Problems on the move how to solve? Methodology for solving motion problems

Mathematics is a rather complicated subject, but absolutely everyone will have to go through it in school. Particular difficulty for students is caused by movement problems. How to solve without problems and the mass 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 is meant by this type of task? These are quite simple and simple tasks, which include the following varieties:

• oncoming traffic;
• after;
• movement in the opposite direction;
• movement on the river.

We suggest that you consider each option individually. Of course, we will disassemble exclusively by examples. But before we get down to the question of how to solve problems on motion, it is worth introducing one formula that we will need when solving absolutely all tasks of this type.

Formula: S = V * t. A little explanation: S is the way, the letter V denotes the speed of movement, and the letter t means time. All quantities can be expressed through this formula. Accordingly, speed is equal to a path divided by time, and time is a path divided by speed.

Movement 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 to meet each other at the same time, 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." We turn to the question of how to solve the problem of oncoming movement of cyclists.

To do this, we need to introduce another term: "approach 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 action in solving the problem. Next, we multiply the approach speed by two, as they moved for two hours: 35 * 2 = 70 km. We found the distance that cyclists would approach in 120 minutes. The last action remained: 100-70 = 30 kilometers. With this calculation, we found the distance between cyclists. Answer: 30 km.

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

Second way

First, we find the path that the first cyclist traveled: 20 * 2 = 40 kilometers. Now the path of the second friend: we multiply fifteen by two, which equals thirty kilometers. Add up the distance traveled by the first and second cyclist: 40 + 30 = 70 kilometers. We learned which path they traveled together, so it remains to subtract the distance traveled from the whole path: 100-70 = 30 km. Answer: 30 km.

We examined the first type of motion problem. How to solve them, now it’s clear, we move on to the next view.

Movement in the opposite direction

Condition: “Two hares galloped from one mink in the opposite direction. The speed of the first is 40 km per hour, and the speed of the second is 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 path of the first hare: 40 * 2 = 80 km.
2. The path of the second hare: 45 * 2 = 90 km.
3. The path they traveled together: 80 + 90 = 170 km. Answer: 170 km.

But another option is possible.

Removal rate

As you 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 in the first place: 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 examined the solution of problems of motion in the traditional way, as well as using the speed of approach and removal.

Chase

Let's look at an example of a problem 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 six minutes at a speed of 80 meters per minute. After how much time did Kostya catch up with Kirill and Anton?"

So, how to solve the problems of following the movement? Here we need the approach speed. Only in this case should not add, but subtract: 80-50 = 30 m per minute. The second step is to find out how many meters the students share before Kostya leaves. For this, 50 * 6 = 300 meters. The last action we find is the time in which Kostya will catch up with Cyril and Anton. To do this, the path of 300 meters must be divided by the convergence speed of 30 meters per minute: 300: 30 = 10 minutes. Answer: in 10 minutes.

conclusions

Based on the foregoing, we can summarize some of the results:

• when solving problems of movement, it is convenient to use the speed of approach and removal;
• if we are talking about oncoming traffic or movement from each other, then these values ​​are found by adding the velocities of the objects;
• if we have the task of moving after it, then we use the action opposite to addition, that is, subtraction.

We examined some traffic problems, how to solve, figured out, got acquainted with the concepts of "convergence rate" and "speed of removal", it remains to consider the last point, namely: how to solve traffic problems on the river?

Flow

Here you can meet again:

• tasks for moving towards each other;
• pursuit movement;
• movement in the opposite direction.

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

Example of a task for moving along a river

Prerequisite: "The jet ski went downstream at a speed of 120 km per hour and returned, while spending less time by two hours than against the stream. What is the speed of the jet ski in still water?" We are given a current velocity equal to one kilometer per hour.

We pass to the solution. We suggest compiling a table for an illustrative example. We take the speed of the motorcycle in standing water for x, then the speed with the stream is x + 1, and against x-1. The round-trip distance is 120 km. It turns out that the time taken to move against the current 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 move on to filling out the table.

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;

We solve the equation:

(x ^ 2) = 121

We notice that there are two possible answers: + -11, since both -11 and +11 give a square of 121. But our answer will be positive, because the speed of the motorcycle cannot have a negative value, therefore, we can write down the answer: 11 km per hour . Thus, we found the required value, namely the speed in still water.

We have considered all the possible options for tasks on the movement, now when solving them you should not have problems and difficulties. To solve them, you need to know the basic formula and concepts such as the "speed of convergence and removal." Be patient, complete these tasks, and success will come.