We have many reasons to thank our God.
Have you noticed how every year, God's organization actively and decisively moves forward with a multitude of gifts!
The heavenly chariot is definitely on the move! At the annual meeting it was said: “If you feel like you can’t keep up with Jehovah’s chariot, buckle up so you don’t get thrown out at the turn!”:)
The prudent servant is seen to ensure continual progress, opening up new territories for preaching, making disciples, and gaining a fuller understanding of God's purposes.
Since the faithful servant does not rely on human strength, but on the guidance of the holy spirit, it is clear that the faithful servant is led by God's spirit!!!
It is evident that when the Governing Body sees a need to clarify any aspect of the truth or to make changes in organizational order, it acts without delay.
Isaiah 60:16 says that God's people will enjoy the milk of the nations, which is advanced technology today.
Today in the hands of the organizationa site that connects and unites us with our brotherhood, and other new products that you probably already know about.
It is only because God sustains and blesses them through his Son and the Messianic Kingdom that these imperfect people can achieve victory over Satan and his wicked system of things.
Compare the 2014, 2015, and 2016 editions of the December and January issues of The Watchtower and Awake.
And below is from 1962.
The Watchtower magazine is in blue and the Awake magazine is in red.
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The Watchtower's circulation has grown to 58,987,000 million since January 2015 and is already translated into 254 languages. On the first page of this magazine, there also appeared a plan for presentation in ministry.
Incredible! And they say that miracles don’t happen! This circulation is a real miracle!
What a success our publications have!
Since August last year (2014), our site's ranking has increased by 552 positions, thus improving by 30 percent.
This is an absolute record for non-commercial sites.A little more and we can enter the top 1000!!!
Why do they do this?
Imagine a sinking ship. There are, among other things, three groups of people.
The first ones are trying to feed the passengers.
The latter offer warm fur coats.
Still others help get into boats and get off the ship.
Everyone seems to be doing good. But what kind of good makes sense in this situation? The answer is obvious! What good is it if you feed and clothe someone, but 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 thing - they do good for people that makes sense.
As this materially focused world languishes with spiritual hunger, let us develop an appetite for spiritual food.
Let us not fall into the trap of materialism!
Paul pointed out THREE TIMES who and how we should show concern?
1Ti 2:1 Prayers should be offered “for people of all kinds”
1Тм 2:4 It is necessary “that men of every kind... come to an accurate knowledge of the truth.”
1Тм 2:6 Christ “gave himself as an adequate ransom for all”
What will help us to deeply care for everyone and reach all kinds of people with our preaching?
To do this, you need one very important quality that Jehovah possesses - impartiality! ( Ac 10:34)
Indeed, Jehovah is “no respecter of persons” (attitude) and “shows no partiality toward anyone” (deeds)
Jesus preached to all kinds of people. Remember, in his examples, Jesus spoke of people of different backgrounds and social status: about the farmer sowing seed, about the housewife making bread, about the man working in the field, about the successful merchant who sells pearls, about the hard-working fishermen who cast their nets. (Matthew 13:31-33, 44-48)
Fact: Jehovah and Jesus desire for “all kinds of people to be saved” and to receive eternal blessings. They don't put some people above others.
Lesson for us: To imitate Jehovah and Jesus, we need to preach to people of all kinds, regardless of their race or life circumstances.
God's organization has already done a lot for those who speak a foreign language, immigrants, students, refugees, those living in nursing homes, in gated communities, entrepreneurs, prisoners, the deaf, the blind, adherents of non-Christian religions and others.
]Currently in Russia, under the supervision of a branch of 578 congregations, they are assigned to take care of preaching the good news in the correctional institutions that are assigned to them. Many of these places hosted congregational meetings, group and personal Bible studies. Preaching in such places helps many to “put on a 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 skillfully use publications issued by a faithful servant, which are designed so as to touch the hearts of people of all kinds. After all, how we teach ourselves will determine how we teach others.
In this way we will show that we are deeply concerned about the “desired treasures from all nations” that still need to be brought.
Surely we, like Peter, have learned the lesson:
“we have nowhere to go” - there is only one place, being in which we will not lag behind the chariot of Jehovah and will be under the protection of God the Creator, Jehovah (John 6:68).
2. BODY SPEED. RIGHT LINEAR UNIFORM MOTION.
Speed is a quantitative characteristic of body movement.
average speed is a physical quantity equal to the ratio of the point’s displacement vector to the time period Δt during which this displacement occurred. The direction of the average speed vector coincides with the direction of the displacement vector. The average speed is determined by the formula:
Instantaneous speed, that is, the speed at a given moment in time is a physical quantity equal to the limit to which the average speed tends with an infinite decrease in the time period Δt:
In other words, instantaneous speed at a given moment in time is the ratio of a very small movement to a very short 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 a body travels a distance of one meter in one second. The unit of speed is indicated by m/s. Speed is often measured in other units. For example, when measuring the speed of a car, train, etc. The unit commonly used 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 the same question will not necessarily be in 5).
The velocities of body movement in different reference systems are connected by the classical law of addition of speeds.
Body speed relative fixed frame of reference equal to the sum of the body's velocities in moving reference system and the most mobile reference system relative to the stationary one.
For example, a passenger train moves along the railway 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 stationary and take it as a reference system, then the speed of a person relative to the reference system (that is, relative to the 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 a person and a 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 he will have to take this angle into account, remembering that speed is vector quantity.
The example + The law of addition of displacement is highlighted in red (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 moves along this road is moving frame of reference. The carriage on which the person is walking is part of the train.
The speed of a person relative to the carriage (relative to the moving frame of reference) is 5 km/h. Let's denote it by the letter H.
The speed of the train (and therefore the carriage) relative to a fixed frame of reference (that is, relative to the railway) is 60 km/h. Let's denote it by the letter B. In other words, the speed of the train is the speed of the moving reference frame relative to the stationary reference frame.
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 the letter .
Let us associate the XOY coordinate system with the stationary 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 stationary reference system, that is, relative to the railroad.
Over a short period of time Δt the following events occur:
Then, during this period of time, the movement of a person relative to the railway is:
This law of addition of displacements. In our example, the movement of a person relative to the railway is equal to the sum of the movements of the person relative to the carriage and the carriage relative to the railway.
Rice. 1.7. The law of addition of displacements.
The law of addition of displacements can be written as follows:
= Δ H Δt + Δ B Δt
The speed of a person relative to the railway is:
Speed of a person relative to the carriage:
Δ H = H / Δt
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 = const) and acceleration or deceleration does not occur (a = 0).
Straight-line movement- this is movement in a straight line, that is, the trajectory of rectilinear movement is a straight line.
Uniform linear movement- this is a movement in which a body makes equal movements at any equal intervals of time. For example, if we divide a certain time interval into one-second intervals, then with uniform motion the body will move the same distance for each of these time intervals.
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 movement of a body over any period of time to the value of this interval t:
Thus, the speed of uniform rectilinear motion shows how much movement a material point makes per unit time.
Moving with uniform linear motion is determined by the formula:
Distance traveled in linear motion is equal to the displacement module. If the positive direction of the OX axis coincides with the direction of movement, then the projection of the velocity onto the OX axis is equal to the magnitude of the velocity and is positive:
v x = v, that is v > 0
The projection of displacement onto the OX axis is equal to:
s = vt = 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)
Equation of motion, that is, the dependence of the body coordinates 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’s velocity onto the OX axis is negative, the speed is less than zero (v< 0), и тогда уравнение движения принимает вид.
Problems involving motion in one direction refer to one of three main types of motion problems.
Now we will talk about problems in which objects have different speeds.
When moving in one direction, objects can both come closer and move away.
Here we consider problems involving movement in one direction, in which both objects leave the same point. Next time we will talk about catch-up movement, when objects move in the same direction from different points.
If two objects leave the same point at the same time, then since they have different speeds, the objects move away from each other.
To find the removal rate, you need to subtract the smaller one from the larger speed:
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If one object leaves one point, and after some time another object leaves in the same direction after it, then they can both approach and move away from each other.
If the speed of an object moving in front is less than the object moving behind it, then the second one catches up with the first one and they get closer.
To find the closing speed, you need to subtract the smaller from the higher speed:
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If the speed of the object that is moving ahead is greater than the speed of the object that is moving behind, then the second one will not be able to catch up with the first one and they will move away from each other.
We find the removal rate in the same way - subtract the smaller one from the higher speed:
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Speed, time and distance are related:
Task 1.
Two cyclists left the same village at the same time in the same direction. The speed of one of them is 15 km/h, the speed of the other is 12 km/h. What distance will be through them after 4 hours?
Solution:
It is most convenient to write the problem conditions in the form of a table:
1) 15-12=3 (km/h) speed of removal of cyclists
2) 3∙4=12 (km) this distance will be between cyclists in 4 hours.
Answer: 12 km.
A bus leaves from point A to point B. 2 hours later a car followed him. At what distance from point A will the car catch up with 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) speed of approach of a car and a bus
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) time after which the car will catch up with the bus
4) 80∙2=160 (km) the distance the car will travel from point A
Answer: at a distance of 160 km.
Problem 3
A pedestrian and a cyclist left the village at the same time at the station. After 2 hours, the cyclist was 12 km ahead of the pedestrian. Find the pedestrian's speed if the cyclist's speed is 10 km/h.
Solution:
1) 12:2=6 (km/h) speed of removal of a cyclist and a pedestrian
2) 10-6=4 (km/h) pedestrian speed.
Answer: 4 km/h.
So, let's say our bodies are moving in the same direction. How many cases do you think there could be for such a condition? That's right, two.
Why does this happen? 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.
Fourth task
Kolya goes to work by car at a speed of km/h. Colleague Kolya Vova is driving at a speed of km/h. Kolya lives kilometers away from Vova.
How long will it take for Vova to catch up with 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 an hour or in minutes.
Let's compare our solutions...
The drawing looks like this:
Similar to yours? Well done!
Since the problem asks how long after the guys met, and they 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). When composing the equations, let's take time for.
So, Vova made his way to the meeting place. Kolya made his way to the meeting place. It's clear. Now let's look at the axis of movement.
Let's start with the path that Kolya took. Its path () is shown in the figure as a segment. 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 took.
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 from this example you understand how important the role is played Well done drawing!
And we smoothly move on, or rather, we have already moved on to the next point of our algorithm - bringing all quantities to the same dimension.
The rule of three "R" - dimension, reasonableness, calculation.
Dimension.
Problems do not always give the same dimension for each participant in the movement (as was the case in our easy problems).
For example, you can find problems where it is said that bodies moved for a certain number of minutes, and their speed of movement is indicated in km/h.
We can’t just take and substitute the values into the formula - the answer will be incorrect. Even in terms of units of measurement, our answer “fails” the reasonableness test. Compare:
Do you see? When multiplying correctly, we also reduce the units of measurement, and, accordingly, we obtain a reasonable and correct result.
What happens if we don’t convert to one measurement system? The answer has a strange dimension and the result is % incorrect.
So, just in case, let me remind you of the meanings of the basic units of length and time.
Length units:
centimeter = millimeters
decimeter = centimeters = millimeters
meter = decimeters = centimeters = millimeters
kilometer = meters
Time units:
minute = seconds
hour = minutes = seconds
day = hours = minutes = seconds
Advice: When converting units of measurement related to time (minutes into hours, hours into seconds, etc.), imagine a clock dial in your head. The naked eye can see that the minutes are a quarter of the dial, i.e. hours, minutes is a third of the dial, i.e. an hour, 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 minutes from an hour (mentally imagined a clock dial, and said that minutes is a quarter of an hour), respectively - min = hours.
Reasonableness.
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 can’t be negative, right? So, rationality, that’s what it’s about)
Calculation.
See if your solution “passes” the dimensions and reasonableness, and only then check the calculations. It is logical - if there is an inconsistency with dimension and rationality, then it is easier to cross out everything and start looking for logical and mathematical errors.
“Love of tables” or “when drawing is not enough”
Movement problems are not always as simple as we solved before. Very often, in order to solve a problem correctly, you need not just draw a competent picture, but also make a table with all the conditions given to us.
First task
A cyclist and a motorcyclist left at the same time from point to point, the distance between them being kilometers. It is known that a motorcyclist travels more kilometers per hour than a cyclist.
Determine the speed of the cyclist if it is known that he arrived at the point minutes later than the motorcyclist.
This is the task. Pull yourself together and read it several times. Have you read it? Start drawing - a straight line, a point, a point, two arrows...
In general, draw, and now we’ll compare what you got.
It's a bit empty, isn't it? Let's draw a table.
As you remember, all movement tasks consist of the following components: speed, time and path. It is these columns that any table in such problems will consist of.
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 there. You remember how important this is, right?
Did you get a table like this?
Now let's analyze everything we have and at the same time enter the data into the table and figure.
The first thing we have is the path that the cyclist and motorcyclist took. It is the same and equal to km. Let's bring it in!
Let's take the speed of the cyclist as, then the speed of the motorcyclist will be...
If with such a variable the solution to the problem does not work, it’s okay, we’ll take another one until we reach the winning one. This happens, the main thing is not to be nervous!
The table has changed. We only have one column left unfilled - time. How to find time when there is a path and speed?
That's right, divide the distance by the speed. Enter this into the table.
Now our table is filled in, now we can enter the data into the drawing.
What can we reflect on it?
Well done. Speed of movement of motorcyclist and cyclist.
Let's re-read the problem again, look at the picture and the completed table.
What data is not reflected in the table or figure?
Right. The time the motorcyclist arrived before the cyclist. We know that the time difference is minutes.
What should we do next? That’s right, convert the time given to us from minutes to hours, because the speed is given to us in km/h.
The magic of formulas: drawing up and solving equations - manipulations leading to the only correct answer.
So, as you may have guessed, now we will make up the equation.
Drawing up the equation:
Look at your table, at the last condition that is not included in it and think, the relationship between what and what can we put into the equation?
Right. We can create an equation based on the time difference!
Logical? The cyclist rode more; if we subtract the motorcyclist’s time from his time, we will get the difference given to us.
This equation is rational. If you don’t know what this is, read the topic “”.
We bring the terms to a common denominator:
Let's open the brackets and present similar terms: Phew! Got it? Try your hand at the following problem.
Solution of the equation:
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. Let's decide!
We received two possible answers. Let's see what we got for? That's right, the speed of the cyclist.
Let us remember the “3P” rule, more specifically “reasonableness”. Do you know what I mean? Exactly! Speed cannot be negative, so our answer is km/h.
Second task
Two cyclists set out on a -kilometer ride at the same time. The first one drove at a speed that was one 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 second to the finish line. Give your answer in km/h.
Let me remind you of the solution algorithm:
- Read the problem a couple of times and understand all the details. Got it?
- Start drawing a picture - in which direction are they moving? how far did they travel? Did you draw it?
- Check that all your quantities are of the same dimension and begin to briefly write out the conditions of the problem, making a table (do you remember what graphs are there?).
- While you are writing all this, think about what to take for? Have you chosen? Write it down in the table! Well, now it’s simple: we make up an equation and solve. Yes, and finally - remember the “3Rs”!
- I've done everything? Well done! I found out that the speed of the cyclist is km/h.
-"What color is your car?" - "She's beautiful!" Correct answers to the questions asked
Let's continue our conversation. So what is the speed of the first cyclist? km/h? I really hope you’re not nodding yes now!
Read the question carefully: “What is the speed of first cyclist?
Do you understand what I mean?
Exactly! Received is not always the answer to the question posed!
Read the questions carefully - perhaps after finding them you will need to perform some more manipulations, for example, add km/h, as in our task.
One more 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.
Watch the dimensions not only during the solution itself, but also when writing down the answers.
Circular movement problems
Bodies in problems can move not necessarily straight, but also in a circle, for example, cyclists can ride along a circular track. Let's look at this problem.
Task No. 1
A cyclist left a point on the circular route. Minutes later, he had not yet returned to the point and the motorcyclist left the point after him. Minutes after leaving, 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 route is km. Give your answer in km/h.
Solution to problem No. 1
Try to draw a picture for this problem and fill out a table for it. Here's what I got:
Between meetings, the cyclist traveled a distance, and the motorcyclist - .
But at the same time, the motorcyclist drove exactly one lap more, as can be seen from the figure:
I hope you understand that they didn't actually drive in a spiral - the spiral just schematically shows that they drive in a circle, passing the same points on the route several times.
Got it? Try to solve the following problems yourself:
Tasks for independent work:
- Two motorcycles start at the same time in one direction of the two dia-metral-but-pro-ti-on- false points of a circular route, the length of which is equal to km. After how many minutes do the cycles become equal for the first time, if the speed of one of them is km/h higher than the speed of the other? ho-ho?
- From one point on a circular highway, the length of which is equal to km, at one time there are two motorcyclists in the same direction. The speed of the first motorcycle is equal to km/h, and minutes after the start it was ahead of the second motorcycle by one lap. Find the speed of the second motorcycle. Give your answer in km/h.
Solutions to problems for independent work:
- Let km/h be the speed of the first motor cycle, then the speed of the second motor cycle is equal to km/h. Let the cycles be equal for the first time in a few hours. In order for the cycles to be equal, the faster one must overcome them from the beginning distance equal to the length of the route.
We get that the time is hours = minutes.
- Let the speed of the second motorcycle be equal to km/h. In an hour, the first motorcycle traveled more kilometers than the second, so we get the equation:
The speed of the second motorcyclist is km/h.
Current problems
Now that you are excellent at solving problems “on land,” let’s move into the water and look at the scary problems associated with the current.
Imagine that you have a raft and you lower it into the lake. What's happening to him? Right. It stands because a lake, a pond, a puddle, after all, is still water.
The current speed in the lake is .
The raft will only move if you start rowing yourself. The speed it acquires will be the raft's own speed. It doesn’t matter where you swim - left, right, the raft will move at the speed with which you row. It's clear? It's logical.
Now imagine that you are lowering a raft onto the river, you turn away to take the rope..., you turn around, and it... floats away...
This happens because the river has a current speed, which carries your raft in the direction of the current.
Its speed is zero (you are standing in shock on the shore and not rowing) - it moves at the speed of the current.
Got it?
Then answer this question: “At what speed will the raft float down the river if you sit and row?” Thinking about it?
There are two possible options here.
Option 1 - you go with the flow.
And then you swim at your own speed + the speed of the current. The flow 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 are making more and more efforts to swim at least meters, respectively, the speed at which you move is equal to your own speed - the speed of the current.
Let's say you need to swim a kilometer. When will you cover this distance faster? When will you go with the flow or against it?
Let's solve the problem and check.
Let's add to our path data on the speed of the current - km/h and the raft's own speed - km/h. How much time will you spend moving with and against the current?
Of course, you coped with this task without difficulty! It takes an hour with the current, and an hour against the current!
This is the whole essence of the tasks at movement with the current.
Let's complicate the task a little.
Task No. 1
The boat with the motor took an hour to travel from point to point, and an hour to return.
Find the speed of the current if the speed of the boat in still water is km/h
Solution to problem No. 1
Let us denote the distance between 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 takes the same path, respectively:
What did we charge for?
Current speed. Then this will be the answer :)
The speed of the current is km/h.
Task No. 2
The kayak left from point to point located km from. After staying at point for an hour, the kayak went back and returned to point c.
Determine (in km/h) the kayak's own speed if it is known that the speed of the river is km/h.
Solution to problem No. 2
So let's get started. Read the problem several times and make a drawing. I think you can easily solve this on your own.
Are all quantities expressed in the same form? No. Our rest time is indicated in both hours and minutes.
Let's convert this into hours:
hour minutes = h.
Now all quantities are expressed in one form. Let's start filling out the table and finding what we'll take for.
Let be the kayak's own speed. Then, the speed of the kayak downstream is equal and against the current is equal.
Let's write down this data, as well as the path (as you understand, it is the same) and 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 journey:
Did she swim for all the hours? Let's reread the task.
No, not all. She had an hour of rest, so from hours we subtract the rest time, which we have already converted into hours:
h the kayak really floated.
Let's bring all the terms to a common denominator:
Let's open the brackets and present similar terms. Next, we solve the resulting quadratic equation.
I think you can handle this on your own too. What answer did you get? I have km/h.
Let's sum it up
ADVANCED LEVEL
Movement tasks. Examples
Let's consider examples with solutionsfor each type of task.
Moving with the flow
Some of the simplest tasks are river navigation problems. 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 again in hours. Find the speed of the current if the speed of the boat in still water is km/h.
Solution #1:
Let us denote the distance between points as AB, and the speed of the current as.
Let's put all the data from the condition into the table:
| Path S | Speed v, km/h |
Time t, hours | |
| A -> B (upstream) | AB | 50-x | 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 row of the table. We see that the distance traveled by the boat back and forth is the same.
This means that we can equate the distance. To do this, we use immediately formula for distance:
Often you have to use formula for time:
Example #2:
A boat travels a distance of kilometers against the current 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 create an equation right away. The time upstream is an hour longer than the time downstream.
It is written like this:
Now, instead of each time, let’s substitute the formula:
We have received an ordinary rational equation, let’s solve it:
Obviously, speed cannot be a 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 bodies move towards each other;
- speed differences if bodies move in the same direction.
Example No. 1
Two cars left points A and B simultaneously towards each other at speeds km/h and km/h. In how many minutes will they meet? If the distance between points is km?
I solution method:
Relative speed of cars km/h. This means that if we are sitting in the first car, it seems motionless to us, but the second car is approaching us at a speed of km/h. Since the distance between the cars is initially km, the time it will take for the second car to pass the first:
Method II:
The time from the start of movement to the meeting of the cars is obviously the same. Let's designate it. Then the first car drove the path, and the second - .
In total they covered all the kilometers. Means,
Other movement tasks
Example #1:
A car left point A to point B. At the same time, another car left with him, which drove exactly half the way at a speed of km/h less than the first, and drove the second half of the way 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 that it is greater than km/h.
Solution #1:
To the left of the equal sign we write down the time of the first car, and to the right - of the second:
Let's simplify the expression on the right side:
Let's divide each term by AB:
The result is an ordinary rational equation. Having solved it, we get two roots:
Of these, only one is larger.
Answer: km/h.
Example No. 2
A cyclist left point A of the circular route. Minutes later, he had not yet returned to point A, and a motorcyclist followed him from point A. Minutes after leaving, 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 route 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 - .
At the same time, they traveled equal distances:
Between meetings, the cyclist traveled a distance, and the motorcyclist - . But at the same time, the motorcyclist drove exactly one lap more, as can be seen from the figure:
I hope you understand that they didn’t actually drive in a spiral; the spiral just schematically shows that they drive in a circle, passing the same points on the route several times.
We solve the resulting equations in the system:
SUMMARY AND BASIC FORMULAS
1. Basic formula
2. Relative motion
- This is the sum of speeds if the bodies move towards each other;
- difference in speed if bodies move in the same direction.
3. Moving 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.
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In previous tasks involving movement in one direction, the movement of bodies began simultaneously from the same point. Let's consider solving problems on movement in one direction, when the movement of bodies begins simultaneously, but from different points.
Let a cyclist and a pedestrian emerge from points A and B, the distance between which is 21 km, and go in the same direction: the pedestrian at a speed of 5 km per hour, the 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 moment they begin to move is 21 km. In an hour of their joint movement in one direction, the distance between them will decrease by 12-5=7 (km). 7 km per hour – speed of approach of a cyclist and a pedestrian:
A B
Knowing the speed of convergence of a cyclist and a pedestrian, it is not difficult to find out how many kilometers the distance between them will decrease after 2 hours or 3 hours of their movement in one direction.
7*2=14 (km) – the distance between a cyclist and a pedestrian will decrease by 14 km in 2 hours;
7*3=21 (km) – the distance between a cyclist and a pedestrian will decrease by 21 km in 3 hours.
With each passing hour, the distance between a cyclist and a pedestrian decreases. After 3 hours, the distance between them becomes 21-21=0, i.e. a cyclist catches up with a pedestrian:
A B
In “catch-up” problems we deal with the following quantities:
1) the distance between points from which simultaneous movement begins;
2) speed of approach
3) the time from the moment the movement begins until the moment when one of the moving bodies catches up with the other.
Knowing the value of two of these three quantities, you can find the value of the third quantity.
The table contains conditions and solutions to problems that can be drawn up for a cyclist to “catch up” with a pedestrian:
|
Closing speed of a cyclist and a pedestrian in km per hour |
Time from the moment the movement begins until the moment the cyclist catches up with the pedestrian, in hours |
Distance from A to B in km | ||
Let us express the relationship between these quantities by the formula. Let us denote by the distance between points and, - the speed of approach, the time from the moment of exit to the moment when one body catches up with the other.
In “catch-up” tasks, the speed of approach is most often not given, but it can be easily found from the task data.
Task. A cyclist and a pedestrian left simultaneously in the same direction from two collective farms, the distance between which was 24 km. The cyclist was traveling at a speed of 11 km per hour, and the pedestrian was walking at a speed of 5 km per hour. How many hours after leaving will the cyclist catch up with the pedestrian?
To find how long after leaving 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 approach speed is equal to the difference in speed between the cyclist and the pedestrian.
Solution formula: =24: (11-5);=4.
Answer. After 4 hours the cyclist will catch up with the pedestrian. The conditions and solutions of inverse problems are written in the table:
|
Cyclist speed in km per hour |
Pedestrian speed in km per hour |
Distance between collective farms in km |
Time per hour | ||
Each of these problems can be solved in other ways, but they will be irrational in comparison with these solutions.