Mathematics is a rather difficult subject, but absolutely everyone will have to pass it in the school course. Movement tasks are especially difficult for students. How to solve without problems and a lot of wasted time, we will consider in this article.
Note that if you practice, these tasks will not cause any difficulties. The solution process can be developed to automatism.
Varieties
What is meant by this type of task? These are quite simple and uncomplicated tasks, which include the following varieties:
- oncoming traffic;
- after;
- travel in the opposite direction;
- river traffic.
We propose to consider each option separately. Of course, we will analyze only on examples. But before we move on to the question of how to solve motion problems, it is worth introducing one formula that we will need when solving absolutely all tasks of this type.
Formula: S=Vt. A little explanation: S is the path, the letter Vdenotes the speed of movement, and the letter t denotes time. All quantities can be expressed through this formula. Accordingly, speed is equal to distance divided by time, and time is distance divided by speed.
Move forward
This is the most common type of task. To understand the essence of the solution, consider the following example. Condition: "Two friends on bicycles set off at the same time towards each other, while the path from one house to another is 100 km. What will be the distance after 120 minutes, if it is known that the speed of one is 20 km per hour, and the second is fifteen. " Let's move on to the question of how to solve the problem of oncoming traffic of cyclists.
To do this, we need to introduce another term: "rapprochement speed". In our example, it will be equal to 35 km per hour (20 km per hour + 15 km per hour). This will be the first step in solving the problem. Next, we multiply the approach speed by two, since they moved for two hours: 352=70 km. We have found the distance that the cyclists will approach in 120 minutes. The last action remains: 100-70=30 kilometers. With this calculation, we found the distance between cyclists. Answer: 30 km.
If you don't understand how to solve the oncoming traffic problem using the approach speed, then use one more option.
Second way
First we find the path traveled by the first cyclist: 202=40 kilometers. Now the path of the 2nd friend: fifteen times two, which equals thirty kilometers. Add updistance traveled by the first and second cyclist: 40+30=70 kilometers. We learned which path they covered together, so it remains to subtract the distance traveled from the entire path: 100-70=30 km. Answer: 30 km.
We have considered the first type of movement task. Now it’s clear how to solve them, let’s move on to the next view.
Movement in the opposite direction
Condition: "Two hares galloped out of the same hole in the opposite direction. The speed of the first is 40 km per hour, and the second is 45 km per hour. How far will they be apart in two hours?"
Here, as in the previous example, there are two possible solutions. In the first, we will act in the usual way:
- Path of the first hare: 402=80 km.
- The path of the second hare: 452=90 km.
- The path they traveled together: 80+90=170 km. Answer: 170 km.
But another option is possible.
Deletion speed
As you may have guessed, in this task, similarly to the first one, a new term will appear. Let's consider the following type of motion problem, how to solve them using the removal velocity.
We will find it first of all: 40+45=85 kilometers per hour. It remains to find out what is the distance separating them, since all other data are already known: 852=170 km. Answer: 170 km. We considered solving motion problems in the traditional way, as well as using the speed of approach and removal.
Following up
Let's look at an example of a problem and try to solve it together. Condition: "Two schoolchildren, Kirill and Anton, left the school and were moving at a speed of 50 meters per minute. Kostya followed them six minutes later at a speed of 80 meters per minute. How long will it take Kostya to catch up with Kirill and Anton?"
So, how to solve the problems of moving after? Here we need the speed of convergence. Only in this case it is worth not adding, but subtracting: 80-50 \u003d 30 m per minute. The second step is to find out how many meters separate the students before Kostya leaves. For this 506=300 meters. The last action is to find the time during which Kostya will catch up with Kirill and Anton. To do this, the path of 300 meters must be divided by the approach speed of 30 meters per minute: 300:30=10 minutes. Answer: in 10 minutes.
Conclusions
Based on what was said earlier, some conclusions can be drawn:
- when solving motion problems, it is convenient to use the speed of approach and removal;
- if we are talking about oncoming movement or movement from each other, then these values are found by adding the speeds of objects;
- if we have a task to move after, then we use the action, the reverse of addition, that is, subtraction.
We have considered some problems on movement, how to solve them, figured it out, got acquainted with the concepts of "speed of approach" and "speed of removal", it remains to consider the last point, namely: how to solve problems on movement along the river?
Current
Heremay occur again:
- tasks to move towards each other;
- moving after;
- travel in the opposite direction.
But unlike the previous tasks, the river has a current speed that should not be ignored. Here the objects will move either along the river - then this speed should be added to the own speed of the objects, or against the current - it must be subtracted from the speed of the object.
An example of a task for moving along a river
Condition: "The jet ski went downstream at a speed of 120 km per hour and returned back, while spending two hours less time than against the current. What is the speed of the jet ski in still water?" We are given a current speed of one kilometer per hour.
Let's move on to the solution. We propose to draw up a table for a good example. Let's take the speed of a motorcycle in still water as x, then the speed downstream is x + 1, and against x-1. The round trip distance is 120 km. It turns out that the time spent moving upstream is 120:(x-1), and downstream 120:(x+1). It is known that 120:(x-1) is two hours less than 120:(x+1). Now we can proceed to filling in the table.
| v | t | s | |
| downstream | x+1 | 120:(x+1) | 120 |
| against the current | x-1 | 120:(x-1) | 120 |
What we have:(120/(x-1))-2=120/(x+1) Multiply each part by (x+1)(x-1);
120(x+1)-2(x+1)(x-1)-120(x-1)=0;
Solving the equation:
(x^2)=121
Note that there are two possible answers here: +-11, since both -11 and +11 give 121 squared. But our answer will be positive, since the speed of a motorcycle cannot have a negative value, therefore, we can write down the answer: 11 km per hour. Thus, we have found the required value, namely the speed in still water.
We have considered all possible variants of tasks for movement, now you should not have any problems and difficulties when solving them. To solve them, you need to learn the basic formula and concepts such as "the speed of approach and removal." Be patient, work through these tasks, and success will come.
