How to solve the problem of motion? Methods of solving traffic problems

Mathematics is a rather complex subject, but in the school course it will have to pass absolutely everything. Particular difficulty in the students cause traffic problems. How to solve without problems and a lot of time spent, we will consider in this article.

Note that if you practice, these tasks will not cause any difficulties. The decision process can be worked out automatically.

Varieties

What do you mean by this type of task? These are rather simple and simple tasks, which include the following varieties:

• oncoming traffic;
• In pursuit;
• Movement in the opposite direction;
• Movement on the river.

We suggest to consider each variant separately. Of course, we will only analyze the examples. But before we proceed to the question of how to solve the problem of motion, it is necessary to introduce one formula that will be necessary for us in solving absolutely all tasks of this type.

Formula: S = V * t. A few explanations: S is the path, the letter V denotes the speed of movement, and the letter t means time. All quantities can be expressed in terms of this formula. Accordingly, the speed is equal to the path divided by time, and time is the path divided by the speed.

Moving towards

This is the most common type of task. To understand the essence of the solution, consider the following example. Condition: "Two friends on bicycles went simultaneously to meet each other, the path from one house to another is 100 km.What will be the distance in 120 minutes if it is known that the speed of one is 20 km per hour and the second is fifteen." We now turn to the question of how to solve the problem of the cyclists' oncoming traffic.

To do this, we need to enter one more term: "speed of rapprochement." 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 action in solving the problem. Further, multiply the rate of convergence by two, since they moved two hours: 35 * 2 = 70 km. We found the distance to which the cyclists will approach in 120 minutes. The last action remains: 100-70 = 30 kilometers. By this calculation, we found the distance between cyclists. Answer: 30 km.

If you do not understand how to solve the problem on the oncoming traffic, using the approach speed, then use one more option.

The second way

First we find the path that the first cyclist passed: 20 * 2 = 40 kilometers. Now the path of the second friend: fifteen is multiplied by two, which equals thirty kilometers. We add the distance covered by the first and second cyclist: 40 + 30 = 70 kilometers. We found out what path they had overcome together, so it left out of the way to subtract what they had covered: 100-70 = 30 km. Answer: 30 km.

We have considered the first type of motion problem. How to solve them, it is now clear, go to the next form.

Movement in the opposite direction

Condition: "Two hares rode out of one mink in the opposite direction: the speed of the first is 40 km per hour, and the second - 45 km per hour." How far will they be from each other in two hours? "

Here, as in the previous example, there are two possible solutions. In the first, we will act in the usual way:

1. The way of the first hare: 40 * 2 = 80 km.
2. The way of the second rabbit: 45 * 2 = 90 km.
3. The path that they shared: 80 + 90 = 170 km. Answer: 170 km.

But another option is possible.

Deletion speed

As you have already guessed, in this task, similarly to the first, a new term will appear. Consider the following type of motion problem, how to solve them using the removal rate.

We will find it first and foremost: 40 + 45 = 85 kilometers per hour. It remains to find out what is the distance separating them, since all other data are already known: 85 * 2 = 170 km. Answer: 170 km. We have considered the solution of tasks for motion in the traditional way, as well as with the speed of convergence and removal.

Movement after

Let's look at an example task and try to solve it together. Condition: "Two schoolchildren, Cyril and Anton, left school and moved at a speed of 50 meters per minute, Kostya came out after them in six minutes at a speed of 80 meters per minute." How long will Kostya catch up with Cyril and Anton? "

So, how to solve the task of moving after? Here we need the speed of rapprochement. Only in this case it is necessary not to add, but to subtract: 80-50 = 30 m per minute. The second action is to find out how many meters the schoolchildren share before the release of Kostya. For this, 50 * 6 = 300 meters. The last action is the time for which Kostya will catch up with Cyril and Anton. To do this, the path of 300 meters must be divided into a rapprochement rate of 30 meters per minute: 300: 30 = 10 minutes. Answer: 10 minutes later.

conclusions

Proceeding from what was said earlier, we can sum up some results:

• When solving problems for motion, it is convenient to use the speed of approach and removal;
• If we are talking about the oncoming movement or movement from one another, then these quantities are found by adding the velocities of the objects;
• If we are faced with the task of moving after, then we use the action opposite to addition, that is, subtraction.

We examined some tasks for the movement, how to solve, sorted out, got acquainted with the concepts of "speed of rapprochement" and "speed of removal", it remains to consider the last point, namely: how to solve the problem of movement on the river?

Flow

Here you can meet again:

• Tasks for moving towards each other;
• Movement after;
• Movement in the opposite direction.

But unlike previous problems, the river has a flow velocity that should not be ignored. Here, objects will move either along the current of the river - then this speed should be added to the own speed of objects, or against the current - it must be subtracted from the speed of the object's movement.

An example of a problem on river traffic

Condition: "The water motorcycle was walking along the current at a speed of 120 km per hour and came back, while spending less time for two hours than against the current." What is the speed of a water motorcycle in standing water? " We are given a current velocity equal to one kilometer per hour.

We now turn to the solution. We propose to create a table for an illustrative example. Let's take the speed of the motorcycle in standing water for x, then the velocity along the stream is x + 1, and against x-1. The distance there and back is 120 km. It turns out that the time spent on motion against the current is 120: (x-1), and along the current 120: (x + 1). It is known that 120: (x-1) is two hours less than 120: (x + 1). Now we can proceed to filling out the table.

That we have: (120 / (x-1)) - 2 = 120 / (x + 1) We multiply each part by (x + 1) (x-1);

120 (x + 1) -2 (x + 1) (x-1) -120 (x-1) = 0;

Solve the equation:

(X ^ 2) = 121

We notice that there are two variants of the answer: + -11, since both -11 and +11 are given in square 121. But our answer will be positive, since the speed of the motorcycle can not have a negative value, therefore, we can write down the answer: 11 km per hour . Thus, we found the necessary value, namely the speed in the standing water.

We have considered all possible variants of traffic tasks, now you should not have problems and difficulties when solving them. To solve them, you need to learn the basic formula and concepts such as "speed of rapprochement and removal." Have patience, work out these tasks, and success will come.