RECTILINEAR MOTION: CHARACTERISTICS, TYPES AND EXAMPLES - PHYSICAL - 2025

The rectilinear movement is one in which the mobile moves along a straight line and therefore occurs in one dimension, hence it is also called one-dimensional movement. This straight line is the path or path followed by the moving object. The cars moving along the avenue of figure 1 follow this type of movement.

It is the simplest model of movement you can imagine. The daily movements of people, animals and things often combine movements in a straight line with movements along curves, but some that are exclusively rectilinear are frequently observed.

Figure 1. Automobiles moving down a straight avenue. Source: Pixabay.

Here are some good examples:

- When running along a rectilinear track of 200 meters.

- Driving a car on a straight road.

- Dropping an object freely from a certain height.

- When a ball is thrown vertically upwards.

Now, the objective of describing a movement is achieved by specifying characteristics such as:

- position

- Displacement

- Speed

- Acceleration

- Weather.

For an observer to detect the motion of an object, he must have a reference point (the origin O) and have established a specific direction in which to move, which can be the x-axis, the y-axis, and any other.

As for the object that moves, it can have an infinite number of shapes. There are no limitations in this regard, however in everything that follows it will be assumed that the mobile is a particle; an object so small that its dimensions are not relevant.

This is known not to be the case for macroscopic objects; however, it is a model with good results in describing the global motion of an object. In this way, a particle can be a car, a planet, a person or any other object that moves.

We will begin our study of rectilinear kinematics with a general approach to motion and then particular cases such as those already named will be studied.

General characteristics of rectilinear motion

The following description is general and applicable to any type of one-dimensional movement. The first thing is to choose a reference system. The line along which the movement takes place will be the x axis. Movement parameters:

Position

Figure 2. Position of a mobile that moves on the x axis. Source: Wikimedia Commons (modified by F. Zapata).

It is the vector that goes from the origin to the point where the object is at a given instant. In figure 2, the vector x ₁ indicates the position of the mobile when it is at the coordinate P ₁ and at time t ₁. The units of the position vector in the international system are meters.

Displacement

The displacement is the vector that indicates the change in position. In figure 3 the car has gone from position P ₁ to position P ₂, therefore its displacement is Δ x = x ₂ - x ₁. The displacement is the subtraction of two vectors, it is symbolized by the Greek letter Δ (“delta”) and it is in turn a vector. Its units in the International System are meters.

Figure 3. Displacement vector. Source: prepared by F. Zapata.

Vectors are denoted in bold in printed text. But being on the same dimension, if you want you can do without the vector notation.

Distance traveled

The distance d traveled by the moving object is the absolute value of the displacement vector:

Being an absolute value, the distance traveled is always greater than or equal to 0 and its units are the same as those of position and displacement. Absolute value notation can be done with modulo bars or simply by removing the bold type in printed text.

Average speed

How fast does the position change? There are slow mobiles and fast mobiles. The key has always been speed. To analyze this factor, the position x is analyzed as a function of time t.

The average speed v _m (see figure 4) is the slope of the secant line (fuchsia) to the curve x vs ty, it provides global information about the movement of the mobile in the time interval considered.

Figure 4. Average speed and instantaneous speed. Source: Wikimedia Commons, modified by F. Zapata.

v _m = (x ₂ - x ₁) / (t ₂ –t ₁) = Δ x / Δ t

Average velocity is a vector whose units in the international system are meters / second (m / s).

Instantaneous speed

Average speed is calculated by taking a measurable time interval, but does not report what happens within that interval. To know the speed at any given moment, you have to make the time interval very small, mathematically equivalent to doing:

The equation above is given for the average speed. In this way the instantaneous speed or simply speed is obtained:

Geometrically, the derivative of the position with respect to time is the slope of the tangent line to the curve x vs t at a given point. In figure 4 the point is orange and the tangent line is green. The instantaneous velocity at that point is the slope of that line.

Speed

Speed ​​is defined as the absolute value or modulus of speed and is always positive (signs, roads and highways are always positive, never negative). The terms "speed" and "velocity" may be used interchangeably on a daily basis, but in physics the distinction between vector and scalar is necessary.

v = Ι v Ι = v

Average acceleration and instantaneous acceleration

The speed can change in the course of the movement and the reality is that it is expected to do so. There is a magnitude that quantifies this change: acceleration. If we note that velocity is the change in position with respect to time, acceleration is the change in velocity with respect to time.

Figure 5. Average acceleration and instantaneous acceleration. Source: Wikimedia Commons, modified by F. Zapata.

The treatment given to the graph of x vs t in the two previous sections can be extended to the corresponding graph of v vs t. Consequently, a mean acceleration and an instantaneous acceleration are defined as:

a _m = (v ₂ - v ₁) / (t ₂ –t ₁) = Δ v / Δ t (Slope of the purple line)

When the acceleration is constant, the average acceleration a _m is equal to the instantaneous acceleration a and there are two options:

- That the acceleration is equal to 0, in which case the speed is constant and there is a Uniform Rectilinear Movement or MRU.

- Constant acceleration other than 0, in which the speed increases or decreases linearly with time (the Uniformly Varied Rectilinear Motion or MRUV):

Where v _f and t _f are final velocity and time respectively, and v _or yt _o are initial velocity and time. If t _o = 0, solving for the final velocity we have the familiar equation for the final velocity:

The following equations are also valid for this movement:

- Position as a function of time: x = x _o + v _o. t + ½ at ²

- Velocity as a function of position: v _f ² = v _o ² + 2a.Δ x (With Δ x = x - x _o)

Horizontal movements and vertical movements

Horizontal movements are those that take place along the horizontal axis or x axis, while vertical movements do so along the y axis. Vertical movements under the action of gravity are the most frequent and interesting.

In the previous equations, we take a = g = 9.8 m / s ² directed vertically downwards, a direction that is almost always chosen with a negative sign.

In this way v _f = v _o + at becomes v _f = v _o - gt and if the initial velocity is 0 because the object was dropped freely, it is further simplified to v _f = - gt. As long as air resistance is not taken into account, of course.

Worked Examples

Example 1

At point A a small package is released to move along the conveyor with sliding wheels ABCD shown in the figure. While it is descending through the inclined sections AB and CD, the package carries a constant acceleration of 4.8 m / s ², while in the horizontal section BC it maintains constant speed.

Figure 6. The package that moves on the sliding track of the resolved example 1. Source: own elaboration.

Knowing that the speed with which the packet reaches D is 7.2 m / s, determine:

a) The distance between C and D.

b) The time required for the package to reach the end.

Solution

The movement of the package is carried out in the three rectilinear sections shown and to calculate what is requested, the speed is required at points B, C and D. Let us analyze each section separately:

Section AB

The time that the packet takes to travel the section AB is:

Section BC

The velocity in section BC is constant, therefore v _B = v _C = 5.37 m / s. The time it takes for the packet to travel this section is:

CD section

The initial velocity of this section is v _C = 5.37 m / s, the final velocity is v _D = 7.2 m / s, through v _D ² = v _C ² + 2. a. d solves the value of d:

Time is calculated as:

The answers to the questions posed are:

a) d = 2.4 m

b) The travel time is t _AB + t _BC + t _CD = 1.19 s +0.56 s +0.38 s = 2.13 s.

Example 2

A person is under a horizontal gate that is initially open and 12 m high. The person vertically throws an object towards the gate with a velocity of 15 m / s.

The gate is known to close 1.5 seconds after the person has thrown the object from a height of 2 meters. Air resistance will not be taken into account. Answer the following questions, justifying:

a) Can the object pass through the gate before it closes?

b) Will the object ever hit the closed gate? If yes, when does it occur?

Figure 7. An object is thrown vertically upward (Worked Example 2). Source: self made.

Answer to)

There are 10 meters between the initial position of the ball and the gate. It is a vertical upward throw, in which this direction is taken as positive.

You can find out the speed it takes to reach this height, with this result the time it would take to do it is calculated and compared with the closing time of the gate, which is 1.5 seconds:

As this time is less than 1.5 seconds, then it is concluded that the object can pass through the gate at least once.

Answer b)

We already know that the object manages to pass through the gate while going up, let's see if it gives it a chance to pass again when going down. The speed, when reaching the height of the gate, has the same magnitude as when it goes uphill, but in the opposite direction. Therefore, we work with -5.39 m / s and the time it takes to reach this situation is:

Since the gate remains open for only 1.5 s, it is evident that it does not have time to pass again before it closes, since it finds it closed. The answer is: the object if it collides with the closed hatch after 2.08 seconds after being thrown, when it is already descending.

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