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 will easily figure out how to display these formulas yourself.
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 - I managed that Vova will catch up with Kolya in an hour or in a minute.
Let's compare our solutions ...
The picture looks like this:
Looks like yours? Well done!
Since the problem asks how many 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 in the figure as a segment. 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, a 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 \u200b\u200bhow 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 \u200b\u200bto 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 it was in our easy tasks).
For example, you can find tasks where it is said that the bodies moved a certain number of minutes, and their speed is indicated in km / h.
We cannot just take and substitute values \u200b\u200bin 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.
And what happens if we don't translate into one measurement system? Strange dimension of the answer and% incorrect result.
So, let me remind you, just in case, the values \u200b\u200bof the basic units of measurement of length and time.
Length units:
centimeter \u003d millimeters
decimeter \u003d centimeters \u003d millimeters
meter \u003d decimetres \u003d centimeters \u003d millimeters
kilometer \u003d meters
Time units:
minute \u003d seconds
hour \u003d minutes \u003d seconds
day \u003d hours \u003d minutes \u003d seconds
Advice: When converting units of measurement related to time (minutes to hours, hours to seconds, etc.), imagine a clock face in your head. The naked eye can see that 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 another minutes from an hour (I mentally imagined the dial of a clock and said that minutes is a quarter of an hour), respectively - min \u003d 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 out everything 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 solve a problem correctly, you need 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 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 - a straight line, point, point, two arrows ...
In general, draw, and now let's compare what you got.
Somehow empty, isn't it? 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 will consist in such tasks.
However, we will add one more column - nameabout whom we write information - motorcyclist and cyclist.
Also indicate in the cap dimension, in which you will enter the values \u200b\u200bthere. You remember how important this is, right?
Have you got this table?
Now let's analyze everything that we have, and simultaneously enter 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!
Let us take the speed of the cyclist as, then the speed of the motorcyclist will be ...
If with such variable solution the task will not work - it's okay, we'll take another 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 either in the table or in 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, 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 is manipulation that leads to the only correct answer.
So, as you 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?
Correctly. We can make an equation based on the time difference!
Is it logical? The cyclist rode more, if we subtract the rider's travel time from his time, we 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 similar terms: Phew! Got it? Try your hand at the next challenge.
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 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 came to the finish line 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 picture - 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! I found 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's the speed of the first cyclist? km / h? I really hope you are not nodding in the affirmative right now!
Read carefully the question: “What is the speed of the first 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.
One more point - often in problems 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 motion tasks
Bodies in tasks may not necessarily move in a straight line, 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 track. 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 did not actually go in a spiral - the spiral just schematically shows that they travel in a circle, passing the same points of the track several times.
Understood? Try to solve the following tasks yourself:
Self-study tasks:
- Two mo-to-cyc-li-a hundred start-to-eut one-time-but-in-one-right-ley out of two dia-metral-but pro-ty-in false points of a circular route, the length of which is equal to km. After how many minutes, the mo-to-cycl-lis-sts will be equal for the first time, if the speed of one of them is by km / h more hoo?
- From one point on 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 in 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-to-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 the chal-but raz-de-la-yu-them distance, equal to lo-vi-not the length of the route.
We get that the time is equal to hours \u003d 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 consider 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? Correctly. 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's 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 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.
Let's add data on the speed of the current - km / h and on the own speed of the raft - km / h to our path. How much time will you spend moving with and against the flow?
Of course, you have 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 designate the distance between points as, and the speed of the current as.
| Path S | Speed \u200b\u200bv, km / h |
Time t, hours |
|
| A -\u003e B (upstream) | 3 | ||
| B -\u003e 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 \u200b\u200bexpressed in one form? No. The rest time is indicated in both hours and minutes.
Let's translate this into hours:
hour minutes \u003d h.
Now all the quantities are expressed in one form. Let's start filling in 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, as you understand, is the same) and the time, expressed through the path and speed, in the table:
| Path S | Speed \u200b\u200bv, 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 for an hour minutes, respectively, from the hours we subtract the rest time, which, we have already translated into hours:
h the kayak really floated.
Let us bring all the terms to a common denominator:
Let us 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 sum up
ADVANCED LEVEL
Movement tasks. Examples of
Consider examples with solutions for each type of task.
Movement with the flow
Some of the most simple tasks - 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 enter all data from the condition into the table:
| Path S | Speed \u200b\u200bv, km / h |
Time t, hours | |
| A -\u003e B (upstream) | AB | 50-x | 5 |
| B -\u003e 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. We see that the distance traveled by the boat back and forth is the same.
This means that we can equate the distance. For this we use immediately formula for 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 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 objects are moving relative to each other, it is often useful to calculate their relative speed. It is equal to:
- the sum of the speeds, if the bodies are moving towards each other;
- the difference in speeds 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 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 vehicles, 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 passed the way, and the second -.
In total, they drove all the kilometers. Hence,
Other traffic tasks
Example # 1:
From point A to point B the car left. 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. Having solved 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 first meeting, the cyclist was on the road for minutes, and the motorcyclist -.
At the same time, they drove equal distances:
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.
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 speeds if the bodies are moving in the same direction.
3. Movement 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!
According to curriculum in mathematics, children should learn to solve movement problems even in primary school... However, tasks of this kind often cause difficulties for students. It is important that the child understands what his own speed, speed currents, speed downstream and speed against the stream. Only under this condition will the student be able to easily solve movement problems.
You will need
- Calculator, pen
Instructions
Own speed - this is speed boats or other means of transportation in still water. Designate it - V proper.
The water in the river is in motion. So she has her speed, which is called speedyu current (V flow)
The speed of the boat along the river, denote - V along the river, and speed upstream - V pr. flow.
Now remember the formulas required to solve motion problems:
V pr. Flow \u003d V proper. - V tech.
V in flow \u003d V proper + V current
So, based on these formulas, we can draw the following conclusions.
If the boat is moving against the stream of the river, then V proper. \u003d V pr. Flow. + V current
If the boat is moving with the current, then V proper. \u003d V on flow. - V tech.
Let's solve several problems on the movement along the river.
Task 1. The speed of the boat against the river flow is 12.1 km / h. Find your own speed boats, knowing that speed river flow 2 km / h.
Solution: 12.1 + 2 \u003d 14.1 (km / h) - own speed boats.
Task 2. The speed of the boat along the river is 16.3 km / h, speed river flow 1.9 km / h. How many meters would this boat go in 1 minute if it was in still water?
Solution: 16.3 - 1.9 \u003d 14.4 (km / h) - own speed boats. We translate km / h into m / min: 14.4 / 0.06 \u003d 240 (m / min.). This means that in 1 minute the boat would have covered 240 m.
Problem 3. Two boats set off simultaneously towards each other from two points. The first boat moved along the river, and the second - against the current. They met three hours later. During this time, the first boat covered 42 km, and the second - 39 km. speed each boat, if it is known that speed river flow 2 km / h.
Solution: 1) 42/3 \u003d 14 (km / h) - speed movement along the river of the first boat.
2) 39/3 \u003d 13 (km / h) - speed movement against the river flow of the second boat.
3) 14 - 2 \u003d 12 (km / h) - own speed the first boat.
4) 13 + 2 \u003d 15 (km / h) - own speed second boat.
Many people find it difficult to solve problems on "movement on water". There are several types of speeds in them, so the decisive ones start to get confused. To learn how to solve problems of this type, you need to know the definitions and formulas. The ability to draw up diagrams makes it very easy to understand the problem, contributes to the correct drawing up of the equation. And a well-formed equation is the most important thing in solving any type of problem.
Instructions
In the tasks "on the movement along the river" there are velocities: own speed (Vс), speed downstream (V downstream), speed upstream (Vpr. Flow), current speed (Vflow). It should be noted that the own speed of a watercraft is the speed in still water. To find the speed with the current, you need to add your own to the speed of the current. In order to find the speed against the current, you need to subtract the speed of the current from your own speed.
The first thing that you need to learn and know "by the teeth" - the formulas. Write down and remember:
Vin flow \u003d Vc + Vflow.
Vpr. flow \u003d Vc-V flow
Vpr. flow \u003d V flow - 2V leak.
Vreq. \u003d Vpr. flow + 2V
Vflow \u003d (Vflow - Vflow) / 2
Vs \u003d (Vcircuit + Vcr.) / 2 or Vc \u003d Vcr
Using an example, we will analyze how to find your own speed and solve problems of this type.
Example 1: The speed of the boat is 21.8 km / h downstream, and 17.2 km / h upstream. Find your own boat speed and the speed of the river.
Solution: According to the formulas: Vc \u003d (Vcr. + Vcr.
Vflow \u003d (21.8 - 17.2) / 2 \u003d 4.62 \u003d 2.3 (km / h)
Vc \u003d Vpr flow + Vflow \u003d 17.2 + 2.3 \u003d 19.5 (km / h)
Answer: Vc \u003d 19.5 (km / h), Vflow \u003d 2.3 (km / h).
Example 2. The steamer passed against the current for 24 km and returned back, spending 20 minutes less on the return journey than when moving against the current. Find its own speed in still water if the current speed is 3 km / h.
For X we will take the steamer's own speed. Let's create a table where we will enter all the data.
Against the flow. With the flow
Distance 24 24
Speed \u200b\u200bX-3 X + 3
time 24 / (X-3) 24 / (X + 3)
Knowing that the steamer spent 20 minutes less time on the return journey than on the way downstream, we will compose and solve the equation.
20 minutes \u003d 1/3 hours.
24 / (X-3) - 24 / (X + 3) \u003d 1/3
24 * 3 (X + 3) - (24 * 3 (X-3)) - ((X-3) (X + 3)) \u003d 0
72X + 216-72X + 216-X2 + 9 \u003d 0
X \u003d 21 (km / h) - own speed of the steamer.
Answer: 21 km / h.
note
The speed of the raft is considered equal to the speed of the reservoir.
Attention, only TODAY!
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This material is a system of tasks on the topic "Movement".
Purpose: to help students more fully master the technologies for solving problems on this topic.
Problems for movement on water.
Very often a person has to move on water: river, lake, sea.
At first he did it himself, then rafts, boats, sailing ships appeared. With the development of technology, steamships, motor ships, nuclear-powered ships came to the aid of man. And he was always interested in the length of the path and the time spent on its overcoming.
Let's imagine it's spring outside. The sun has melted the snow. Puddles appeared and streams ran. Let's make two paper boats and put one of them into a puddle, and the other into a stream. What will happen to each of the ships?
In a puddle the boat will stand still, and in a brook it will float, since the water in it "runs" to a lower place and carries it with it. The same will happen with a raft or boat.
In the lake they will stand still, and in the river they will swim.
Consider the first option: a puddle and a lake. The water in them does not move and is called standing.
The ship will float in a puddle only if we push it or if the wind blows. And the boat will start moving in the lake using oars or if it is equipped with a motor, that is, due to its speed. This movement is called movement in still water.
Is it different from driving on the road? The answer is no. This means that you and I know how to act in this case.
Problem 1. The speed of the boat on the lake is 16 km / h.
What the way will pass boat in 3 hours?
Answer: 48 km.
It should be remembered that the speed of the boat in still water is called own speed.
Problem 2. The motor boat sailed 60 km across the lake in 4 hours.
Find your own speed boat.
Answer: 15 km / h.
Problem 3. How long will it take for a boat whose own speed
is 28 km / h to swim 84 km on the lake?
Answer: 3 hours.
So, to find the distance traveled, you need to multiply the speed by time.
To find the speed, the path length must be divided by the time.
To find the time, the path length must be divided by the speed.
What is the difference between driving on a lake and driving on a river?
Let's remember a paper boat in a stream. He swam because the water moves in him.
This movement is called downstream... And in the opposite direction - upstream.
So, the water in the river moves, which means it has its own speed. And they call her river speed... (How to measure it?)
Problem 4. The speed of the river is 2 km / h. How many kilometers does the river carry
any object (sliver, raft, boat) in 1 hour, in 4 hours?
Answer: 2 km / h, 8 km / h.
Each of you swam in the river and remembers that it is much easier to swim with the current than against the current. Why? Because in one direction the river "helps" to swim, and in the other - it "interferes".
Those who do not know how to swim can imagine a situation when a strong wind is blowing. Consider two cases:
1) the wind blows in the back,
2) the wind blows in the face.
And in either case it is difficult to go. The wind in the back makes us run, which means that the speed of our movement increases. The wind in our face knocks us down, slows down. In this case, the speed decreases.
Let's dwell on the movement along the river. We have already talked about a paper boat in a spring stream. The water will carry it along with it. And the boat, launched into the water, will float at the speed of the current. But if it has its own speed, then it will float even faster.
Therefore, to find the speed of movement along the river, it is necessary to add the boat's own speed and the speed of the current.
Problem 5. The own speed of the boat is 21 km / h, and the speed of the river is 4 km / h. Find the speed of the boat along the river.
Answer: 25km / h.
Now let's imagine that the boat must sail against the current of the river. Without a motor, or at least a paddle, the current will carry her in the opposite direction. But, if you give the boat its own speed (start the engine or land the rower), the current will continue to push it back and prevent it from moving forward at its own speed.
therefore To find the speed of the boat against the current, it is necessary to subtract the speed of the current from its own speed.
Problem 6. The speed of the river is 3 km / h, and the boat's own speed is 17 km / h.
Find the speed of the boat upstream.
Answer: 14 km / h.
Problem 7. The own speed of the ship is 47.2 km / h, and the speed of the river is 4.7 km / h. Find the speed of the boat upstream and upstream.
Answer: 51.9 km / h; 42.5 km / h.
Problem 8. The speed of the motor boat downstream is 12.4 km / h. Find your own boat speed if the river speed is 2.8 km / h.
Answer: 9.6 km / h.
Problem 9. The speed of the boat against the current is 10.6 km / h. Find your own boat speed and downstream speed if the river speed is 2.7 km / h.
Answer: 13.3 km / h; 16 km / h.
The relationship between downstream speed and upstream speed.
Let us introduce the following notation:
V c. - own speed,
V tech. - current speed,
V on tech. - downstream speed,
V pr. Leak. - speed upstream.
Then you can write the following formulas:
V no flow \u003d V c + V flow;
V np. flow \u003d V c - V flow;
Let's try to depict this graphically:
Conclusion: the difference between the velocities upstream and upstream is equal to the doubled current velocity.
Vno tech - Vnp. flow \u003d 2 V flow.
Vflow \u003d (Vflow - Vnp.flow): 2
1) The speed of the boat against the current is 23 km / h, and the speed of the current is 4 km / h.
Find the speed of the boat downstream.
Answer: 31 km / h.
2) The speed of the motor boat along the river is 14 km / h / and the speed of the current is 3 km / h. Find the boat speed against the current
Answer: 8 km / h.
Task 10. Determine the speeds and fill in the table:
* - when solving clause 6, see Fig. 2.
Answer: 1) 15 and 9; 2) 2 and 21; 3) 4 and 28; 4) 13 and 9; 5) 23 and 28; 6) 38 and 4.
According to the mathematics curriculum, children are required to learn how to solve movement problems in their original school. However, tasks of this type often cause difficulties for students. It is important for the child to realize what his own speed , speed currents, speed downstream and speed contrary to the current. Only under this condition will the student be able to easily solve movement problems.
You will need
- Calculator, pen
Instructions
1. Own speed - this is speed boats or other means of transportation in static water. Designate it - V proper. The water in the river is in motion. So she has her speed which is called speed current (V current) Speed \u200b\u200bof the boat along the river, denote - V along the current, and speed opposite the current - V pr. flow.
2. Now remember the formulas needed to solve traffic problems: V pr. Flow \u003d V proper. - V current, V current \u003d V own. + V current
3. It turns out, based on these formulas, it is allowed to make the following results: If the boat moves against the river flow, then V proper. \u003d V pr. Flow. + V current. If the boat is moving with the current, then V proper. \u003d V on flow. - V tech.
4. Let's solve several problems on the movement along the river. Problem 1. The speed of the boat against the river flow is 12.1 km / h. Discover your own speed boats, knowing that speed river flow 2 km / h Solution: 12.1 + 2 \u003d 14.1 (km / h) - own speed boats. Task 2. The speed of the boat along the river is 16.3 km / h, speed river flow 1.9 km / h. How many meters would this boat go in 1 minute if it was in still water? Solution: 16.3 - 1.9 \u003d 14.4 (km / h) - own speed boats. We translate km / h into m / min: 14.4 / 0.06 \u003d 240 (m / min.). This means that in 1 minute the boat would have covered 240 m. Problem 3. Two boats set off at the same time opposite each other from 2 points. The 1st boat was moving along the river, and the 2nd - against the current. They met three hours later. During this time, the 1st boat covered 42 km, and the 2nd - 39 km. speed any boat, if it is known that speed river flow 2 km / h Solution: 1) 42/3 \u003d 14 (km / h) - speed movement along the river of the first boat. 2) 39/3 \u003d 13 (km / h) - speed movement against the river flow of the second boat. 3) 14 - 2 \u003d 12 (km / h) - own speed the first boat. 4) 13 + 2 \u003d 15 (km / h) - own speed second boat.
Movement problems seem difficult only at first glance. To discover, say, speed movement of the vessel in spite of currents , it is enough to imagine the situation expressed in the problem. Take your child on a small trip along the river, and the student will learn to “click puzzles like nuts”.
You will need
- Calculator, pen.
Instructions
1. According to the current encyclopedia (dic.academic.ru), speed is the collation of the translational motion of a point (body), which is numerically equal to the ratio of the distance traveled S to the intermediate time t in uniform motion, i.e. V \u003d S / t.
2. In order to detect the speed of movement of a vessel against the current, you need to know the own speed of the vessel and the speed of the current. Self speed is the speed of the vessel in still water, say, in a lake. Let's designate it - V proper. The speed of the current is determined by how far the river carries the object per unit of time. Let's denote it - V tech.
3. In order to find the speed of the vessel against the current (V pr. Flow), it is necessary to subtract the speed of the current from the vessel's own speed. It turns out that we got the formula: V pr. Flow \u003d V own. - V tech.
4. Let us find the speed of the vessel's movement contrary to the river flow, if it is known that the own speed of the vessel is 15.4 km / h, and the speed of the river flow is 3.2 km / h. 15.4 - 3.2 \u003d 12.2 (km / h ) - the speed of the vessel opposite the river flow.
5. In driving tasks, it is often required to convert km / h to m / s. In order to do this, it is necessary to remember that 1 km \u003d 1000 m, 1 h \u003d 3600 s. Consequently, x km / h \u003d x * 1000 m / 3600 s \u003d x / 3.6 m / s. It turns out that in order to convert km / h to m / s, you must divide by 3.6. Let's say 72 km / h \u003d 72: 3.6 \u003d 20 m / s. To convert m / s to km / h, you must multiply by 3, 6. Let's say 30 m / s \u003d 30 * 3.6 \u003d 108 km / h.
6. Let's translate x km / h into m / min. To do this, remember that 1 km \u003d 1000 m, 1 h \u003d 60 minutes. Hence, x km / h \u003d 1000 m / 60 min. \u003d x / 0.06 m / min. Consequently, in order to convert km / h to m / min. must be divided by 0.06. Let's say 12 km / h \u003d 200 m / min. to translate m / min. in km / h must be multiplied by 0.06, say 250 m / min. \u003d 15 km / h
Helpful advice
Do not forget about the units in which you measure the speed.
Note!
Do not forget about the units in which you measure the speed. To convert km / h to m / s, divide by 3.6. To convert m / s to km / h, multiply by 3.6. To convert km / h to m / min. must be divided by 0.06. In order to translate m / min. in km / h must be multiplied by 0.06.
Helpful advice
Drawing helps to solve the problem of movement.