We have many reasons to thank our God.
Have you noticed how in every year, actively and decisively, God's organization accelerates the course, providing many gifts!
The heavenly chariot is definitely in motion! At the annual meeting, it was reported: "If you think that you are not keeping up with Jehovah's chariot, buckle up so as not to fly out at the bend!":)
The prudent servant is seen to keep going, opening up new territories for preaching, making disciples, and gaining a fuller understanding of God's purposes.
Since the faithful slave does not rely on human strength, but on the guidance of the holy spirit, it is clear that the faithful slave is guided by God's spirit!!!
It can be seen that when the Governing Body sees the need to clarify any aspect of the truth or to make changes in the organizational order, it acts without delay.
Isaiah 60:16 says that God's people will enjoy the milk of the nations, which is cutting edge technology today.
Today in the hands of the organizationa site that connects and unites us with our brotherhood, and other novelties that you probably already know about.
It is only because God supports and blesses them through his Son and the Messianic Kingdom that these imperfect people can triumph over Satan and his wicked system of things.
Compare circulations and numbers of languages for the December and January 2014, 2015, 2016 Watchtower and Awake.
And below is 1962.
The blue is the Watchtower magazine and the red is the Awake magazine.
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The circulation of The Watchtower has grown to 58,987,000 million since January 2015 and is already being translated into 254 languages. On the front page of this magazine, there is also an outline for ministry presentation.
Incredible! And they say there are no miracles! This edition is a real miracle!
What about our publications!
Since August last year (2014), our website ranking has increased by 552 positions, thus improving by 30 percent.
For non-commercial sites, this is an absolute record.A little more and we will be able to enter the top 1000!!!
Why do they do this?
Imagine a sinking ship. There are, among other things, three groups of people.
The first try to feed the passengers.
The second offer warm fur coats.
Still others help to board the boats and get off the ship.
Everyone seems to be doing good. But what kind of good in this situation makes sense? The answer is obvious! What's the point if someone is fed, clothed, and he still dies. First you need to transfer from the sinking ship and get to a safe place, and then feed and warm.
Jehovah's Witnesses do the same - they do good to people that makes sense.
While this materialistic world languishes with spiritual hunger, let's develop an appetite for spiritual food.
Let's not fall into the trap of materialism!
Paul pointed out THREE TIMES whom and how we should show concern?
1Tm 2:1 Prayers are to be offered “for every kind of people”
1tm 2:4 It is necessary "that people of every kind... come to an accurate knowledge of the truth"
1tm 2:6 Christ "gave himself as an appropriate ransom for all"
What will help us to have a deep concern for all and to reach people of all kinds by preaching?
This requires one very important quality that Jehovah possesses - impartiality! ( De 10:34 )
Indeed, Jehovah is “no respecter of persons” (attitude) and “shows no partiality to anyone” (deeds)
Jesus preached to all kinds of people. Remember, in his illustrations, Jesus spoke of people of all backgrounds and social backgrounds: a farmer sowing seed, a housewife making bread, a man working in the field, a successful merchant selling pearls, hard-working fishermen casting nets. (Matthew 13:31-33, 44-48)
Fact: Jehovah and Jesus desire that “every kind of people be saved” and receive eternal blessings. They do not put some people above others.
The lesson for us is that in order to imitate Jehovah and Jesus, we need to preach to people of all kinds, regardless of their race or life circumstances.
Much has already been done by God's organization for those who speak foreign language, immigrants, students, refugees, those who live in nursing homes, in gated communities, entrepreneurs, prisoners, the deaf, the blind, adherents of non-Christian religions and others.
]At present, on the territory of Russia, under the supervision of a branch in 578 congregations, they are assigned to take care of the preaching of the good news in the correctional institutions assigned to them. Many of these places hosted congregation meetings, group and individual Bible studies. Preaching in such places helps many to “put on the new personality” and serve the true God, Jehovah. Yes, it is important to continue to sanctify the name of God!
Therefore, let us appreciate everything that happens in God's organization. Let us learn to make good use of the publications issued by the faithful slave, which are designed to touch the hearts of all kinds of people. After all, how we educate ourselves will depend on how we educate others.
In this way we will show that we care deeply for the "desired treasures from all nations" that still need to be brought.
Surely we, like Peter, have learned our lesson:
"we have nowhere to go" - there is only one place where we will not fall behind the chariot of Jehovah and will be under the protection of God the Creator, Jehovah (John 6:68).
2. SPEED OF THE BODY. RECTILINEAR UNIFORM MOVEMENT.
Speed is a quantitative characteristic of the movement of the body.
average speed- This physical quantity, equal to the ratio of the point displacement vector to the time interval Δt, during which this displacement occurred. The direction of the average velocity vector coincides with the direction of the displacement vector . average speed is determined by the formula:
Instant Speed, that is, the speed in this moment time is a physical quantity equal to the limit to which the average speed tends with an infinite decrease in the time interval Δt:
In other words, instantaneous speed at a given point in time is the ratio of a very small movement to a very small period of time during which this movement occurred.
The instantaneous velocity vector is directed tangentially to the trajectory of the body (Fig. 1.6).
Rice. 1.6. Instantaneous velocity vector.
In the SI system, speed is measured in meters per second, that is, the unit of speed is considered to be the speed of such uniform rectilinear motion, in which in one second the body travels a distance of one meter. The unit of speed is denoted m/s. Often speed is measured in other units. For example, when measuring the speed of a car, train, etc. The commonly used unit of measure is kilometers per hour:
1 km/h = 1000 m / 3600 s = 1 m / 3.6 s
1 m/s = 3600 km / 1000 h = 3.6 km/h
Addition of speeds (perhaps not necessarily the same question will be in 5).
The velocities of the body in different reference systems are connected by the classical law of addition of speeds.
body speed relative to fixed frame of reference is equal to the sum of the velocities of the body in moving frame of reference and the most mobile frame of reference relative to the fixed one.
For example, a passenger train is moving along a railroad at a speed of 60 km/h. A person is walking along the carriage of this train at a speed of 5 km/h. If we consider the railway to be motionless and take it as a frame of reference, then the speed of a person relative to the frame of reference (that is, relative to railway), will be equal to the addition of the speeds of the train and the person, that is
60 + 5 = 65 if the person is walking in the same direction as the train
60 - 5 = 55 if the person and the train are moving in different directions
However, this is only true if the person and the train are moving along the same line. If a person moves at an angle, then this angle will have to be taken into account, remembering that speed is vector quantity.
An example is highlighted in red + The law of displacement addition (I think this does not need to be taught, but for general development you can read it)
Now let's look at the example described above in more detail - with details and pictures.
So, in our case, the railway is fixed frame of reference. The train that is moving along this road is moving frame of reference. The car on which the person is walking is part of the train.
The speed of a person relative to the car (relative to the moving frame of reference) is 5 km/h. Let's call it C.
The speed of the train (and hence the wagon) relative to a fixed frame of reference (that is, relative to the railway) is 60 km/h. Let's denote it with the letter B. In other words, the speed of the train is the speed of the moving reference frame relative to the fixed frame of reference.
The speed of a person relative to the railway (relative to a fixed frame of reference) is still unknown to us. Let's denote it with a letter.
We will associate the XOY coordinate system with the fixed reference system (Fig. 1.7), and the X P O P Y P coordinate system with the moving reference system. Now let's try to find the speed of a person relative to the fixed reference system, that is, relative to the railway.
For a short period of time Δt, the following events occur:
Then for this period of time the movement of a person relative to the railway:
This displacement addition law. In our example, the movement of a person relative to the railway is equal to the sum of the movements of a person relative to the wagon and the wagon relative to the railway.
Rice. 1.7. The law of addition of displacements.
The law of addition of displacements can be written as follows:
= ∆ H ∆t + ∆ B ∆t
The speed of a person relative to the railroad is:
The speed of a person relative to the car:
Δ H \u003d H / Δt
The speed of the car relative to the railway:
Therefore, the speed of a person relative to the railway will be equal to:
This is the lawspeed addition:
Uniform movement- this is movement at a constant speed, that is, when the speed does not change (v \u003d const) and there is no acceleration or deceleration (a \u003d 0).
Rectilinear motion- this is movement in a straight line, that is, the trajectory of rectilinear movement is a straight line.
Uniform rectilinear motion is a movement in which the body makes the same movements for any equal intervals of time. For example, if we divide some time interval into segments of one second, then with uniform motion the body will move the same distance for each of these segments of time.
The speed of uniform rectilinear motion does not depend on time and at each point of the trajectory is directed in the same way as the movement of the body. That is, the displacement vector coincides in direction with the velocity vector. In this case, the average speed for any period of time is equal to the instantaneous speed:
Speed of uniform rectilinear motion is a physical vector quantity equal to the ratio of the displacement of the body for any period of time to the value of this interval t:
Thus, the speed of uniform rectilinear motion shows what movement a material point makes per unit of time.
moving with uniform rectilinear motion is determined by the formula:
Distance traveled in rectilinear motion is equal to the displacement modulus. If the positive direction of the OX axis coincides with the direction of movement, then the projection of the velocity on the OX axis is equal to the velocity and is positive:
v x = v, i.e. v > 0
The projection of displacement onto the OX axis is equal to:
s \u003d vt \u003d x - x 0
where x 0 is the initial coordinate of the body, x is the final coordinate of the body (or the coordinate of the body at any time)
Motion equation, that is, the dependence of the body coordinate on time x = x(t), takes the form:
If the positive direction of the OX axis is opposite to the direction of motion of the body, then the projection of the body velocity on the OX axis is negative, the velocity is less than zero (v< 0), и тогда уравнение движения принимает вид.
So let's say our bodies move in the same direction. How many cases do you think there might be for such a condition? That's right, two.
Why is it so? I am sure that after all the examples you will easily figure out how to derive these formulas.
Got it? Well done! It's time to solve the problem.
The fourth task
Kolya goes to work by car at a speed of km/h. Colleague Kolya Vova travels at a speed of km/h. Kolya lives at a distance of km from Vova.
How long will it take Vova to overtake Kolya if they left the house at the same time?
Did you count? Let's compare the answers - it turned out that Vova will catch up with Kolya in hours or minutes.
Let's compare our solutions...
The drawing looks like this:
Similar to yours? Well done!
Since the problem asks how long the guys met and left at the same time, the time they traveled will be the same, as well as the meeting place (in the figure it is indicated by a dot). Making equations, take the time for.
So, Vova made his way to the meeting place. Kolya made his way to the meeting place. It's clear. Now we deal with the axis of movement.
Let's start with the path that Kolya did. Its path () is shown as a segment in the figure. And what does Vova's path () consist of? That's right, from the sum of the segments and, where is the initial distance between the guys, and is equal to the path that Kolya did.
Based on these conclusions, we obtain the equation:
Got it? If not, just read this equation again and look at the points marked on the axis. Drawing helps, doesn't it?
hours or minutes minutes.
I hope that in this example you understand how important the role of well crafted drawing!
And we are smoothly moving on, or rather, we have already moved on to the next step in our algorithm - bringing all quantities to the same dimension.
The rule of three "P" - dimension, reasonableness, calculation.
Dimension.
Not always in tasks the same dimension is given for each participant in the movement (as it was in our easy tasks).
For example, you can meet tasks where it is said that the bodies moved a certain number of minutes, and the speed of their movement is indicated in km / h.
We can't just take and substitute the values in the formula - the answer will be wrong. Even in terms of units of measurement, our answer “will not pass” the test for reasonableness. Compare:
See? With proper multiplication, we also reduce the units of measurement, and, accordingly, we get a reasonable and correct result.
And what happens if we do not translate into one system of measurement? The answer has a strange dimension and % is an incorrect result.
So, just in case, let me remind you the meanings of the basic units of measurement of length and time.
Length units:
centimeter = millimeters
decimeter = centimeters = millimeters
meter = decimeters = centimeters = millimeters
kilometer = meters
Time units:
minute = seconds
hour = minutes = seconds
days = hours = minutes = seconds
Advice: When converting units of measurement related to time (minutes to hours, hours to seconds, etc.), imagine a clock face in your head. It can be seen with the naked eye that minutes is a quarter of the dial, i.e. hours, minutes is a third of the dial, i.e. hours, and a minute is an hour.
And now a very simple task:
Masha rode her bicycle from home to the village at a speed of km/h for minutes. What is the distance between the car house and the village?
Did you count? The correct answer is km.
minutes is an hour, and another minute from an hour (mentally imagined a clock face, and said that minutes is a quarter of an hour), respectively - min \u003d h.
Intelligence.
Do you understand that the speed of a car cannot be km/h, unless, of course, we are talking about a sports car? And even more so, it cannot be negative, right? So, reasonableness, that's about it)
Calculation.
See if your solution "passes" the dimension and reasonableness, and only then check the calculations. It is logical - if there is an inconsistency with dimension and reasonableness, then it is easier to cross out everything and start looking for logical and mathematical errors.
"Love for tables" or "when drawing is not enough"
Far from always, the tasks for movement are as simple as we solved before. Very often, in order to correctly solve a problem, you need to not just draw a competent drawing, but also make a table with all the conditions given to us.
First task
From point to point, the distance between which is km, a cyclist and a motorcyclist left at the same time. It is known that a motorcyclist travels more miles per hour than a cyclist.
Determine the speed of the cyclist if it is known that he arrived at the point a minute later than the motorcyclist.
Here is such a task. Pull yourself together and read it several times. Read? Start drawing - straight line, point, point, two arrows ...
In general, draw, and now let's compare what you got.
Kind of empty, right? We draw a table.
As you remember, all movement tasks consist of components: speed, time and path. It is from these graphs that any table in such problems will consist.
True, we will add one more column - Name about whom we write information - a motorcyclist and a cyclist.
Also indicate in the header dimension, in which you will enter the values \u200b\u200bin there. You remember how important this is, right?
Do you have a table like this?
Now let's analyze everything that we have, and in parallel enter the data into a table and into a figure.
The first thing we have is the path that the cyclist and motorcyclist have traveled. It is the same and equal to km. We bring in!
Let us take the speed of the cyclist as, then the speed of the motorcyclist will be ...
If the solution of the problem does not work with such a variable, it's okay, we'll take another one until we reach the victorious one. This happens, the main thing is not to be nervous!
The table has changed. We have left not filled only one column - time. How to find the time when there is a path and speed?
That's right, divide the path by the speed. Enter it in the table.
So our table has been filled, now you can enter data into the figure.
What can we reflect on it?
Well done. The speed of movement of a motorcyclist and a cyclist.
Let's read the problem again, look at the figure and the completed table.
What data is not shown in the table or in the figure?
Right. The time by which the motorcyclist arrived earlier than the cyclist. We know that the time difference is minutes.
What should we do next? That's right, translate the time given to us from minutes to hours, because the speed is given to us in km / h.
The magic of formulas: writing and solving equations - manipulations that lead to the only correct answer.
So, as you already guessed, now we will make up the equation.
Compilation of the equation:
Look at your table, at the last condition that was not included in it, and think about the relationship between what and what can we put into the equation?
Right. We can make an equation based on the time difference!
Is it logical? The cyclist rode more, if we subtract the time of the motorcyclist from his time, we will just get the difference given to us.
This equation is rational. If you don't know what it is, read the topic "".
We bring the terms to a common denominator:
Let's open the brackets and give like terms: Phew! Got it? Try your hand at the next task.
Equation solution:
From this equation we get the following:
Let's open the brackets and move everything to the left side of the equation:
Voila! We have a simple quadratic equation. We decide!
We received two responses. Look what we got for? That's right, the speed of the cyclist.
We recall the rule "3P", more specifically "reasonableness". Do you understand what I mean? Exactly! Speed cannot be negative, so our answer is km/h.
Second task
Two cyclists set out on a 1-kilometer run at the same time. The first one was driving at a speed that was 1 km/h faster than the second one, and arrived at the finish line hours earlier than the second one. Find the speed of the cyclist who came to the finish line second. Give your answer in km/h.
I recall the solution algorithm:
- Read the problem a couple of times - learn all the details. Got it?
- Start drawing the drawing - in which direction are they moving? how far did they travel? Did you draw?
- Check if all the quantities you have are of the same dimension and start writing out the condition of the problem briefly, making up a table (do you remember what columns are there?).
- While writing all this, think about what to take for? Chose? Record in the table! Well, now it’s simple: we make an equation and solve it. Yes, and finally - remember the "3P"!
- I've done everything? Well done! It turned out that the speed of the cyclist is km / h.
-"What color is your car?" - "She's beautiful!" Correct answers to the questions
Let's continue our conversation. So what is the speed of the first cyclist? km/h? I really hope you're not nodding in the affirmative right now!
Read the question carefully: "What is the speed of first cyclist?
Got what I mean?
Exactly! Received is not always the answer to the question!
Read the questions carefully - perhaps, after finding it, you will need to perform some more manipulations, for example, add km / h, as in our task.
Another point - often in tasks everything is indicated in hours, and the answer is asked to be expressed in minutes, or all the data is given in km, and the answer is asked to be written in meters.
Look at the dimension not only during the solution itself, but also when writing down the answers.
Tasks for movement in a circle
The bodies in the tasks may not necessarily move in a straight line, but also in a circle, for example, cyclists can ride along a circular track. Let's take a look at this problem.
Task #1
A cyclist left the point of the circular track. In minutes he had not yet returned to the checkpoint, and a motorcyclist followed him from the checkpoint. Minutes after departure, he caught up with the cyclist for the first time, and minutes after that he caught up with him for the second time.
Find the speed of the cyclist if the length of the track is km. Give your answer in km/h.
Solution of problem No. 1
Try to draw a picture for this problem and fill in the table for it. Here's what happened to me:
Between meetings, the cyclist traveled the distance, and the motorcyclist -.
But at the same time, the motorcyclist drove exactly one lap more, this can be seen from the figure:
I hope you understand that they didn't actually go in a spiral - the spiral just schematically shows that they go in a circle, passing the same points of the track several times.
Got it? Try to solve the following problems yourself:
Tasks for independent work:
- Two mo-to-tsik-li-hundreds start-to-tu-yut one-but-time-men-but in one-right-le-ni from two dia-met-ral-but pro-ty-in-po- false points of a circular route, the length of a swarm is equal to km. After how many minutes, mo-the-cycle-lists are equal for the first time, if the speed of one of them is by km / h more than the speed of the other th?
- From one point of the circle-howl of the highway, the length of some swarm is equal to km, at the same time, in one right-le-ni, there are two motorcyclists. The speed of the first motorcycle is km / h, and minutes after the start, he was ahead of the second motorcycle by one lap. Find the speed of the second motorcycle. Give your answer in km/h.
Solving problems for independent work:
- Let km / h be the speed of the first mo-to-cycle-li-hundred, then the speed of the second mo-to-cycle-li-hundred is km / h. Let the first time mo-the-cycle-lists be equal in hours. In order for mo-the-cycle-li-stas to be equal, the faster one must overcome them from the beginning distance, equal in lo-vi-not to the length of the route.
We get that the time is equal to hours = minutes.
- Let the speed of the second motorcycle be km/h. In an hour, the first motorcycle traveled a kilometer more than the second swarm, respectively, we get the equation:
The speed of the second motorcyclist is km/h.
Tasks for the course
Now that you're good at solving problems "on land", let's move on to the water and look at the scary problems associated with the current.
Imagine that you have a raft and you lower it into a lake. What is happening to him? Right. It stands because a lake, a pond, a puddle, after all, is stagnant water.
The current velocity in the lake is .
The raft will only move if you start rowing yourself. The speed he gains will be own speed of the raft. No matter where you swim - left, right, the raft will move at the same speed with which you row. It's clear? It's logical.
Now imagine that you are lowering the raft onto the river, turn away to take the rope ..., turn around, and he ... floated away ...
This happens because the river has a flow rate, which carries your raft in the direction of the current.
At the same time, its speed is equal to zero (you are standing in shock on the shore and not rowing) - it moves with the speed of the current.
Got it?
Then answer this question - "How fast will the raft float on the river if you sit and row?" Thinking?
Two options are possible here.
Option 1 - you go with the flow.
And then you swim at your own speed + the speed of the current. The current seems to help you move forward.
2nd option - t You are swimming against the current.
Hard? That's right, because the current is trying to "throw" you back. You make more and more efforts to swim at least meters, respectively, the speed with which you move is equal to your own speed - the speed of the current.
Let's say you need to swim a mile. When will you cover this distance faster? When will you move with the flow or against?
Let's solve the problem and check.
Let's add to our path data on the speed of the current - km/h and on the own speed of the raft - km/h. How much time will you spend moving with and against the current?
Of course, you easily coped with this task! Downstream - an hour, and against the current as much as an hour!
This is the whole essence of the tasks on flow with the flow.
Let's complicate the task a little.
Task #1
A boat with a motor sailed from point to point in an hour, and back in an hour.
Find the speed of the current if the speed of the boat in still water is km/h
Solution of problem No. 1
Let's denote the distance between the points as, and the speed of the current as.
| Path S | speed v, km/h |
time t, hours |
|
| A -> B (upstream) | 3 | ||
| B -> A (downstream) | 2 |
We see that the boat makes the same path, respectively:
What did we charge for?
Flow speed. Then this will be the answer :)
The speed of the current is km/h.
Task #2
The kayak went from point to point, located km away. After staying at point for an hour, the kayak set off and returned to point c.
Determine (in km/h) the own speed of the kayak if it is known that the speed of the river is km/h.
Solution of problem No. 2
So let's get started. Read the problem several times and draw a picture. I think you can easily solve this on your own.
Are all quantities expressed in the same form? No. The rest time is indicated both in hours and in minutes.
Converting this to hours:
hour minutes = h.
Now all quantities are expressed in one form. Let's start filling out the table and looking for what we'll take for.
Let be the own speed of the kayak. Then, the speed of the kayak downstream is equal, and against the current is equal.
Let's write this data, as well as the path (as you understand, it is the same) and the time expressed in terms of path and speed, in a table:
| Path S | speed v, km/h |
time t, hours |
|
| Against the stream | 26 | ||
| With the flow | 26 |
Let's calculate how much time the kayak spent on its trip:
Did she swim all hours? Rereading the task.
No, not all. She had a rest of an hour of minutes, respectively, from the hours we subtract the rest time, which we have already translated into hours:
h kayak really floated.
Let's bring all the terms to a common denominator:
We open the brackets and give like terms. Next, we solve the resulting quadratic equation.
With this, I think you can also handle it on your own. What answer did you get? I have km/h.
Summing up
ADVANCED LEVEL
Movement tasks. Examples
Consider examples with solutionsfor each type of task.
moving with the flow
One of the most simple tasks - tasks for the movement on the river. Their whole essence is as follows:
- if we move with the flow, the speed of the current is added to our speed;
- if we move against the current, the speed of the current is subtracted from our speed.
Example #1:
The boat sailed from point A to point B in hours and back in hours. Find the speed of the current if the speed of the boat in still water is km/h.
Solution #1:
Let's denote the distance between the points as AB, and the speed of the current as.
We will enter all the data from the condition in the table:
| Path S | speed v, km/h |
Time t, hours | |
| A -> B (upstream) | AB | 50s | 5 |
| B -> A (downstream) | AB | 50+x | 3 |
For each row of this table, you need to write the formula:
In fact, you don't have to write equations for each of the rows in the table. We see that the distance traveled by the boat back and forth is the same.
So we can equate the distance. To do this, we immediately use distance formula:
Often it is necessary to use formula for time:
Example #2:
A boat travels a distance in km against the current for an hour longer than with the current. Find the speed of the boat in still water if the speed of the current is km/h.
Solution #2:
Let's try to write an equation. The time upstream is one hour longer than the time downstream.
It is written like this:
Now, instead of each time, we substitute the formula:
We got the usual rational equation, we solve it:
Obviously the speed cannot be negative number so the answer is km/h.
Relative motion
If some bodies are moving relative to each other, it is often useful to calculate their relative speed. It is equal to:
- the sum of velocities if the bodies move towards each other;
- speed difference if the bodies are moving in the same direction.
Example #1
From points A and B, two cars left simultaneously towards each other with speeds of km/h and km/h. In how many minutes will they meet? If the distance between points is km?
I solution way:
Relative speed of cars km/h. This means that if we are sitting in the first car, it seems to be stationary, but the second car is approaching us at a speed of km/h. Since the distance between cars is initially km, the time after which the second car will pass the first:
Solution 2:
The time from the start of the movement to the meeting at the cars is obviously the same. Let's designate it. Then the first car drove the way, and the second -.
In total, they traveled all km. Means,
Other motion tasks
Example #1:
A car left point A for point B. Simultaneously with it, another car left, which traveled exactly half the way at a speed of km/h less than the first one, and the second half of the way it drove at a speed of km/h.
As a result, the cars arrived at point B at the same time.
Find the speed of the first car if it is known to be greater than km/h.
Solution #1:
To the left of the equal sign, we write the time of the first car, and to the right - the second:
Simplify the expression on the right side:
We divide each term by AB:
It turned out the usual rational equation. Solving it, we get two roots:
Of these, only one is larger.
Answer: km/h.
Example #2
A cyclist left point A of the circular track. After a few minutes, he had not yet returned to point A, and a motorcyclist followed him from point A. Minutes after departure, he caught up with the cyclist for the first time, and minutes after that he caught up with him for the second time. Find the speed of the cyclist if the length of the track is km. Give your answer in km/h.
Solution:
Here we will equate the distance.
Let the speed of the cyclist be, and the speed of the motorcyclist -. Until the moment of the first meeting, the cyclist was on the road for minutes, and the motorcyclist -.
In doing so, they traveled equal distances:
Between meetings, the cyclist traveled the distance, and the motorcyclist -. But at the same time, the motorcyclist drove exactly one lap more, this can be seen from the figure:
I hope you understand that they didn't actually go in a spiral - the spiral just schematically shows that they go in a circle, passing the same points of the track several times.
We solve the resulting equations in the system:
SUMMARY AND BASIC FORMULA
1. Basic formula
2. Relative motion
- This is the sum of the speeds if the bodies are moving towards each other;
- speed difference if the bodies are moving in the same direction.
3. Move with the flow:
- If we move with the current, the speed of the current is added to our speed;
- if we move against the current, the speed of the current is subtracted from the speed.
We have helped you deal with the tasks of movement...
Now it's your turn...
If you carefully read the text and solved all the examples yourself, we are ready to argue that you understood everything.
And this is already half way.
Write below in the comments if you figured out the tasks for movement?
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Do you understand that tasks for "work" are almost the same thing?
Write to us and good luck on your exams!
Tasks on movement in one direction belong to one of the three main types of tasks on movement.
Now we will talk about tasks in which objects have different speeds.
When moving in one direction, objects can both approach and move away.
Here we consider problems for movement in one direction, in which both objects leave the same point. Next time we will talk about the movement in pursuit, when objects move in the same direction from different points.
If two objects left the same point at the same time, then, since they have different speeds, the objects move away from each other.
To find the speed of removal, it is necessary to subtract the smaller from the greater speed:
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If one object left one point, and after some time another object left it in the same direction, then they can both approach and move away from each other.
If the speed of the object moving in front is less than the object moving after it, then the second catches up with the first and they approach each other.
To find the speed of approach, subtract the smaller speed from the larger one:
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If the speed of the object that goes ahead is greater than the speed of the object that moves behind, then the second one will not be able to catch up with the first one and they move away from each other.
We find the removal rate in the same way - subtract the smaller one from the larger one:
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Speed, time and distance are related:
Task 1.
Two cyclists left the same village in the same direction at the same time. The speed of one of them is 15 km/h, the speed of the other is 12 km/h. How far will they be in 4 hours?
Solution:
The condition of the problem is most conveniently written in the form of a table:
1) 15-12=3 (km/h) removal speed of cyclists
2) 3∙4=12 (km) this distance will be between cyclists after 4 hours.
Answer: 12 km.
A bus leaves from point A to point B. After 2 hours, a car left behind him. At what distance from point A will the car overtake the bus if the speed of the car is 80 km/h and the speed of the bus is 40 km/h?
1) 80-40=40 (km/h) vehicle and bus approach speed
2) 40∙2=80 (km) at this distance from point A there is a bus when the car leaves A
3) 80:40=2 (h) the time after which the car will overtake the bus
4) 80∙2=160 (km) the distance that the car will cover from point A
Answer: at a distance of 160 km.
Task 3
A pedestrian left the village and a cyclist left the station at the same time. After 2 hours, the cyclist was ahead of the pedestrian by 12 km. Find the speed of the pedestrian if the speed of the cyclist is 10 km/h.
Solution:
1) 12:2=6 (km/h) removal speed of cyclist and pedestrian
2) 10-6=4 (km/h) walking speed.
Answer: 4 km/h.
In the previous tasks for movement in one direction, the movement of bodies began simultaneously from the same point. Consider solving problems for movement in one direction, when the movement of bodies begins at the same time, but from different points.
Let a cyclist and a pedestrian depart from points A and B, the distance between which is 21 km, and go in the same direction: a pedestrian at a speed of 5 km per hour, a cyclist at 12 km per hour
12 km per hour 5 km per hour
A B
The distance between a cyclist and a pedestrian at the start of their movement is 21 km. For an hour of their joint movement in one direction, the distance between them will decrease by 12-5=7 (km). 7 km per hour - the speed of convergence of a cyclist and a pedestrian:
A B
Knowing the speed of approach of a cyclist and a pedestrian, it is easy to find out how many kilometers the distance between them will decrease after 2 hours, 3 hours of their movement in one direction.
7*2=14 (km) - the distance between the cyclist and the pedestrian will decrease by 14 km after 2 hours;
7*3=21 (km) - the distance between the cyclist and the pedestrian will decrease by 21 km after 3 hours.
Every hour the distance between the cyclist and the pedestrian decreases. After 3 hours, the distance between them becomes equal to 21-21=0, i.e. the cyclist overtakes the pedestrian:
A B
In tasks “to catch up” we deal with quantities:
1) the distance between the points from which the simultaneous movement begins;
2) approach speed
3) the time from the moment the movement begins to the moment when one of the moving bodies overtakes the other.
Knowing the value of two of these three quantities, you can find the value of the third quantity.
The table contains the conditions and solutions to problems that can be compiled to “catch up” with a pedestrian cyclist:
|
Approach speed of cyclist and pedestrian in km per hour |
Time from the start of the movement to the moment when the cyclist catches up with the pedestrian, in hours |
Distance from A to B in km | ||
We express the relationship between these quantities by the formula. Denote by the distance between the points and, - the speed of approach, the time from the moment of exit to the moment when one body catches up with another.
In catch-up problems, the convergence rate is most often not given, but it can be easily found from the problem data.
Task. A cyclist and a pedestrian left simultaneously in the same direction from two collective farms, the distance between which is 24 km. A cyclist was traveling at a speed of 11 km per hour, and a pedestrian was walking at a speed of 5 km per hour. In how many hours after his exit will the cyclist overtake the pedestrian?
To find how long after his exit the cyclist will catch up with the pedestrian, you need to divide the distance that was between them at the beginning of the movement by the speed of approach; the speed of approach is equal to the difference between the speeds of the cyclist and the pedestrian.
Solution formula: =24: (11-5);=4.
Answer. In 4 hours the cyclist will overtake the pedestrian. Conditions and solutions of inverse problems are written in the table:
|
The speed of the cyclist in km per hour |
Pedestrian speed in km per hour |
Distance between collective farms in km |
Time per hour | ||
Each of these tasks can be solved in other ways, but they will be irrational compared to these solutions.
