# Straight movement: characteristics, types and examples

The **rectilinear movement** is one in which the mobile moves along a straight line and, therefore, passes in one dimension; therefore, 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 in Figure 1 follow this type of movement.

It’s the simplest movement model you can imagine. The daily movements of people, animals and things generally combine straight-line transfers with movements along curves, but it is often observed that they are exclusively straight.

Here are some good examples:

– When traveling a straight line of 200 meters.

– Driving a car on a straight road.

– Drop an object freely from a certain height.

– When a ball is thrown vertically upwards.

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

– Position

– Displacement

– Speed

– Acceleration

– Time

For an observer to detect the movement of an object, it must have a reference point (the O origin) and have established a specific direction in which to move, which can be the *x-* axis, *y-* axis, or anything else.

As for the moving object, it can have infinite shapes. There are no limitations in this respect, however, in all that follows it will be assumed that the cell phone is a particle; an object so small that its dimensions are not relevant.

It is known that this is not the case with macroscopic objects; However, it is a model with good results in describing the global movement of an object. In this way, a particle can be a car, a planet, a person, or any other moving object.

We will start our study of rectilinear kinematics with a general approach to motion and then we will study particular cases like the ones mentioned above.

__General characteristics of rectilinear movement__

__General characteristics of rectilinear movement__

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

**Position**

It is the vector that goes from the origin to the point where the object is at a given moment. In figure 2, the vector *x ** _{1}* indicates the position of the rover when it is at coordinate

*P*and at time

_{1}*t*. The units of the position vector in the international system are

_{1}*meters*.

**Displacement**

The displacement is the vector that indicates the change of position. In Figure 3, the car has moved from position *P _{1}* to position

*P*, therefore, its displacement is Δ

_{2}*. Displacement is the subtraction of two vectors, is symbolized by the Greek letter Δ (“delta”) and, in turn, is a vector. Its units in the International System are*

**x**=**x**_{2}–**x**_{1}*meters*.

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

**Travelled distance**

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

*d =* ΙΔ * x* Ι = Δ

*x*

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

**Average speed**

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

The average velocity *v ** _{m}* (see figure 4) is the inclination of the secant line (fuchsia) in relation to the

*x*vs

*t*curve

*and*provides global information about the displacement of the rover in the considered time interval.

*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 velocity**

Average speed is calculated over a measurable time span, but does not tell you what happens within that span. To know the speed at any time, it is necessary to reduce the time interval, mathematically, it is equivalent to:

*Δt →*

The above equation is given for the average speed. That way you get instant speed or just speed:

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

**Fast**

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 “velocity” and “velocity” can be used interchangeably on a daily basis, but in physics the distinction between vector and scalar is necessary.

v *=* Ι ** v** Ι =

*v*

**Medium Acceleration and Instant Acceleration**

Velocity can change in the course of movement and the reality is that it is expected to happen. There is a magnitude that quantifies this change: acceleration. If we notice that velocity is the change of position with respect to time, acceleration is the change of velocity with respect to time.

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

*a *_{m} = ( **v **_{2} – **v **_{1} ) / (t _{2} –t _{1} ) = Δ **v** / Δ *t* (Purple line gradient)

In one-dimensional motion, vectors by convention have plus or minus signs as they go one way or another. When acceleration has the same direction as velocity, its magnitude increases, but when it has the opposite direction and velocity decreases its magnitude. The movement is then said to be retarded.

__Types__

__Types__

The classification of rectilinear movements is generally made according to:

– Whether the acceleration is constant or not.

– Movement takes place along a horizontal or vertical line.

**Motion with constant acceleration**

When the acceleration is constant, the average acceleration *a ** _{m}* is equal to the instantaneous acceleration

**and there are two options:**

*a*– That the acceleration is 0; in this case, the velocity is constant and there is uniform rectilinear motion or MRU.

– Constant acceleration other than 0, in which the velocity increases or decreases linearly with time (uniformly varied rectilinear motion or MRUV):

Where *v _{f}* and

*t*are the final velocity and time, respectively, and

_{f}*v*and

_{o}*t*are the initial velocity and time. If

_{o}*t*, clearing the final velocity is the familiar equation for the final velocity:

_{or}= 0*v _{f} = v _{o} + in*

The following equations are also valid for this move:

– Position as a function of time: *x = x _{or} + v _{o. }t + ½ to ^{2}*

– Velocity depending on position: *v _{f }^{2} = v _{or }^{2} + 2a.Δ *

*x*(With Δ

*x = x – x*)

_{o}**Horizontal and vertical movements**

Horizontal moves are those that run along the horizontal or x-axis, while vertical moves are along the y-axis. Vertical movements under gravity are the most frequent and interesting.

In the previous equations, *a = g = 9.8 m / s ^{2 is taken}* vertically downward, a direction which is almost always chosen with a minus sign.

In this way, *v _{f} = v _{or} + at* is transformed into

*v*and if the initial velocity is 0 because the object was released freely, it will be further simplified to

_{f}= v_{o}– gt*v*. As long as air resistance is not taken into account, of course.

_{f}= – gt__Solved Examples__

__Solved Examples__

**Example 1**

At point A, a small package is released to move along the conveyor with the ABCD sliding wheels shown in the figure. As it descends along sloping sections AB and CD, the package has a constant acceleration of 4.8 m/s ^{2} , while on the horizontal section BC it maintains a constant velocity.

Knowing that the speed at which the package 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**

Packing movement is performed on the three straight sections shown and to calculate the requested speed is required at points B, C and D. Let’s look at each section separately:

**Section AB**

As time is not available in this section, *v _{f }^{2} = v _{or }^{2} + 2a.Δ *

*x*with vo = 0 will be used:

*v _{f }^{2} = 2a.Δ *

*x →*

*v*

_{f }^{2}= 2. 4.8 m / s^{2}. 3 m = 28.8 m^{2}/ s^{2}→*v*

_{f}^{ }= 5.37 m / s = v_{B}The time it takes for the package to travel in the AB section is:

*t _{AB} = (v _{f} – v _{o} ) / a = 5.37 m / s / 4.8 m / s ^{2}* =

*1.19 s*

**BC section**

The velocity in the BC section is constant, so *v _{B} = v _{C} = 5.37 m / s* . The time it takes for the package to travel in this section is:

*t _{BC} = distance _{BC} / v _{B} = 3 m / 5.37 m / s = 0.56 s*

**CD stretch**

The initial velocity of this section is *v _{C}* =

*5.37 m / s*, the final velocity is

*v*= 7.2 m / s, using

_{D}*v*

_{D }

^{2}*= v*

_{C }

^{2}*+ 2. a.*

*d*the value of

*d*is cleared :

*d =* ( *v _{D }^{2} – v _{C }^{2} ) /2.a =* (

*7.2*

^{2}– 5.37^{2}) / 2*x*

*4.8 m = 2.4 m*

Time is calculated as:

*t _{CD =} (v _{D} – v _{C} ) / a =* (

*7.2 – 5.37) / 4.8 s = 0.38 s.*

The answers to the questions 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**

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

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

a) Does the object pass through the gate before it is closed?

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

**Answer a)**

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

You can figure out the speed needed to reach that height. With this result, the time needed to calculate it is calculated and compared with the gate closing time, which is 1.5 seconds:

*v _{f }^{2} = v _{or }^{2} – 2.g. Δ *

*y →*

*v*

_{f}= (15^{2}– 2*x*

*9.8*

*x*

*10)*

^{1/2}m = 5.39 m / s*t = (v _{f} – v _{o} ) / g = (5.39 – 15) / (-9.8) s = 0.98 s*

As this time is less than 1.5 seconds, 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 as it goes up, let’s see if it gives you a chance to pass again when you go down. The velocity, when reaching the height of the gate, has the same magnitude as when it goes up, but in the opposite direction. Therefore, he works with -5.39 m / s and the time needed to reach this situation is:

*t = (v _{f} – v _{o} ) / g = (-5.39 – 15) / (-9.8) s = 2.08 s*

As the gate remains open for only 1.5 s, it is clear that you do not have time to pass again before closing, as it is closed. The answer is: if the object hits the closed gate after 2.08 seconds after launch, when it is already descending.