the body:

v = v 0 2 + 2 a s .

For v 0 = 0 s = v 2 2 a and v = 2 a s

Important!

The values v , v 0 , a , y 0 , s included in the expressions are algebraic quantities. Depending on the nature of the movement and the direction of the coordinate axes in a particular task, they can take both positive and negative values.

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