How to build a graph of movement of uniformly accelerated motion. Graphical representation of uniformly accelerated rectilinear motion
Lesson plan on the topic " »
The date:
Topic: Graphs of path and speed for uniform rectilinear motion
Goals:
Educational: formation of knowledge and representations of path and speed graphs in uniform rectilinear motion;
Developing: development and formation of practical skillsuse physical concepts and quantities to describe a uniform rectilinear motion; develop cognitive interest;
Educational: to instill a culture of mental work, accuracy, to teach to see the practical benefits of knowledge, to continue the formation of communication skills, to cultivate attentiveness, observation.
Lesson type: a lesson in learning new knowledge
Equipment and sources of information:
Isachenkova, L. A. Physics: textbook. for 7 cells. institutions of general avg. education with Russian lang. education / L. A. Isachenkova, G. V. Palchik, A. A. Sokolsky; ed. A. A. Sokolsky. Minsk: Narodnaya Asveta, 2017.
Lesson structure:
Organizing time(5 minutes)
Updating of basic knowledge (5min)
Learning new material (14 min)
Physical education (3 min)
Consolidation of knowledge (13min)
Lesson summary (5 min)
Lesson content
Organizing time (checking the presence in the class, checking the execution homework, voicing the topic and the main objectives of the lesson)
Updating of basic knowledge
№ 1. Finish the phrases.
Speed in uniform rectilinear motion over time __________________________________________________________________
Speed in SI is measured ____________________________________________
Distance traveled with uniform motion over time _______________________________________________________________
№ 2. There is a way to obtain formulas using the "memory triangle" (Fig. 1). If you close the symbol of the value to be determined, then in the triangle (open part) there remains a formula for its calculation. Get and write formulas for path calculations, speed and time intervalt.
Learning new material
Is it possible to express the relationship of the pathsand timet not through formulas, but in some other way? Charts are used for this.
Let's explain the essence graphic method on the specific example. Let the plane move uniformly and in a straight line with a speedv = 900 (Fig. 96). Let us describe the movement of the aircraft graphically, i.e., we will construct graphs of the dependence of the path and speed of the aircraft on the time of movement.
Pathsfrom the start timet 0 up to the point in timet equalsv ( t - t 0 ). Start timet 0 take for zero( t 0 = 0). Then the path formula becomes simpler:s = vt .
Let's find the path values for different meanings time interval and put them in the table1.
Now let's plot the path vs. time graph. On the abscissa, on a certain scale (for example, 1 cm - 1 hour), we will plot the time intervals of movement, and along the ordinate (on a scale of 1 cm - 900 km) - the path (Fig. 97).
The straight line I expresses the graphical dependence of the path on the time of uniform motion of the aircraft. This line is calledpath schedule. The path graph resembles the function graph you know from mathematics.at = kx , expressing a direct proportional relationshipat fromX.
The value of the path graph is that it, like the ratios = vt , allows you to solve main task- find a ways, passed by the body in an arbitrary period of timet .
For example, we are interested in the path of an aircraft over a period of timet = 4 hours. To do this, from a point on the horizontal axis corresponding to the timet = 4 hours (see Fig. 97), we draw a perpendicular to the intersection with the graph (pointTO). From the found pointTo we lower the perpendicular to the y-axis and get the answer without calculations. Paths = 3600 km.
And what isspeed graph ? It expresses the dependence of speed on time. Since the speed does not change over time, the same speed value corresponds to different points in time. Let's compile table 2 and build a straight line expressing the dependence of speed on time, plotting time along the abscissa and speed along the ordinate (Fig. 98).
The graph of the speed of uniform rectilinear motion is a straight line parallel to the time axis.
Straight line II depicts a graph of the speed of the aircraft. What is a speed graph? It not only shows the value of the speed, but also allows you to find the distance traveled. Calculate the path of the aircraft over a period of timet = 2 hours. According to the formulas = vt this ways= 900 2 h = 1800 km. Let's look at this work from the point of view of geometry. The first factor (900 expresses one side of the filled rectangle (see Fig. 98), the second (2 hours) - the other. From mathematics, you already know that by multiplying the sidesa and b find the areaSrectangle (Fig. 99).
Of course, the square is not a path, we are talking only about numerical equality.Distance traveled numerically equal to area figures under the graph of speed.
The area of \u200b\u200bthe figure under the velocity graph determines the path not only for uniform rectilinear, but also for any other movement. For example, the path over a period of time (see figure) is numerically equal to the area of the shaded figure:
s =
Physical education minute
Consolidation of knowledge
And now let's work with cards on the topic "Path and speed graphs for uniform rectilinear motion" (Appendix 1)
№ 1.
Answer: in the 4th movement, more time was spent on the passage of the same path.
№ 2.
Answer: in movement 1, a greater distance has been covered in the same period of time, sinces = v/ t(in movement 1, the speed is greater than in case 2, so the path will be longer in case 1)
№ 3. t= 2.0 h?
Answer:
the bus traveled a distance of 10 km in 15 minutes;
The bus drove for 15 minutes without stopping, and then made a stop for the duration: 1 hour 15 minutes - 30 minutes = 40 minutes;
until the stop, the bus was moving at a speed:
and after stopping was driving at a speed of:.
The bus traveled 60 km in 2 hours.
№ 4. For a period of timet
Answer:
a) graph 1 corresponds to the movement of Nadia;
b)
Consequently, the speed of Nadia's movement is several times less than that of Igor.
№ 5.
Answer:
a) the beetle first moved, then rested, and then moved again;
b) at the end of the 3rd second, the speed of movement is 2, and at the end of the 11th second, the speed of movement is 3;
in)s= v* t = 3 = 36 m.
No, because beetle moves slower
№ 6. t= 4 s?
Answer:
The movement of the cyclist was uniform rectilinear. He was moving at a speed of 8. s = v* t = 8 * 4c = 32m.
№ 7.
Answer:
The movement is uniform and straight. For the entire time of the movement, the athlete ranpaths= 6 km. In 15 minutes he ran the path .
Lesson summary
So let's sum it up:
The path graph expresses the dependence of the distance traveled on the time of movement of the body.
The path for uniform rectilinear motion can be determined by the formulas= vt , according to the path graph or using the speed graph.
Homework organization
§17, answer control questions.
Reflection
Continue the phrases:
Today in class I learned...
It was interesting…
The knowledge that I received in the lesson will come in handy.
Attachment 1
Card on the topic "Graphs of path and speed for uniform rectilinear motion"
Complete tasks and solve problems
№ 1.
In which of the movements (Fig. 2.) more time was spent on the passage of the same path?
№ 2.
In which of the movements, the speed graphs of which are shown in Figure 3, has the greater distance been covered in the same period of time?
№ 3. According to the graph (Fig. 4) of the dependence of the path on the time of the bus, determine which path the bus traveled over the period of time. Determine the time interval for the bus to stop and the stop time. How fast was the bus before and after the stop? How far did the bus travel in the timet= 2.0 h?
№ 4. For a period of timet= 4 s Nadia cycled the wayand Igor for the same period of time - the path Determine:
a) which of the graphs of the path versus time (Fig. 5) corresponds to the movement of the Nadia;
b) how many times the speeds of movement of Nadia and Igor differ.
№ 5.
A graph of the speed of the beetle is given. According to the schedule (Fig. 6), determine:
a) the nature of the movement; b) the speed of the beetle at the end of the 3rd and 11th seconds of movement; c) the path traveled by the beetle in timet= 12 s. Can the graph describe the real movement of the beetle?
№ 6. Figure 7 shows a graph of the speed of a cyclist on a straight section of the road as a function of time. What was the movement of the cyclist? How fast was he moving? How far did the cyclist cover the timet= 4 s?
№ 7.
According to the graph (Fig. 8) of the dependence of the path on time, determine the speed and time of movement of the athlete. What is this movement? What distance did the athlete run during the entire movement? How long did it take him to runBuild a graph of the athlete's speed versus time.
Physics problems - it's easy!
Don't forget that problems must always be solved in the SI system!
And now to the tasks!
Elementary tasks from the course of school physics in kinematics.
The task of compiling a description of the movement and compiling the equation of motion for this schedule movements
Given: body movement chart
Find:
1. write a description of the movement
2. draw up an equation of motion of the body.
We determine the projection of the velocity vector according to the graph, choosing any time interval convenient for consideration.
Here it is convenient to take t=4c
Compiling body motion equation:
We write down the formula for the equation of rectilinear uniform motion.
We substitute the found coefficient V x into it (do not forget about the minus!).
The initial coordinate of the body (X o) corresponds to the beginning of the graph, then X o \u003d 3
Compiling body movement description:
It is advisable to make a drawing, this will help not to be mistaken!
Let's not forget that everything physical quantities have units of measurement, they must be specified!
The body moves in a straight line and uniformly from the starting point X o = 3 m at a speed of 0.75 m / s opposite to the direction of the X axis.
The task of determining the place and time of the meeting of two moving bodies (with rectilinear uniform motion)
The movement of bodies is given by the equations of motion for each body.
Given:
1. equation of motion of the first body
2. equation of motion of the second body
Find:
1. meeting point coordinate
2. moment in time (after the start of movement) when the bodies meet
According to the given equations of motion, we build graphs of motion for each body in one coordinate system.
Intersection point two motion schedules defines:
1. on the t axis - the time of the meeting (how long after the start of the movement the meeting will occur)
2. on the X axis - the coordinate of the meeting place (relative to the origin)
As a result:
Two bodies will meet at a point with a coordinate of -1.75 m 1.25 seconds after the start of movement.
To check the received graphically answers, you can solve a system of equations from two given
equations of motion:
Everything was right!
For those who somehow forgot how to plot a rectilinear uniform motion graph:
The motion graph is a linear relationship (straight line), built on two points.
We choose any two values t 1 and t 2 convenient for ease of calculation.
For these values of t, we calculate the corresponding values of the coordinates X 1 and X 2 .
Set aside 2 points with coordinates (t 1 , X 1) and (t 2 , X 2) and connect them with a straight line - the graph is ready!
Tasks for compiling a description of the motion of a body and plotting motion graphs according to a given equation of rectilinear uniform motion
Task 1
Given: body motion equation
Find:
We compare the given equation with the formula and determine the coefficients.
Do not forget to make a drawing to once again pay attention to the direction of the velocity vector.
Task 2
Given: body motion equation
Find:
1. write a description of the movement
2. build a motion schedule
Task 3
Given: body motion equation
Find:
1. write a description of the movement
2. build a motion schedule
Task 4
Given: body motion equation
Find:
1. write a description of the movement
2. build a motion schedule
Movement description:
The body is at rest at a point with coordinate X=4m (the state of rest is special case movement when the velocity of the body is zero).
Task 5
Given:
initial coordinate of the moving point xo=-3 m
velocity vector projection Vx=-2 m/s
Find:
1. write down the equation of motion
2. build a motion schedule
3. show the velocity and displacement vectors on the drawing
4. find the coordinate of the point 10 seconds after the start of the movement
§ 14. GRAPHS OF PATH AND SPEED
Determination of the path according to the speed graph
In physics and mathematics, three ways of presenting information about the relationship between different quantities are used: a) in the form of a formula, for example, s = v ∙ t; b) in the form of a table; c) in the form of a graph (figure).
Velocity versus time v(t) - the velocity graph is depicted using two mutually perpendicular axes. We will plot time along the horizontal axis, and speed along the vertical axis (Fig. 14.1). It is necessary to think over the scale in advance so that the drawing is not too large or too small. At the end of the axis, a letter is indicated, which is a designation numerically equal to the area of \u200b\u200bthe shaded rectangle abcd of the value that is deposited on it. Near the letter indicate the unit of measurement of this value. For example, near the time axis indicate t, s, and near the velocity axis v (t), months. Choose a scale and put divisions on each axis.
Rice. 14.1. Graph of the speed of a body moving uniformly at a speed of 3 m/s. The path traveled by the body from the 2nd to the 6th second,
Image of uniform movement by table and graphs
Consider the uniform motion of a body with a speed of 3 m/s, that is, the numerical value of the speed will be constant throughout the entire time of motion. In short, this is written as follows: v = const (constant, that is, a constant value). In our example, it is equal to three: v = 3 . You already know that information about the dependence of one quantity on another can be presented in the form of a table (an array, as they say in computer science):
It can be seen from the table that at all indicated times the speed is 3 m/s. Let the scale of the time axis be 2 cells. \u003d 1 s, and the velocity axis is 2 cells. = 1 m/sec. A graph of speed versus time (abbreviated to say: speed graph) is shown in Figure 14.1.
Using the speed graph, you can find the path that the body travels in a certain time interval. To do this, we need to compare two facts: on the one hand, the path can be found by multiplying the speed by the time, and on the other hand, the product of the speed by the time, as can be seen from the figure, is the area of a rectangle with sides t and v.
For example, from the second to the sixth second the body moved for four seconds and passed 3 m/s ∙ 4 s = 12 m. segment ab along the vertical). The area, however, is somewhat unusual, since it is measured not in m 2, but in g. Therefore, the area under the speed graph is numerically equal to the distance traveled.
Path chart
The graph of the path s(t) can be depicted using the formula s = v ∙ t, that is, in our case, when the speed is 3 m/s: s = 3 ∙ t. Let's build a table:
Time (t, s) is again plotted along the horizontal axis, and the path along the vertical axis. Near the axis of the path we write: s, m (Fig. 14.2).
Determination of speed according to the path schedule
Let us now depict two graphs in one figure, which will correspond to movements with speeds of 3 m / s (straight line 2) and 6 m / s (straight line 1) (Fig. 14.3). It can be seen that the greater the speed of the body, the steeper the line of points on the graph.
There is also an inverse problem: having a motion schedule, you need to determine the speed and write down the equation of the path (Fig. 14.3). Consider straight line 2. From the beginning of the movement to the moment of time t = 2 s, the body has traveled a distance s = 6 m. Therefore, its speed is: v = = 3 . Choosing another time interval will not change anything, for example, at the moment t = 4 s, the path traveled by the body from the start of movement is s = 12 m. The ratio is again 3 m/sec. But this is how it should be, since the body is moving at a constant speed. Therefore, it would be easiest to choose a time interval of 1 s, because the path traveled by the body in one second is numerically equal to the speed. The path traveled by the first body (graph 1) in 1 s is 6 m, that is, the speed of the first body is 6 m/s. The corresponding path-time dependencies in these two bodies will be:
s 1 \u003d 6 ∙ t and s 2 \u003d 3 ∙ t.
Rice. 14.2. Path schedule. The remaining points, except for the six indicated in the table, were set in the task that the movement was uniform throughout the entire time
Rice. 14.3. Path graph in case of different speeds
Summing up
In physics, three methods of presenting information are used: graphical, analytical (by formulas) and table (array). The third method is more suitable for solving on a computer.
The path is numerically equal to the area under the speed graph.
The steeper the graph s(t), the greater the speed.
Creative tasks
14.1. Draw graphs of speed and path when the speed of the body increases or decreases uniformly.
Exercise 14
1. How is the path determined on the speed graph?
2. Is it possible to write a formula for the dependence of the path on time, having a graph of s (t)?
3. Or will the slope of the path graph change if the scale on the axes is halved?
4. Why is the graph of the path of uniform motion depicted as a straight line?
5. Which of the bodies (Fig. 14.4) has the highest speed?
6. What are the three ways of presenting information about the movement of the body, and (in your opinion) their advantages and disadvantages.
7. How can you determine the path according to the speed graph?
8. a) What is the difference between path graphs for bodies moving at different speeds? b) What do they have in common?
9. According to the graph (Fig. 14.1), find the path traveled by the body from the beginning of the first to the end of the third second.
10. What is the distance traveled by the body (Fig. 14.2) in: a) two seconds; b) four seconds? c) Indicate where the third second of the movement begins and where it ends.
11. Draw on the speed and path graphs the movement at a speed of a) 4 m/s; b) 2 m/sec.
12. Write down the formula for the dependence of the path on time for the movements shown in fig. 14.3.
13. a) Find the velocities of the bodies according to the graphs (Fig. 14.4); b) write down the corresponding equations of path and velocity. c) Plot the velocity graphs of these bodies.
14. Build graphs of the path and speed for bodies whose movements are given by the equations: s 1 = 5 ∙ t and s 2 = 6 ∙ t. What are the speeds of the bodies?
15. According to the graphs (Fig. 14.5), determine: a) the speed of the body; b) the paths they traveled in the first 5 seconds. c) Write down the path equation and plot the corresponding graphs for all three movements.
16. Draw a path graph for the movement of the first body relative to the second (Fig. 14.3).
Lesson Objectives:
educational: consider and form the skills of constructing graphs of the dependence of kinematic quantities on time with uniform and uniformly accelerated motion; teach students to analyze these graphs; by solving problems to consolidate the acquired knowledge in practice;
developing: development of the ability to observe, analyze specific situations; highlight certain signs;
nurturing: education of discipline and norms of behavior, creative attitude to the subject being studied; encourage student activity.
Methods:
verbal- conversation;
visual- video lesson, notes on the board;
controlling- testing or oral (written) survey, problem solving).
Connections:
interdisciplinary: mathematics - linear dependence, linear function graph; quadratic function and its graph;
intra-subject: Uniform and uniformly accelerated motion.
move lesson:
1. Organizational stage.
Good afternoon. Before we start the lesson, I would like each of you to tune in to the working mood.
2. Actualization of knowledge.
3. Explanation of new material.
You and I know that mechanical movement is a change in the position of a body (or body parts) in space relative to other bodies over time.
In turn, mechanical motion can be of two types - uniform, in which the body makes the same movements for any equal time intervals, and uneven, in which the body makes different movements for any equal time intervals.
Let's remember the basic formulas that we learned for uniform and non-uniform motion.
If the motion is uniform, then:
1. The speed of the body does not change over time;
2. To find the speed of the body, it is necessary to divide the path that the body has traveled for a certain period of time by this period of time;
3. The displacement equation has the form:
4. And - kinematic equation of uniform motion.
For uniformly accelerated:
1. The acceleration of the body does not change over time;
2. Acceleration is a value equal to the ratio of the change in the speed of the body to the time interval during which this change occurred
3. The equation of speed for uniformly accelerated motion has the form:
4. - displacement equation for uniformly accelerated motion;
5. - kinematic equation of uniformly accelerated motion.
For greater clarity, the movement can be described using graphs.
Consider the dependence of the acceleration that a body can have as a result of its motion on time.
If on the horizontal axis (abscissa) the time elapsed since the beginning of time is plotted on a certain scale, and on the vertical axis (y-axis) - also on the appropriate scale - the values of the acceleration of the body, the resulting graph will express the dependence of the acceleration of the body on time.
For uniform rectilinear motion, the graph of acceleration versus time has the form of a straight line, which coincides with the time axis, since acceleration in uniform motion is zero.
For uniformly accelerated motion, the acceleration graph also has the form of a straight line parallel to the time axis. In this case, the graph is located above the time axis, if the body is moving rapidly, and below the time axis, if the body is moving slowly.
If on the horizontal axis (abscissa) we plot time on a certain scale, and on the vertical axis of ordinates - also on the appropriate scale - the values of the body's speed, then we will get a speed graph.
For uniform motion, the velocity graph has the form of a straight line parallel to the time axis. In this case, the velocity graph is located above the time axis if the body moves along the axis X, and under the time axis, if the body moves against the axis X.
Such graphs show how the speed changes over time, i.e. how the speed depends on time. In the case of rectilinear uniform motion, this "dependence" is that the speed does not change over time. Therefore, the velocity graph is a straight line parallel to the time axis.
From the velocity graph, you can also find out the absolute value of the movement of the body for a given period of time. It is numerically equal to the area of the shaded rectangle: the upper one, if the body moves in the positive direction, and the lower one, if the body moves in the negative direction.
Indeed, the area of a rectangle is equal to the product of its sides: S=ab, where a and b sides of the rectangle.
But one of the parties in a certain scale is equal to time, and the other - to speed. And their product is just equal to the absolute value of the displacement of the body. In this case, the displacement will be positive if the projection of the velocity vector is positive, and negative if the projection of the velocity vector is negative.
With uniformly accelerated motion of the body along the coordinate axis X, the speed does not remain constant over time, but changes with time according to the formula v = v 0+ at, i.e., the speed is a linear function, and therefore the speed graphs look like a straight line, inclined to the time axis. Moreover, the greater the angle of inclination, the greater the speed of the body. On our graph, line 1 corresponds to movement with positive acceleration (speed increases) and some initial speed, line 2 corresponds to movement with negative acceleration (speed decreases) and initial speed equal to zero.
According to the velocity graph for uniformly accelerated motion, you can also find out the absolute value of the body's displacement over a given period of time. It is numerically equal to the area of the shaded trapezoid for body 1, and right triangle- in the opposite case. Indeed, for example, the area of a trapezoid is equal to the product of half the sum of its bases and its height. In our case, on a certain scale, the height of the trapezoid is equal to time, and the base is equal to the initial and final speed.
In this case, the displacement projection for the first body will be positive.
For the second body, a right triangle - half the product of its legs. In our case, the legs are the time and the final speed of the body.
The displacement projection is negative.
Now consider the dependence of the distance traveled on time.
As in previous cases, along the abscissa we will set aside the time from the moment the movement began, and along the ordinate - the path.
For uniform motion, the graph of the path versus time is a straight line, because dependence is linear.
In this case, the slope of the graph to the time axis depends on the speed modulus: the greater the speed, the greater the angle of inclination and the greater the speed of the body.
With uniformly accelerated movement, the graph will be a parabola branch, because the dependence, in this case, will be quadratic. And the greater the acceleration with which the body moves, the stronger the graph will be pressed against the y-axis.
Now let's move on to considering the dependence of displacement on time.
Consider uniform motion.
Because with uniform motion, the displacement linearly depends on time ( sx = υ xt), then the graph will be a straight line. The direction and angle of inclination of the graph to the time axis will depend on the projection of the velocity vector onto the coordinate axis.
So, in our case, bodies 2 and 3 move in the positive direction of the axis X, while the speed of the third body is greater than the speed of the second.
And body 1 - in the direction opposite to the direction of the axis X, so the graph is located under the time axis.
For uniformly accelerated motion, the displacement graph is a parabola, the position of the vertex of which depends on the directions of the initial velocity and acceleration.
For the 1st body, the acceleration is less than zero, the initial speed is zero.
For the 2nd body, the acceleration and initial velocity of the body are greater than zero.
For the 3rd body, the acceleration is greater than zero, the initial velocity is less than zero.
The 4th body has an initial velocity and acceleration less than zero.
For the 5th body, the acceleration is greater than zero, and the initial velocity is zero.
And, finally, the 6th body moves slowly, but with some initial speed.
And the last thing we will consider is the dependence of the body coordinate on time.
If on the horizontal axis (abscissa) to plot on a certain scale the time that has elapsed since the beginning of the time reference, and on the vertical axis (ordinate) - also on the appropriate scale - the values of the body coordinate, the resulting graph will express the dependence of the body coordinate on time (it is also called a motion schedule).
For uniformly accelerated motion, the motion graph, as in the case of displacement, is a parabola, the position of the vertex of which also depends on the directions of the initial velocity and acceleration.
The uniform motion graph is a straight line. This means that the coordinate depends linearly on time.
In the case of rectilinear motion of the body, the motion graphs give complete solution problems of mechanics, since they allow you to find the position of the body at any moment of time, including at the moments of time preceding the initial moment (assuming that the body was moving at the same speed before the start of the time reference).
With the help of the motion graph, you can determine:
1. body coordinates at any time;
2. the path traveled by the body for a certain period of time;
3. the time for which a path has been traveled;
4. the shortest distance between bodies at any given time;
5. moment and place of meeting, etc.
By the form of graphs of dependence of coordinates on time, one can also judge the speed of movement. It is clear that the greater the speed, the steeper the graph, i.e., the greater the angle between it and the time axis (the greater this angle, the greater the change in the coordinate for the same time).
At the same time, it must be remembered that the graph of the dependence of the body's coordinate on time should not be confused with the trajectory of the body's movement - a straight line, at all points of which the body has visited during its movement.
4. The stage of generalization and consolidation of new material
And so, let's make the main conclusion.
For greater clarity, mechanical movement can be described using graphs:
1) Dependences of speed on time;
2) Dependences of acceleration on time;
3) Dependence of the body coordinate on time;
4) And the dependence of the movement of the body on the time during which this movement occurred.
5. Reflection
I would like to hear your feedback about today's lesson: what did you like, what did you not like, what else would you like to know.
6. Homework.
Graphical representation of uniformly accelerated rectilinear motion.
Movement with uniformly accelerated motion.
Ilevel.
Many physical quantities that describe the motion of bodies change over time. Therefore, for greater clarity, the description of the movement is often depicted graphically.
Let us show how the time dependences of the kinematic quantities describing a rectilinear uniformly accelerated motion are graphically depicted.
Uniformly accelerated rectilinear motion- this is a movement in which the speed of the body changes in the same way for any equal time intervals, i.e., this is a movement with constant acceleration in magnitude and direction.
a=const - acceleration equation. That is, a has a numerical value that does not change with time.
By definition of acceleration
From here we have already found equations for the dependence of speed on time: v = v0 + at.
Let's see how this equation can be used to graphically represent uniformly accelerated motion.
Let us graphically depict the dependences of kinematic quantities on time for three bodies
.
1 the body moves along the 0X axis, while increasing its speed (the acceleration vector a is co-directed with the velocity vector v). vx >0, ax > 0
2 the body moves along the 0X axis, while reducing its speed (the acceleration vector and not co-directed with the velocity vector v). vx >0, ax< 0
2 the body moves against the 0X axis, while reducing its speed (the acceleration vector and not co-directed with the velocity vector v). vx< 0, ах > 0
Acceleration Graph
Acceleration is by definition a constant. Then, for the presented situation, the graph of the dependence of acceleration on time a(t) will look like:
From the acceleration graph, you can determine how the speed changed - increased or decreased, and by what numerical value the speed changed and for which body the speed changed more.
Speed Graph
If we compare the dependence of the coordinate on time for uniform motion and the dependence of the projection of velocity on time for uniformly accelerated motion, we can see that these dependencies are the same:
x= x0 + vx t vx = v 0 x + a X t
This means that the dependency graphs have the same form.
To build this graph, the time of movement is plotted on the abscissa axis, and the speed (velocity projection) of the body is plotted on the ordinate axis. In uniformly accelerated motion, the speed of a body changes over time.
Movement with uniformly accelerated motion.
With uniformly accelerated rectilinear motion, the speed of the body is determined by the formula
vx = v 0 x + a X t
In this formula, υ0 is the speed of the body at t = 0 (starting speed ), a= const - acceleration. On the velocity graph υ ( t), this dependence has the form of a straight line (Fig.).
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. The acceleration is numerically equal to the ratio of the sides of the triangle ABC:MsoNormalTable">
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 plot I: υ0 = –2 m/s, a= 1/2 m/s2.
For graph II: υ0 = 3 m/s, a= –1/3 m/s2.
The velocity graph also allows you to determine the displacement projection s body for a while t. Let us allocate on the time axis some small time interval Δ t. If this time interval is small enough, then the change in speed over this interval is small, i.e., the movement during this time interval can be considered uniform with a certain average speed, which is equal to the instantaneous velocity υ of the body in the middle of the interval Δ t. Therefore, displacement Δ s in time Δ t will be equal to Δ s = υΔ t. This displacement is equal to the area of the shaded strip (Fig.). Breaking down the time span from 0 to some point t for small intervals Δ t, we get that the displacement s for a given time t with uniformly accelerated rectilinear motion is equal to the area of the trapezoid ODEF. Corresponding constructions are made for graph II in fig. 1.4.2. Time t taken equal to 5.5 s.
Since υ – υ0 = at, the final formula for moving s bodies with uniformly accelerated motion over a time interval from 0 to t will be written in the form:
To find the coordinate y body at any given time. t y t: https://pandia.ru/text/78/516/images/image008_63.gif" width="84" height="48 src=">
To find the x coordinate of a body at any time t to the starting coordinate x 0 add displacement over time t:
When analyzing uniformly accelerated motion, sometimes the problem arises of determining the displacement of a body according to given values of the initial υ0 and final υ velocities and acceleration a. This problem can be solved using the equations written above by eliminating time from them. t. The result is written as
If the initial velocity υ0 is equal to zero, these formulas take the form MsoNormalTable">
It should once again be noted that the quantities υ0, υ, s, a, y 0 are algebraic quantities. Depending on the specific type of movement, each of these quantities can take both positive and negative values.
An example of solving the problem:
Petya moves down the mountain slope from rest with an acceleration of 0.5 m/s2 in 20 s and then moves along the horizontal section. Having traveled 40 m, he crashes into a gaping Vasya and falls into a snowdrift, reducing his speed to 0 m/s. With what acceleration did Petya move along the horizontal surface to the snowdrift? What is the length of the slope of the mountain from which Petya so unsuccessfully slid down?
Given: | |
a 1 = 0.5 m/s2 t 1 = 20 s s 2 = 40 m | Petya's movement consists of two stages: at the first stage, descending from the slope of the mountain, he moves with an increasing speed in absolute value; at the second stage, when moving along a horizontal surface, its speed decreases to zero (collided with Vasya). The values related to the first stage of the movement will be written with index 1, and for the second stage with index 2. |
Stage 1.
The equation for Petit's speed at the end of the descent from the mountain:
v 1 = v 01 + a 1t 1.
In projections on the axis X we get:
v 1x = a 1xt.
Let's write an equation relating the projections of Petya's speed, acceleration and displacement at the first stage of movement:
or because Petya was driving from the very top of the hill with the initial speed V01=0
(if I were Petya, I would be careful not to ride from such high hills)
Considering that Petya's initial speed at this 2nd stage of movement is equal to his final speed at the first stage:
v 02 x = v 1 x, v 2x = 0, where v1 is the speed with which Petya reached the bottom of the hill and started moving towards Vasya. V2x - Petya's speed in a snowdrift.
We use the equation and find the speed v1
On the horizontal section of the road, the path of Petit ramen:
BUT!!! it is more expedient to use another equation, since we do not know the time of Petya's movement to Vasya t2
The acceleration will turn out to be negative - this means that Petya tried very hard to slow down not about Vasya, but a little earlier.
Answer: a 2 = -1.25 m/s2; s 1 = 100 m.
IIlevel. Solve problems in writing.
1. Using the graphs shown in the figure, write down the equations for the dependence of speed on time. How the bodies moved at each stage of their movement (make according to the model, see example).
2. According to this acceleration graph, tell how the speed of the body changes. Write down the equations of the dependence of the speed on time, if at the moment of the beginning of the movement (t=0) the speed of the body is v0х =0. Please note that each subsequent segment of the movement, the body begins to pass at some speed (which was achieved in the previous time!).
3. A subway train leaving a station can reach a speed of 72 km/h in 20 seconds. Determine with what acceleration the bag forgotten in the subway car is moving away from you. What path will she take?
4. A cyclist moving at a speed of 3 m/s starts downhill with an acceleration of 0.8 m/s2. Find the length of the mountain if the descent took 6 s.
5. Starting braking with an acceleration of 0.5 m/s2, the train went to a stop 225 m. What was its speed before braking?
6. Starting to move, the soccer ball reached a speed of 50 m / s, traveled a distance of 50 m and crashed into the window. Determine the time it took the ball to travel this path and the acceleration with which it moved.
7. Uncle Oleg's neighbor's reaction time = 1.5 minutes, during which time he will figure out what happened to his window and have time to run out into the yard. Determine what speed young football players should develop so that the joyful owners of the window do not catch up with them if they need to run 350 m to their entrance.
8. Two cyclists are going towards each other. The first, having a speed of 36 km/h, began to climb uphill with an acceleration of 0.2 m/s2, and the second, having a speed of 9 km/h, began to descend the mountain with an acceleration of 0.2 m/s2. After how much time and in what place will they collide because of their absent-mindedness, if the length of the mountain is 100 m?