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 rolling down a hill; a stone thrown at an angle to the horizontal. Uniform motion is a special case of uniformly accelerated motion with acceleration equal to zero.
Let us consider the case of free fall (a body thrown at an angle to the horizontal) in more detail. Such movement can be represented as the sum of movements relative to the vertical and horizontal axes.
At any point of the trajectory, the body is affected by the acceleration of gravity g →, which does not change in magnitude and is always directed in one direction.
Along the X axis the movement 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 during uniformly accelerated motion:
Here v 0 is the initial velocity 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 by the slope of the velocity graph. In the figure above, the acceleration modulus is equal to the ratio of the sides of triangle ABC.
a = v - v 0 t = B C A C
The larger the angle β, the greater the slope (steepness) of the graph relative to the time axis. Accordingly, the greater the acceleration of the body.
For the first graph: v 0 = - 2 m s; a = 0.5 m s 2.
For the second graph: v 0 = 3 m s; a = - 1 3 m s 2 .
Using this graph, you can also calculate the displacement of the body during time t. How to do it?
Let us highlight a small period of time ∆ t on the graph. We will assume that it is so small that the movement during 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 us divide the entire time t into infinitesimal intervals ∆ t. The displacement s during 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 take the form:
s = v 0 t + a t 2 2
In order to find the coordinate of the body at a given 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
Law of uniformly accelerated motiony = y 0 + v 0 t + a t 2 2 .
Another common kinematics problem that arises when analyzing uniformly accelerated motion is finding the coordinate for given values of the initial and final velocities and acceleration.
Eliminating t from the equations written above and solving them, we obtain:
s = v 2 - v 0 2 2 a.
Using the known initial velocity, acceleration and displacement, the final velocity of the body can be found:
v = v 0 2 + 2 a s .
For v 0 = 0 s = v 2 2 a and v = 2 a s
Important!
The quantities 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 under the conditions of a specific task, they can take on both positive and negative values.
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Let us consider the motion of a body thrown horizontally and moving under the influence of gravity alone (we neglect 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 speed directed horizontally (Fig. 174).
Let's project the movement of the ball onto the vertical axis and onto the horizontal axis. The motion of the projection of the ball onto the axis is motion without acceleration with speed ; the movement of the projection of the ball onto the axis is a free fall with acceleration greater than the initial speed under the influence of gravity. We know the laws of both movements. The velocity component remains constant and equal to . The component grows in proportion to time: . The resulting speed can be easily found using the parallelogram rule, as shown in Fig. 175. It will be inclined downward, and its inclination will increase over time.
Rice. 174. Movement of a ball rolling off a table
Rice. 175. A ball thrown horizontally with speed has an instantaneous velocity
Let us find the trajectory of a body thrown horizontally. The coordinates of the body at the moment of time have meaning
To find the trajectory equation, we express time from (112.1) through and substitute this expression into (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 abscissa. We know that such curves are called parabolas. The graph of the path of uniformly accelerated motion was depicted as a parabola (§ 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, at a high horizontal initial speed, the parabola along which the body falls is more elongated in the horizontal direction. If a stream of water is released from a horizontal 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 speed of the water and the further from the tap the stream reaches the bottom of the cuvette. By placing a screen with pre-drawn parabolas behind the jet, you can make sure that the water jet really has the shape of a parabola.
In this lesson, we will look at an important characteristic of uneven motion - acceleration. In addition, we will consider uneven motion with constant acceleration. Such movement is also called uniformly accelerated or uniformly decelerated. Finally, we will talk about how to graphically depict the dependence of the speed of a body on time during uniformly accelerated motion.
Homework
Having solved the problems for this lesson, you will be able to prepare for questions 1 of the State Examination and questions A1, A2 of the Unified State Exam.
1. Problems 48, 50, 52, 54 sb. problems A.P. Rymkevich, ed. 10.
2. Write down the dependence of 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 turning points on the graphs, if any.
3. Consider the following questions and their answers:
Question. Is the acceleration due to gravity an acceleration as defined above?
Answer. Of course it is. The acceleration of gravity is the acceleration of a body that is freely falling from a certain height (air resistance must be neglected).
Question. What will happen if the acceleration of the body is directed perpendicular to the speed of the body?
Answer. The body will move uniformly around the circle.
Question. Is it possible to calculate the tangent of an angle using a protractor and a calculator?
Answer. No! Because the acceleration obtained in this way will be dimensionless, and the dimension of acceleration, as we showed earlier, should have the dimension m/s 2.
Question. What can be said about motion if the graph of speed versus time is not straight?
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 increased by n once:
Speed decreased in n once:
Speed increased by 2 m/s:
How many times did the speed increase?
How many times did the speed decrease?
How did the speed change?
How much has the speed increased?
How much did the speed decrease?
The body has reached its greatest height:
The body has traveled half the distance:
A body is thrown from the ground: (the last condition often escapes sight - if the body has zero speed, for example, a pen lying on a table, can it fly upward by itself?), the initial speed is directed upward.
The body is thrown down: the initial speed is directed downward.
The body is thrown upward: the initial speed is directed upward.
At the moment of falling to the ground:
A body falls out of an aerostat (balloon): the initial speed is equal to the speed of the aerostat (balloon) and is directed in the same direction.
3.2.2. How to determine acceleration from a velocity graph?
The law of speed change has the form:
The graph of this equation is a straight line. Since - coefficient before t, then is the slope of the 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 means a change in the direction of movement to the opposite.
To determine and, we select points on the graph at which the values can be accurately determined; 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 expressed as the area under the graph of speed versus acceleration. A simple case is shown in paragraph 3.1.6. Let's consider a more complex option, when the speed graph intersects the time axis.
Let us recall that the path can only increase, so the path traveled by the body in the example in Figure 9 is equal to:
where and are the areas of the figures shaded in the figure.
To determine the movement, you need to notice that at the points and the body changes the direction of movement. As the body travels along the path, it moves in the positive direction of the axis Ox, since the graph lies above the time axis. When traveling a path, the body moves in the opposite direction, in the negative direction of the axis Ox since the graph lies under the time axis. While traveling a 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 us pay attention once 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 a “+” sign in the definition of the distance traveled, but with a “−” sign in the definition of displacement.
3.2.4. How to determine the dependence of speed on time and coordinates on time from a graph of acceleration versus time?
In order to determine the required dependencies, initial conditions are required - velocity values 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 problem statement.
IN in this example We will try to present all the arguments in letters, so that in a particular example (when substituting numbers) we do not lose the essence of the actions.
Let at the moment of time the speed of the body be 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 movement is uniformly accelerated and the law of change of speed has the form:
By the end of this period of time (), the speed () and coordinate () will be equal (instead of time in the formulas, you need to substitute ):
The initial value of the speed in this interval must be equal to the final value in the previous interval, the initial value of the coordinate is equal to the final value of the coordinate in the previous interval, and the acceleration is determined from the graph:
therefore, the movement is uniformly accelerated and the law of change of speed has the form:
By the end of this period of time (), the speed () and coordinate () will be equal (instead of time in the formulas, you need to substitute ):
For a better understanding, let’s plot the results obtained on a graph (see figure)
On the speed graph:
1) From 0 to a straight line, “rising upward” (since);
2) From to is a horizontal straight line (since);
3) From to: a straight line “going down” (since).
Coordinates on the graph:
1) From 0 to : a parabola whose branches are directed upward (since );
2) From to: a straight line rising upward (since);
3) From to: a parabola whose branches are directed downwards (since).
3.2.5. How to write down the analytical formula of the law of motion from the graph of the law of motion?
Let a graph of uniformly alternating motion be given.
There are three unknown quantities 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 vertices of the cells. We get the system:
At the same time, we believe that we already know. Let's multiply the 1st equation of the system by and the 2nd equation by:
Subtract the 2nd from the 1st equation, after which we get:
We substitute the value obtained from this expression into any of the equations of system (3.67) and solve the resulting equation for:
3.2.6. How to determine the law of change of speed using the known law of motion?
The law of uniformly alternating motion has the form:
This is its standard appearance for this type of movement and it cannot look any other way, so it is worth remembering.
In this law the coefficient before t- this is the value of the initial speed, the pre coefficient is the acceleration divided in half.
For example, let the law be given:
And the speed equation looks like:
Thus, to solve such problems, it is necessary to accurately remember the form of the law of uniform motion and the meaning of the coefficients included in this equation.
However, you can go another 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 meeting, the bodies find themselves in the same coordinate, that is, it is necessary to solve the equation:
Let's rewrite it in the form:
This is a quadratic equation, the general solution of which will not be given, due to its cumbersomeness. The quadratic equation either has no solutions, which means the bodies have not met; or 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: that is, the time of the meeting must be positive.
3.2.8. How to determine the path in the th second?
Let a body begin to move from a state of rest and cover a path in the th second. We need to find which path the body covers in n-th second.
To solve this problem, you need to use formula (3.25):
Let us denote Then
Divide the equation by and we get:
3.2.9. How does a body move when thrown upward from a height? h?
Body thrown upward from a height h with speed
Coordinate equation y
The time of ascent to the highest point of the flight is determined from the condition:
H necessary in must be substituted:
Speed at the time of fall:
3.2.10. How does a body move when thrown from a height? h?
Body thrown upward from a height h with speed
Coordinate equation y at an arbitrary point in time:
The equation :
The entire flight time 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 needs to be “removed”. The main screening criterion is that the flight time cannot be negative:
Speed at the time of fall:
3.2.11. How does a body thrown upward from the surface of the earth move?
A body is thrown upward from the surface of the earth with a speed
Coordinate equation y at an arbitrary point in time:
The equation for the projection of velocity at an arbitrary moment in time:
The time of ascent to the highest point of the flight is determined from the condition
To find the maximum height H necessary in (3.89) necessary to substitute
The entire flight time is determined from the condition We obtain the equation:
Speed at the time of fall:
Note that this means that the time of ascent is equal to the time of falling to the same height.
We also got: that is, at what speed they threw it, at the same speed the body fell. The “−” sign in the formula indicates that the speed at the moment of 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 end up at the same height twice - the first time when moving up, the second time when falling down.
1) When the body is at a height h?
For a body thrown upward from the surface of the earth, the law of motion is valid:
When the body is on top h its coordinate will be equal to We obtain the equation:
the solution of which is:
2) The times and when the body was at its height are known h. When will the body be at its maximum height?
Flight time from altitude h back to height h equals As has already been shown, the time of ascent is equal to the time of falling to the same height, so the flight time depends on the height h to maximum height is:
Then the flight time from the start of movement to the maximum altitude:
3) The times and when the body was at its height are known h. What is the time of flight of the body?
The entire flight time is equal to:
4) The times and when the body was at its height are known h. What is the maximum lift height?
3.2.13. How does a body thrown horizontally from a height move? h?
A body thrown horizontally from a height h with speed
Acceleration projections:
Projections of velocity at an arbitrary moment in time t:
t:
t:
The flight time is determined from the condition
To determine the flight range it is necessary to enter the equation for the coordinates x instead of t substitute
To determine the speed of a body at the moment of falling, it is necessary to use the equation instead 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 from a height move? h?
A body thrown at an angle α to the horizontal from a height h with speed
Projections of initial velocity on the axis:
Acceleration projections:
Projections of velocity at an arbitrary moment in time t:
Velocity module at an arbitrary moment in time t:
Body coordinates at an arbitrary moment in time t:
Maximum height H
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 needs to be “removed”. The main screening criterion is that the flight time cannot be negative:
x L:
Speed at the moment 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 horizontal from the surface of the earth with a speed
Projections of initial velocity on the axis:
Acceleration projections:
Projections of velocity at an arbitrary moment in time t:
Velocity module at an arbitrary moment in time t:
Body coordinates at an arbitrary moment in time t:
The flight time to the highest point is determined from the condition
Speed in highest point flight
Maximum height H is determined by substituting into the law of change of coordinate y time
The entire flight time is found from the condition we obtain the equation:
We get
Once again we got that, that is, they showed once again that the time of rise is equal to the time of fall.
If we substitute into the law of coordinate changes x time then we get the flight range L:
Speed at the moment of fall
The angle that the velocity vector makes with the horizontal at an arbitrary moment in time:
Angle of incidence:
3.2.16. What are flat and mounted trajectories?
Let us 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 throwing point?
Flight range is determined by the formula:
From physical considerations it is clear that the angle α cannot be more than 90°, therefore, from a series of solutions to the equation, two roots are suitable:
The trajectory of movement for which is called the flat trajectory. The trajectory of movement for which is called a hinged trajectory.
3.2.17. How to use the speed triangle?
As was said in 3.6.1, the velocity triangle in each problem will have its own form. Let's look at a specific example.
The body was thrown from the top of the tower at a speed so that the flight range was 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 figure). Let us draw a height in it, which is obviously equal to Then the area of the velocity triangle is equal to:
Here we used formula (3.121).
Let's find the area of the same triangle using another formula:
Since these are the areas of the same triangle, we equate the formulas and:
Where do we get it from?
As can be seen from the formulas for final speed obtained in the previous paragraphs, the final speed does not depend on the angle at which the body was thrown, but depends only on the values of the initial speed and initial height. 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 speed triangle will be rectangular, therefore, the Pythagorean theorem is satisfied:
Where do we get it from?
The property of the velocity triangle, which has just been proven, can be used to solve other problems: the velocity triangle is rectangular in the maximum flight range problem.
3.2.18. How to use the displacement triangle?
As mentioned in 3.6.2, the displacement triangle in each problem 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 having an angle of inclination α. At what speed must a body be thrown so that it falls exactly at a distance? L from the throwing point?
Let's construct a triangle of displacements - this is a triangle ABC(see Fig. 19). Let's draw the 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:
Let's equate and:
How do we find the flight time:
Let's express AD from a triangle ABD:
Let's express the side DC from a triangle BCD:
But we get it
Let us substitute into this equation the resulting expression for flight time:
Finally we get
3.2.19. How to solve problems using the law of motion? (horizontally)
As a rule, in school, when solving problems involving uniformly alternating motion, formulas are used
However, this solution approach is difficult to apply to many problems. Let's look at a specific example.
A late passenger approached the last carriage of the train at the moment when the train started moving with constant acceleration. The only open door in one of the carriages was at a distance from the passenger. What is the lowest constant speed he must develop in order to board the train in time?
Let's introduce the axis Ox, directed along the movement of a person and a train. Let us take the initial position of the person (“2”) as the zero position. Then the initial coordinate of the open door ("1") L:
The door (“1”), like the entire train, have an initial speed of zero. Man (“2”) starts moving at speed
The door (“1”), like the entire train, moves with acceleration a. Man (“2”) moves at a constant speed:
The law of movement of both the door and the person has the form:
Let us 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 we will use the already known algorithm to find the place and time of 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 of his solutions have a physical meaning - the smallest root is the first meeting of a person and a door (a person can run quickly from a standstill, but the train will not immediately pick up speed, so the person can overtake the door), the second root is the second meeting (when the train has already accelerated and caught up with the man). But the presence of both roots means that a person can run slower. 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 understand the conditions of the problem: what the initial coordinate, initial velocity and acceleration are equal to. After this, we draw up an equation of motion and think about how to further solve the problem.
3.2.20. How to solve problems using the law of motion? (vertically)
Let's look at 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. Neglect air resistance.
For a freely falling body, the law of motion is valid:
In our case:
starting coordinate:
starting speed:
Let's substitute the conditions into the law of motion:
Substituting the required time values into the equation of motion, we will obtain the coordinates of the body at these moments.
At the moment of fall, the coordinate of the body
For s before the moment of fall, that is, at the coordinate of the body
The equations 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, we obtain the law of motion for this specific problem. Then, by substituting the required time values, we obtain the corresponding coordinate values. And we solve the problem!
