Motion problems in one direction are one of three main types of motion problems.
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 will consider tasks for movement in one direction, in which both objects leave the same point. Next time we will talk about chasing, 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 removal rate, subtract the lower rate from the higher rate:
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If one object left one point, and after a while another object followed 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 approach speed, subtract the lower one from the higher speed:
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If the speed of the object that is in front is greater than the speed of the object that is following, then the second will not be able to catch up with the first and they move away from each other.
We find the removal speed in the same way - subtract the lower one from the higher speed:
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Speed, time and distance are related:
Objective 1.
Two cyclists left one village in one direction at the same time. The speed of one of them is 15 km / h, the speed of the other is 12 km / h. What is the distance through them in 4 hours?
Decision:
The problem condition is most conveniently written in the form of a table:
1) 15-12 = 3 (km / h) cyclist removal speed
2) 3 ∙ 4 = 12 (km) this distance will be between cyclists in 4 hours.
Answer: 12 km.
A bus departed from point A to point B. After 2 hours a car drove out after him. At what distance from point A will the car overtake the bus if the car's speed is 80 km / h and the bus speed is 40 km / h?
1) 80-40 = 40 (km / h) the speed of convergence of a car and a bus
2) 40 ∙ 2 = 80 (km) at this distance from point A is the 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) distance that 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. After 2 hours, the cyclist was 12 km ahead of the pedestrian. Find the speed of the pedestrian if the speed of the cyclist is 10 km / h.
Decision:
1) 12: 2 = 6 (km / h) speed of removal of the cyclist and 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 can be for such a condition? That's right, two.
Why does this happen? I am sure that after all the examples, you can easily figure out how to display these formulas.
Understood? Well done! It's time to solve the problem.
The fourth task
Kolya is driving to work at a speed of km / h. Kolya's colleague Vova is driving at a speed of km / h. Kolya lives from Vova at a distance of km.
How long will it take for Vova to catch up with Kolya if they left the house at the same time?
Have you counted? 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 picture looks like this:
Does it look like yours? Well done!
Since the problem asks how long 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). Composing 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 we are dealing with the axis of movement.
Let's start with the path taken by Kolya. 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 made.
Based on these conclusions, we get the equation:
Understood? 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 this example gives you an idea of how important well-composed drawing!
And we are smoothly moving on, or rather, have already passed to the next point of our algorithm - bringing all values to the same dimension.
The rule of three "R" - dimension, rationality, calculation.
Dimension.
It is far from always that the tasks give the same dimension for each participant in the movement (as was the case in our easy tasks).
For example, you can find tasks where it is said that the bodies moved for a certain number of minutes, and their speed is indicated in km / h.
We cannot just take and substitute values in the formula - the answer will be wrong. Even in units of measurement, our answer "will not pass" the reasonableness test. Compare:
See? With correct multiplication, we also reduce units of measurement, and, accordingly, a reasonable and correct result is obtained.
What happens if we do not translate into one measurement system? Strange dimension of the answer and% incorrect result.
So, let me remind you, just in case, the values of the basic units of measurement of length and time.
Length units:
centimeter = millimeters
decimeter = centimeters = millimeters
meter = decimetres = centimeters = millimeters
kilometer = meters
Time units:
minute = seconds
hour = minutes = seconds
day = hours = minutes = seconds
Advice: When converting time units (minutes to hours, hours to seconds, etc.), imagine a clock face 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. hours, and a minute is an hour.
And now a very simple task:
Masha rode her bike from home to the village at a speed of km / h for minutes. What is the distance between the car's house and the village?
Have you counted? The correct answer is km.
minutes is an hour, and more minutes from an hour (I mentally imagined a clock face and said that minutes is a quarter of an hour), respectively - min = h.
Reasonableness.
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, rationality, that's about it)
Payment.
See if your solution "passes" for dimension and rationality, and only then check the calculations. It is logical - if there is an inconsistency with dimension and rationality, then it is easier to cross everything out and start looking for logical and mathematical errors.
"Love for tables" or "when drawing is not enough"
Movement problems are not always as simple as we solved earlier. Very often, in order to correctly solve a problem, you need not just draw a competent drawing, but also draw up 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 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.
Here is such a task. Pull yourself together and read it several times. Have you read it? Start drawing - straight line, point, point, two arrows ...
In general, draw, and now let's compare what you got.
It's kind of empty, isn't it? We draw a table.
As you remember, all movement tasks consist of components: speed, time and path... It is these graphs that any table will consist of in such tasks.
However, we will add one more column - name about whom we write information - motorcyclist and cyclist.
Also indicate in the cap dimension, in which you will enter the values there. You remember how important this is, right?
Have you got this table?
Now let's analyze everything that we have, and in parallel enter the data into the table and into the figure.
The first thing we have is the path that the cyclist and motorcyclist have done. It is the same and equal to km. We bring in!
Take the speed of the cyclist as, then the speed of the motorcyclist will be ...
If the solution to the problem does not work with such a variable, it's okay, we'll take another one until we reach the victorious one. It happens, the main thing is not to be nervous!
The table has changed. We have only one column left unfilled - time. How to find the time when there is a path and speed?
That's right, divide the path into speed. Put it on the table.
So our table has been filled in, now you can enter the data on the figure.
What can we reflect on it?
Well done. The speed of movement of the motorcyclist and cyclist.
Let's re-read the problem again, look at the figure and the completed table.
What data is not reflected in either the table or the figure?
Right. The time at which the motorcyclist arrived earlier than the cyclist. We know that the difference in time 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: drawing up and solving equations are manipulations that lead to the only correct answer.
So, as you already guessed, now we will make up the equation.
Equation drawing:
Take a look at your table, at the last condition that was not included in it and think, the relationship between what and what can we take 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 motorcyclist's travel time from his time, we will get the difference given to us.
This equation is rational. If you do not know what it is, read the topic "".
We bring the terms to a common denominator:
Let's open the brackets and give similar terms: Oof! Got it? Try your hand at the next challenge.
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. We decide!
We received two options for an answer. See what we got for? That's right, the speed of the cyclist.
We recall the rule "3P", more specifically "rationality". Do you understand what I mean? Exactly! The speed cannot be negative, therefore our answer is km / h.
Second task
Two cyclists set off for a mile-long run at the same time. The first was driving at a speed that is km / h higher than the speed of the second, and arrived at the finish line hours earlier than the second. Find the speed of the cyclist who finished second. Give your answer in km / h.
I remind the solution algorithm:
- Read the problem a couple of times - learn all the details. Got it?
- Start drawing a drawing - in which direction are they moving? how far did they go? Drew?
- Check if all your quantities are of the same dimension and start writing out briefly the condition of the problem, drawing up 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 an equation and solve it. Yes, and finally - remember about "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 posed
Let's continue our conversation. So what is the speed of the first cyclist? km / h? I really hope that you are not nodding in the affirmative right now!
Carefully read the question: “What is the speed of the first a cyclist? "
Do you understand what I mean?
Exactly! Received is not always the answer to the question posed!
Read the questions thoughtfully - 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.
Watch the dimension not only during the solution itself, but also when you write down the answers.
Circular tasks
Bodies in tasks may not necessarily move straight, but also in a circle, for example, cyclists can ride on a circular track. Let us examine such a problem.
Problem number 1
A cyclist left the point of the circular route. In minutes he had not yet returned to the point and a motorcyclist followed him from the point. Minutes after departure, he caught up with the cyclist for the first time, and minutes after that he caught up with him a second time.
Find the speed of the cyclist if the length of the track is km. Give your answer in km / h.
Solution to problem number 1
Try to draw a picture for this problem and fill in the table for it. Here's what I got:
Between the meetings, the cyclist traveled the distance, and the motorcyclist -.
But at the same time, the motorcyclist drove exactly one more lap, 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.
Understood? Try to solve the following tasks yourself:
Tasks for independent work:
- Two mo-to-cyc-li-a hundred start-to-go one-time-but in one on-right-ley out of two dia-metral-but pro-ti-in-po false points of a circular route, the length of which is equal to km. After how many minutes, mo-to-cycl-lis-sts will equalize for the first time, if the speed of one of them is greater than the speed of the other by km / h ho-ho?
- From one point of a steep track, the length of which is equal to km, one-n-time-but in one on-the-right-ley there are two motorcyclists. The speed of the first motorcycle is equal to km / h, and minutes after the start, he operated the second motorcycle for one lap. Nai-di-te speed of the second-ro-th motorcycle. Give your answer in km / h.
Solving problems for independent work:
- Let km / h be the speed of the first mo-to-cycle-leaf, then the speed of the second mo-to-cycle-leaf is equal to km / h. Let the first time my-that-cycl-lis-sts will be equal in hours. In order for the mo-to-tsik-lis-sts to be equal, the faster-wives must overcome from-chal-but-de-la-yu distance, equal to lo-vi-not the length of the route.
We get that the time is equal to 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, respectively, we get the equation:
The speed of the second rider is km / h.
Tasks for the course
Now that you are excellent at solving problems "on land", let's go into the water and look at the daunting problems associated with the current.
Imagine that you have a raft, and you lowered it into the lake. What's going on with him? Right. It stands because a lake, a pond, a puddle, after all, is stagnant water.
The speed of the current in the lake is .
The raft will only go if you start rowing yourself. The speed that he gains will be own speed of the raft. It doesn't matter where you sail - to the left, to the right, the raft will move as fast as you paddle. It's clear? It is logical.
Now imagine that you are lowering the raft onto the river, turning away to take the rope ..., turning, and he ... swam away ...
This is because the river has a current speed, 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 do not row) - it moves with the speed of the current.
Understood?
Then answer this question - "How fast will the raft float on the river if you are sitting and rowing?" Thinking?
There are two possibilities here.
Option 1 - you go with the flow.
And then you swim at your own speed + current speed. The flow, as it were, helps you to move forward.
2nd option - t You are swimming against the tide.
Heavy? Correct, because the current is trying to "throw" you back. You make more and more effort 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 km. When will you cover this distance faster? When will you go with the flow or against?
Let's solve the problem and check it out.
Add to our route 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 flow?
Of course, you easily coped with this task! Downstream - an hour, and upstream as much as an hour!
This is the whole essence of the tasks for movement with the flow.
Let's complicate the task a little.
Problem number 1
The boat with a motor sailed from point to point in an hour, and back - in an hour.
Find the current speed if the boat speed in still water is km / h
Solution to problem number 1
Let's 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 travels the same path, respectively:
What did we take for?
Current speed. Then this will be the answer :)
The current speed is equal to km / h.
Problem number 2
The kayak went from point to point located in km from. After staying at the point for an hour, the kayak went back and returned to point c.
Determine (in km / h) your own speed of the kayak if you know that the speed of the river is km / h.
Solution to problem number 2
So let's get started. Read the problem several times and draw a drawing. I think you can easily solve this on your own.
Are all values expressed in one form? Not. The rest time is indicated in both hours and minutes.
Let's translate this into hours:
hour minutes = h.
Now all values are expressed in one form. Let's start filling out the table and finding what we will take for.
Let be the kayak's own speed. Then, the speed of the kayak downstream is equal, and upstream is equal.
Let's write this data, as well as the path (it is, as you understand, the same) and the time, expressed in terms of the 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 has spent on its journey:
Did she swim all the hours? We reread the problem.
No, not all. She had a rest of an hour minutes, respectively, from the hours we subtract the rest time, which, we have already converted into hours:
h the kayak really floated.
Let us bring all the terms to a common denominator:
Let's expand the brackets and present similar terms. Next, we solve the resulting quadratic equation.
With this, I think you can handle it yourself. What answer did you get? I have km / h.
Let's summarize
ADVANCED LEVEL
Movement tasks. Examples of
Consider examples with solutionsfor each type of task.
Movement with the flow
Some of the simplest tasks are - river driving tasks... Their whole point is as follows:
- if we move with the current, the speed of the current is added to our speed;
- if we move against the current, the current velocity is subtracted from our speed.
Example # 1:
The boat sailed from point A to point B for hours and back - for hours. Find the current speed if the speed of the boat in still water is km / h.
Solution # 1:
Let's denote the distance between points as AB, and the speed of the current as.
We will enter 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 in the table. After all, 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, use immediately the formula for the distance:
You often have to use and formula for time:
Example # 2:
Against the current, the boat sails a distance in km for an hour longer than downstream. Find the speed of the boat in still water if the current speed is km / h.
Solution # 2:
Let's try to make an equation right away. The upstream time is one hour longer than the downstream time.
It is written like this:
Now, instead of each time, we substitute the formula:
We got the usual rational equation, let's solve it:
Obviously, the 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 the velocities, if the bodies are moving towards each other;
- the difference in velocities if the bodies are moving in the same direction.
Example # 1
Two cars drove out of points A and B at the same time towards each other at the speeds of km / h and km / h. In how many minutes they will meet. If the distance between points is km?
Solution I:
The relative speed of the vehicles is km / h. This means that if we are sitting in the first car, then it seems motionless to us, but the second car approaches us at a speed of km / h. Since the distance between the cars is initially km, the time after which the second car will pass the first:
Solution II:
The time from the start of the movement to the meeting of the cars is obviously the same. Let's designate it. Then the first car drove through the path, and the second -.
In total, they drove all the kilometers. Means,
Other traffic tasks
Example # 1:
A car drove from point A to point B. Simultaneously with him, another car drove out, which drove exactly half the way at a speed km / h less than the first, and the second half of the way it traveled 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 down the time of the first car, and to the right of the second:
Let's simplify the expression on the right side:
We divide each term by AB:
The result is the usual rational equation. Solving it, we get two roots:
Of these, only one is more.
Answer: km / h.
Example No. 2
A cyclist left point A of the circular track. In minutes he had not yet returned to point A and from point A a motorcyclist followed him. Minutes after departure, he caught up with the cyclist for the first time, and minutes after that he caught up with him a second time. Find the speed of the cyclist if the length of the track is km. Give your answer in km / h.
Decision:
Here we will equate distance.
Let the speed of the cyclist be, and 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 drove equal distances:
Between meetings, the cyclist traveled the distance, and the motorcyclist -. But at the same time, the motorcyclist drove exactly one more lap, 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 FORMULAS
1. Basic formula
2. Relative motion
- This is the sum of the velocities if the bodies are moving towards each other;
- the difference in velocities if the bodies are moving in the same direction.
3. Driving 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 current velocity is subtracted from the velocity.
We helped you deal with traffic problems ...
Now it's your turn ...
If you have carefully read the text and solved all the examples yourself, we are ready to argue that you understood everything.
And this is already half the way.
Write down in the comments if you figured out the tasks for the movement?
Which ones cause the greatest difficulties?
Do you understand that tasks for "work" are almost the same?
Write to us and good luck with your exams!
In the previous problems on movement in one direction, the movement of bodies began simultaneously from the same point. Let us consider the solution of problems on movement in one direction, when the movement of bodies begins simultaneously, but from different points.
Let a cyclist and a pedestrian leave points A and B, the distance between which is 21 km, and walk 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 moment of their start 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 - speed of convergence of a cyclist and a pedestrian:
A B
Knowing the speed of convergence of a cyclist and a pedestrian, it is easy to find out how many kilometers the distance between them will decrease in 2 hours, 3 hours of their movement in the same direction.
7 * 2 = 14 (km) - the distance between a cyclist and a 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.
The distance between the cyclist and the pedestrian decreases with each passing hour. After 3 hours, the distance between them becomes equal to 21-21 = 0, i.e. the cyclist will catch up with the pedestrian:
A B
In catch-up problems, we deal with the quantities:
1) the distance between the points from which the simultaneous movement begins;
2) the speed of convergence
3) the time from the moment the movement starts until the moment when one of the moving bodies overtakes the other.
Knowing the meaning of two of these three quantities, one can find the meaning of the third quantity.
The table contains the conditions and solutions of problems that can be compiled to “catch up” by a cyclist a pedestrian:
|
The speed of convergence of a cyclist and a pedestrian in km per hour |
Time from the moment of starting the movement until the moment when the cyclist overtakes the pedestrian, in hours |
Distance from A to B in km | ||
Let us express the relationship between these values by the formula. Let us 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 overtakes another.
In “catch-up” problems, the approach speed is most often not given, but it can be easily found from the problem data.
A task. The cyclist and the 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. How many hours after leaving the cyclist will 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. After 4 hours the cyclist will overtake the pedestrian. 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 tasks can be solved in other ways, but they will be irrational in comparison with these solutions.
§ 1 Formula of simultaneous movement
We come across formulas of simultaneous motion when solving problems of simultaneous motion. The ability to solve a particular movement problem depends on several factors. First of all, it is necessary to distinguish between the main types of tasks.
Tasks for simultaneous movement are conventionally divided into 4 types: tasks for oncoming movement, tasks for movement in opposite directions, tasks for movement in pursuit and tasks for movement with a lag.
The main components of these types of tasks are:
distance traveled - S, speed - ʋ, time - t.
The relationship between them is expressed by the formulas:
S = ʋ t, ʋ = S: t, t = S: ʋ.
In addition to the named main components, when solving problems on movement, we can encounter such components as: the speed of the first object - ʋ1, the speed of the second object - ʋ2, the speed of approach - ʋsbl., The speed of removal - sp., Meeting time - tintr., Initial distance - S0, etc.
§ 2 Tasks for oncoming traffic
When solving problems of this type, the following components are used: the speed of the first object - ʋ1; the speed of the second object is ʋ2; approach speed - ʋsbl .; time before meeting - tvstr .; path (distance) traveled by the first object - S1; the path (distance) traveled by the second object - S2; the entire path traveled by both objects - S.
The relationship between the components of oncoming traffic tasks is expressed by the following formulas:
1. The initial distance between objects can be calculated using the following formulas: S = ʋsbl. · Tvr. or S = S1 + S2;
2.the approach speed is found by the formulas: ʋsbl. = S: tintr. or ʋsbl. = ʋ1 + ʋ2;
3.The meeting time is calculated as follows:
Two ships sail towards each other. The speed of the motor ships is 35 km / h and 28 km / h. How long will it take for them to meet if the distance between them is 315 km?
ʋ1 = 35 km / h, ʋ2 = 28 km / h, S = 315 km, tintr. =? h
To find the time of the meeting, you need to know the initial distance and the speed of approach, since tvr. = S: ʋsbl. Since the distance is known by the problem statement, we find the approach speed. ʋsbl. = ʋ1 + ʋ2 = 35 + 28 = 63 km / h. Now we can find the required meeting time. tint = S: ʋsbl = 315: 63 = 5 hours. We got that the ships will meet in 5 hours.
§ 3 Tasks for the movement in pursuit
When solving problems of this type, the following components are used: the speed of the first object - ʋ1; the speed of the second object is ʋ2; approach speed - ʋsbl .; time before meeting - tvstr .; the path (distance) traveled by the first object - S1; the path (distance) traveled by the second object - S2; the initial distance between objects is S.
The scheme for tasks of this type is as follows:
The relationship between the components of the pursuit tasks is expressed by the following formulas:
1.The initial distance between objects can be calculated using the following formulas:
S = ʋsbl. Tintr. Or S = S1 - S2;
2.the approach speed is found by the formulas: ʋsbl. = S: tintr. or ʋsbl. = ʋ1 - ʋ2;
3.The meeting time is calculated as follows:
tint = S: sbl., Tintr. = S1: ʋ1 or tintr. = S2: ʋ2.
Let's consider the application of these formulas on the example of the following problem.
The tiger chased after the deer and caught up with him after 7 minutes. What is the initial distance between them if the speed of the tiger is 700 m / min and the speed of the deer is 620 m / min?
ʋ1 = 700 m / min, ʋ2 = 620 m / min, S =? m, tvstr. = 7 minutes
To find the initial distance between a tiger and a deer, it is necessary to know the time of meeting and the speed of convergence, since S = tstr. · Sbl. Since the meeting time is known from the problem statement, we find the approach speed. ʋsbl. = ʋ1 - ʋ2 = 700 - 620 = 80 m / min. Now we can find the desired initial distance. S = tintr. · Sbl = 7 · 80 = 560 m. We found that the initial distance between the tiger and the deer was 560 meters.
§ 4 Problems for movement in opposite directions
When solving problems of this type, the following components are used: the speed of the first object - ʋ1; the speed of the second object is ʋ2; removal rate - ud; travel time - t .; the path (distance) traveled by the first object - S1; the path (distance) traveled by the second object - S2; initial distance between objects - S0; the distance that will be between objects after a certain time is S.
The scheme for tasks of this type is as follows:
The relationship between the components of tasks for movement in opposite directions is expressed by the following formulas:
1.The final distance between objects can be calculated using the following formulas:
S = S0 + ʋsp. T or S = S1 + S2 + S0; and the initial distance - according to the formula: S0 = S - sp. · T.
2.The rate of removal is found by the formulas:
ʋud. = (S1 + S2): t orʋud. = ʋ1 + ʋ2;
3. Travel time is calculated as follows:
t = (S1 + S2): sp., t = S1: ʋ1 or t = S2: ʋ2.
Let's consider the application of these formulas on the example of the following problem.
Two cars left the car parks simultaneously in opposite directions. The speed of one is 70 km / h, the other is 50 km / h. What is the distance between them in 4 hours if the distance between the fleets is 45 km?
ʋ1 = 70 km / h, ʋ2 = 50 km / h, S0 = 45 km, S =? km, t = 4 h.
To find the distance between cars at the end of the path, you need to know the travel time, the initial distance and the speed of removal, since S = sp. · T + S0 Since the time and the initial distance are known from the problem statement, we find the removal rate. ʋud. = ʋ1 + ʋ2 = 70 + 50 = 120 km / h. Now we can find the required distance. S = ud. T + S0 = 120 4 + 45 = 525 km. We got that in 4 hours there will be a distance of 525 km between cars
§ 5 Tasks for movement with a lag
When solving problems of this type, the following components are used: the speed of the first object - ʋ1; the speed of the second object is ʋ2; removal rate - ud; travel time - t .; initial distance between objects - S0; the distance that will become between objects after a certain amount of time - S.
The scheme for tasks of this type is as follows:
The relationship between the components of lagging tasks is expressed by the following formulas:
1. The initial distance between objects can be calculated by the following formula: S0 = S - sp. · T; and the distance that will become between objects after a certain time - according to the formula: S = S0 + sp. · T;
2.The speed of removal is found by the formulas: sp. = (S - S0): t or ʋsp. = ʋ1 - ʋ2;
3. Time is calculated as follows: t = (S - S0): sp.
Let's consider the application of these formulas using the example of the following problem:
Two cars drove out of two cities in the same direction. The speed of the first is 80 km / h, the speed of the second is 60 km / h. In how many hours will there be 700 km between cars if the distance between cities is 560 km?
ʋ1 = 80 km / h, ʋ2 = 60 km / h, S = 700 km, S0 = 560 km, t =? h
To find the time, you need to know the initial distance between objects, the distance at the end of the path and the speed of removal, since t = (S - S0): sp. Since both distances are known by the condition of the problem, we find the speed of removal. ʋud. = ʋ1 - ʋ2 = 80 - 60 = 20 km / h. Now we can find the required time. t = (S - S0): ʋsp = (700 - 560): 20 = 7h. We got that in 7 hours there will be 700 km between cars.
§ 6 Brief summary on the topic of the lesson
With the simultaneous oncoming movement and movement in pursuit, the distance between two moving objects decreases (before the meeting). For a unit of time, it decreases by ʋsbl., And for the entire time of movement before the meeting, it will decrease by the initial distance S. Hence, in both cases, the initial distance is equal to the speed of approach multiplied by the time of movement before the meeting: S = ʋsbl. · Tvstr .. The only difference is that with oncoming traffic sbl. = ʋ1 + ʋ2, and when moving after ʋsbl. = ʋ1 - ʋ2.
When moving in opposite directions and with a lag, the distance between the objects increases, so the meeting will not occur. For a unit of time it increases by ud., And for the entire time of movement it will increase by the value of the product ʋud. · T. This means that in both cases the distance between the objects at the end of the path is equal to the sum of the initial distance and the product ʋsp · t. S = S0 + sp. T. The only difference is that with the opposite motion sp. = ʋ1 + ʋ2, and when moving with a lag ʋsp. = ʋ1 - ʋ2.
List of used literature:
- Peterson L.G. Mathematics. 4th grade. Part 2. / L.G. Peterson. - M .: Juventa, 2014 .-- 96 p.: Ill.
- Mathematics. 4th grade. Methodical recommendations for the textbook of mathematics "Learning to learn" for grade 4 / L.G. Peterson. - M .: Juventa, 2014 .-- 280 p.: Ill.
- Zak S.M. All tasks for the mathematics textbook for grade 4 L.G. Peterson and a set of independent and control works. FSES. - M .: YUNVES, 2014.
- CD-ROM. Mathematics. 4th grade. Scripts of lessons for the textbook for part 2 Peterson L.G. - M .: Juventa, 2013.
Images used:
We have many reasons to thank our God.
Have you noticed how in every year, actively and decisively, God's organization is accelerating its pace by providing many gifts!
The celestial chariot is definitely on the move! At the annual meeting, it was reported: "If you think that you are not able to keep up with Jehovah's chariot, buckle up so that you do not fly out at the bend!" :)
It can be seen how the discreet slave ensures continuous movement, opening new territories for preaching, preparing disciples, and gaining a more complete understanding of God's purposes.
Since the faithful slave relies not 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 can be seen that when the Governing Body sees a need to clarify an aspect of the truth or to make organizational changes, it acts without delay.
Isaiah 60:16 says that God's people will enjoy the milk of the nations, which is the leading 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 humans can triumph over Satan and his wicked system of things.
Compare the 2014, 2015, 2016 circulation and language counts of the December and January issues of The Watchtower and Awake magazines.
And below 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 million since January 2015 and is already being translated into 254 languages. On the front page of this magazine, there was also an outline for the ministry presentation.
Incredible! And they say that miracles do not happen! Such a circulation is a real miracle!
What our publications have!
Since August of last year (2014), the rating of our site has grown by 552 positions, thus improving by 30 percent.
For non-commercial sites, this is an unconditional record.A little more and we will be able to enter the top 1000 !!!
Why are they doing 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 get into the boats and get out of the ship.
Everyone seems to be doing good. But what kind of goodness makes sense in this situation? The answer is obvious! What's the point if someone is fed, clothed, and he will die anyway. First you need to transfer from the sinking ship and get to a safe place, and then feed and heat.
Jehovah's Witnesses do the same - they do good to people that makes sense.
While this material-centered world is wasting away from spiritual hunger, let us develop an appetite for spiritual food.
Let's not fall into the trap of materialism!
Paul pointed out THREE TIMES to whom and how we should show concern?
1Tm 2: 1 Prayers should be offered up “for people of all kinds”
1tm 2: 4 It is necessary "that people of all kinds ... come to an accurate knowledge of the truth."
1tm 2: 6 Christ "gave himself as a ransom for all"
What will help us to show deep concern for all and reach people of all kinds by preaching?
This requires one very important quality that Jehovah possesses - impartiality! ( Ac 10:34)
Truly, Jehovah is "impartial" (attitude) and "shows no respect for anyone" (actions)
Jesus preached to people of all kinds. Remember, in his examples, Jesus spoke of people of different backgrounds and social backgrounds: a farmer sowing seed, a housewife making bread, a man working in a field, a prosperous merchant who sells pearls, hard-working fishermen who cast their nets. (Mt 13: 31-33, 44-48)
Fact: Jehovah and Jesus want “people of all kinds to be saved” and receive eternal blessings. They don't put some people above others.
A lesson for us: To imitate Jehovah and Jesus, we need to preach to people of all kinds, regardless of race or life circumstances.
The Organization of God has already done a lot for those who speak a foreign language, immigrants, students, refugees, those who live in nursing homes, in gated complexes, entrepreneurs, prisoners, deaf, blind, adherents of non-Christian religions and others.
] At present, in Russia, under the supervision of a branch, 578 congregations are assigned to take care of the preaching of the good news in the correctional institutions that are assigned to them. Many of these locations have hosted congregation 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 sanctifying the name of God!
Therefore, let us appreciate everything that happens in God's organization. Let us learn to skillfully use the publications issued by the faithful slave, which are designed in such a way as to touch the hearts of people of all kinds. After all, how we teach ourselves, it will depend on how we teach others.
In this way, we will show that we show deep concern for the “coveted treasures of 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 in which we will keep up with Jehovah's chariot and will be under the protection of God the Creator, Jehovah (John 6:68).
