Movement is the way of existence of everything that a person sees around himself. Therefore, the tasks of moving different objects in space are typical problems that are proposed to solve for students. In this article, we will consider in detail the following movement and the formulas that you need to know in order to be able to solve problems of this type.
What is movement?
Before proceeding to the consideration of the formulas for the motion to follow, it is necessary to understand this concept in more detail.
By motion is meant a change in the spatial coordinates of an object over a certain period of time. For example, a car that moves along a road, an airplane that flies in the sky, or a cat that runs along the grass are all examples of movement.
It is important to note that the moving object in question (car, plane, cat) is considered immense, that is, its dimensions have absolutely no significance for solving the problem, therefore they are neglected. This is a kind of mathematical idealization, or model. For such an object there is a name: material point.
The movement after it and its features
Now we turn to the consideration of popular school problems on the movement to follow and the formulas for it. This type of movement means the movement of two or more objects in the same direction, which depart from different points (material points have different initial coordinates) or / and at different times, but from the same point. That is, a situation is created in which one material point tries to catch up with the other (others), so these tasks are called.
According to the definition, the following are the features of the movement:
- The presence of two or more moving objects. If only one material point will move, then she will have “no one” to catch up with.
- Rectilinear movement in one direction. That is, objects move along the same trajectory and in the same direction. Moving towards each other is not among the tasks considered.
- The departure point plays an important role. The idea is that at the beginning of the movement of the objects are separated in space. Such a separation will take place if they start at the same time, but from different points or from one point, but at different times. The start of two material points from one point and at the same time does not apply to tasks after pursuit, since in this case one object will constantly move away from the other.
Follow-up formulas
In the fourth grade of a comprehensive school, such problems are usually considered. This means that the formulas that are necessary for the solution should be as simple as possible. Such a case is satisfied by a uniform rectilinear motion, in which three physical quantities appear: speed, distance traveled and time of movement:
- Speed is a value that shows the distance that a body travels per unit of time, that is, it characterizes the speed of change of coordinates of a material point. The speed is indicated by the Latin letter V and is measured, as a rule, in meters per second (m / s) or in kilometers per hour (km / h).
- The path is the distance that the body travels during its movement. It is indicated by the letter S (D) and is usually expressed in meters or kilometers.
- Time - the period of movement of a material point, which is indicated by the letter T and is given in seconds, minutes or hours.
Having described the main quantities, we give the following movement formulas:
- s = v * t;
- v = s / t;
- t = s / v.
The solution to any problem of the type in question is based on the application of these three expressions that every schoolchild needs to remember.
An example of solving the problem №1
Let us give an example of the task of following the motion and the solution (the formulas necessary for it are given above). The problem is formulated as follows: “A truck and a passenger car leave points A and B at a speed of 60 km / h and 80 km / h, respectively. Both vehicles move in the same direction so that the car approaches point A and the truck moves away from both points. After what time does the car catch up with the truck if the distance between A and B is 40 km? "
Before solving a problem, it is necessary to teach children to determine the essence of the problem. In this case, it lies in the unknown time that both vehicles will spend on the road. Suppose this time is t hours. That is, after time t, the car will catch up with the truck. Find this time.
We calculate the distance that each moving object will travel in time t, we have: s 1 = v 1 * t and s 2 = v 2 * t, here s 1 , v 1 = 60 km / h and s 2 , v 2 = 80 km / h - the distance traveled and the speed of the truck and the car until the second catches up with the first. Since the distance between points A and B is 40 km, the car, having caught up with the truck, will travel 40 km more, that is, s 2 - s 1 = 40. Substituting the formulas for the paths s 1 and s 2 in the last expression, we get: v 2 * t - v 1 * t = 40 or 80 * t - 60 * t = 40, whence t = 40/20 = 2 hours
Note that this answer can be obtained if we use the concept of the speed of approach between moving objects. In the problem, it is 20 km / h (80-60). That is, with this approach, a situation arises when one object moves (a car), and the second stands still relative to it (truck). Therefore, it is enough to divide the distance between points A and B by the approach speed to solve the problem.
An example of solving problem No. 2
Let’s give one more example of problems in pursuit of movement (the formulas for solving are the same): “A cyclist leaves a point, and after 3 hours a car leaves in the same direction. How long after the start of its movement the car will catch up with the cyclist, if it is known that is he moving 4 times faster? "
This problem should be solved in the same way as the previous one, that is, it is necessary to determine which path each participant of the movement will go until one catches up with the other. Suppose that the car caught up with the cyclist in time t, then we get the following paths traveled: s 1 = v 1 * (t + 3) and s 2 = v 2 * t, here s 1 , v 1 and s 2 , v 2 are the paths and the speed of the cyclist and the car, respectively. Note that before the car caught up with the cyclist, the latter was on the way t + 3 hours, since he had left 3 hours earlier.
Knowing that both participants left from the same point, and the paths traveled by them will be equal, we get: s 2 = s 1 or v 1 * (t + 3) = v 2 * t. The speeds v 1 and v 2 are not known to us, however, the condition of the problem says that v 2 = 4 * v 1 . Substituting this expression in the formula for equal paths, we obtain: v 1 * (t + 3) = 4 * v 1 * t or t + 3 = 4 * t. Solving the latter, we come to the answer: t = 3/3 = 1 h.
Some tips
The catch-up formulas are simple, however, students in grade 4 are important to learn to think logically, to understand the meaning of the quantities they deal with, and to be aware of the problem they are facing. Guys are encouraged to call for reasoning out loud, as well as for teamwork. In addition, for clarity of tasks, you can use a computer and a projector. All this contributes to the development of abstract thinking, communication skills, as well as mathematical abilities.