*Notes to a video lecture on http://www.unizor.com*

__Uniform Motion__

Let's consider a uniform motion from the position of Kinematics and

derive simple equations that mathematically describe this motion in

coordinate system.

The uniform motion along a straight line is characterized by two factors: its

*trajectory*is a straight line in space and its

*velocity vector*is constant, that is each component of the

*velocity vector*does not change with time to assure the direction and magnitude of

*velocity*are constant.

Actually, the second condition (

*velocity vector*is constant) is sufficient for the first, which we will prove.

So, assume that

*velocity vector*

**is constant with coordinates**

*V***,**

*a***,**

*b***}.**

*c*It means that, if

**,**

*x(t)***and**

*y(t)***are coordinate functions defining the**

*z(t)**position vector*

*of our object, the derivatives of these functions are constant:*

**P**

*x'(t) = a*

*y'(t) = b*

*z'(t) = c*where

**,**

*a***and**

*b***are constants**

*c*and at least one of these constants is not zero (otherwise, we deal

with a trivial case of object at rest considered in one of the earlier

lectures).

These three equations of motion for each coordinate can be written in vector form as

*=*

**P'**

**V**,where derivative of a vector

*means a vector of derivatives*

**P'****}**

*x'(t), y'(t), z'(t)*The only function, whose derivative is constant, is a linear function.

Hence, the position of our object can be described by linear functions

of time:

*x(t) = a·t + x*_{0}

*y(t) = b·t + y*_{0}

*z(t) = c·t + z*_{0}where

**,**

*x*_{0}**and**

*y*_{0}**are some constants.**

*z*_{0}Obviously, the derivatives of these three functions representing the

*position*correspondingly equal to

**,**

*a***and**

*b***- components of our constant**

*c**velocity vector*of motion.

How to determine constants

**,**

*x*_{0}**and**

*y*_{0}**that participate in the equations for**

*z*_{0}*position*of our object?

Well, we cannot without additional information.

Notice that for

*(that is, at the beginning of motion)*

**t=0****,**

*x(0) = a·0 + x*_{0}= x_{0}**,**

*y(0) = b·0 + y*_{0}= y_{0}**.**

*z(0) = c·0 + z*_{0}= z_{0}Therefore, we can say that

**,**

*x*_{0}**,**

*y*_{0}**}**

*z*_{0}*initial position*is needed to recreate the trajectory of an object moving uniformly in addition to the knowledge about its

*velocity vector*.

Let vector

*represent the initial position*

**P**_{0}**,**

*x*_{0}**,**

*y*_{0}**}**

*z*_{0}

**P(t) = t·V + P**_{0}By itself,

*velocity*does not define the motion, we need an

*initial position*as well.

Knowing both, an

*initial position*and a constant

*vector of velocity*, we can find equations of motion for

**,**

*x(t)***and**

*y(t)***.**

*z(t)*Let's now prove that the motion described by equations

*x(t) = a·t + x*_{0}

*y(t) = b·t + y*_{0}

*z(t) = c·t + z*_{0}goes along a straight line.

To prove this, we will prove that any two vectors from the point where the motion started

**{**

*P*_{0}**,**

*x*_{0}**,**

*y*_{0}**}**

*z*_{0}**{**

*P*_{1}**,**

*x(t*_{1})**,**

*y(t*_{1})**}**

*z(t*_{1})**{**

*P*_{2}**,**

*x(t*_{2})**,**

*y(t*_{2})**}**

*z(t*_{2})In other words, we will prove that for some real number

*k*the following is true:

**=**

*P*_{0}P_{1}*k·*

*P*_{0}P_{2}Vector

**has coordinates equal to a difference between coordinates of**

*P*_{0}P_{1}**{**

*P*_{1}**,**

*x(t*_{1})**,**

*y(t*_{1})**}**

*z(t*_{1})**{**

*P*_{0}**,**

*x*_{0}**,**

*y*_{0}**}**

*z*_{0}**=**

*P*_{0}P_{1}={

**,**

*x(t*_{1})−x_{0}**,**

*y(t*_{1})−y_{0}**}=**

*z(t*_{1})−z_{0}= {

**,**

*a·t*_{1}**,**

*b·t*_{1}**}**

*c·t*_{1}Vector

**has coordinates equal to a difference between coordinates of**

*P*_{0}P_{2}**{**

*P*_{2}**,**

*x(t*_{2})**,**

*y(t*_{2})**}**

*z(t*_{2})**{**

*P*_{0}**,**

*x*_{0}**,**

*y*_{0}**}**

*z*_{0}**=**

*P*_{0}P_{2}={

**,**

*x(t*_{2})−x_{0}**,**

*y(t*_{2})−y_{0}**}=**

*z(t*_{2})−z_{0}= {

**,**

*a·t*_{2}**,**

*b·t*_{2}**}**

*c·t*_{2}Obviously, vectors

**= {**

*P*_{0}P_{1}**,**

*a·t*_{1}**,**

*b·t*_{1}**} =**

*c·t*_{1}=

**·{**

*t*_{1}**,**

*a***,**

*b***} =**

*c***·**

*t*_{1}

*V*and

**= {**

*P*_{0}P_{2}**,**

*a·t*_{2}**,**

*b·t*_{2}**} =**

*c·t*_{2}=

**·{**

*t*_{2}**,**

*a***,**

*b***} =**

*c***·**

*t*_{2}

*V*are collinear to each other:

**= (**

*P*_{0}P_{1}*t*)·

_{1}/t_{2}

*P*_{0}P_{2}In addition, both these vectors are collinear to

*velocity vector*

**= {**

*V***,**

*a***,**

*b***}**

*c***, is the same as the direction of the**

*P*_{0}P_{1}*velocity vector*

**,**

*a***,**

*b***}.**

*c**velocity vector*.

End of proof.

Since uniform motion is a motion along a straight line, it's very

convenient to choose a system of coordinates, where X-axis coincides

with this line and the positive direction of the X-axis coincides with

the direction of the motion.

Then

**and**

*y(t)=z(t)=0***for all moments of time**

*y'(t)=z'(t)=0***. In particular,**

*t**velocity vector*looks now as

**,**

*a***,**

*0***}.**

*0*For these cases traditionally the letter

**is used for the X-component of the**

*v**velocity vector*.

The only non-trivial equation of motion then will be about

**:**

*x(t)*

*x(t) = v·t + x*_{0}If the origin of the system of coordinates coincides with the initial position of the object,

**and the equation of motion looks really simple:**

*x*_{0}=0

*x(t) = v·t*Since the positive direction of the X-axis is the same as the direction of the movement,

*velocity vector's*direction plays minimal role, and only its magnitude (positive) is important, in which case we can use the term

*speed*for

**in the above equation of motion.**

*v*CONCLUSION

An object moving from the

*initial position*

**= {**

*P*_{0}**,**

*x*_{0}**,**

*y*_{0}**}**

*z*_{0}*velocity vector*

**= {**

*V***,**

*a***,**

*b***}**

*c**velocity vector*and has the following equations of motion:

*x(t) = a·t + x*_{0}

*y(t) = b·t + y*_{0}

*z(t) = c·t + z*_{0}Or in vector form

**·**

*P(t) = t*

*V + P*_{0}If the X-axis coincides with the line of motion and a

*vector of velocity*has X-component

**(Y- and Z-comonents are, of course, equal to zero), the equations are simplified:**

*v*

*x(t) = v·t + x*_{0}

*y(t) = 0*

*z(t) = 0*If, in addition, the origin of coordinates coincides with the point of beginning of motion, that is

**, the equations are simplified even further:**

*x*_{0}=0

*x(t) = v·t*

*y(t) = 0*

*z(t) = 0*
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