Monday, April 23, 2018

Unizor - Physics - Mechanics - Kinematics - Uniform Motion

Notes to a video lecture on

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 V is constant with coordinates {a, b, c}.

It means that, if x(t), y(t) and z(t) are coordinate functions defining the position vector P of our object, the derivatives of these functions are constant:

x'(t) = a

y'(t) = b

z'(t) = c

where a, b and c are constants
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

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

P' = V,

where derivative of a vector P' means a vector of derivatives {x'(t), y'(t), z'(t)} of each of its components.

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 + x0

y(t) = b·t + y0

z(t) = c·t + z0

where x0, y0 and z0 are some constants.

Obviously, the derivatives of these three functions representing the position correspondingly equal to a, b and c - components of our constant velocity vector of motion.

How to determine constants x0, y0 and z0 that participate in the equations for position of our object?

Well, we cannot without additional information.

Notice that for t=0 (that is, at the beginning of motion)

x(0) = a·0 + x0 = x0,

y(0) = b·0 + y0 = y0,

z(0) = c·0 + z0 = z0.

Therefore, we can say that {x0, y0, z0} is a point, where the motion starts. Knowledge of this initial position is needed to recreate the trajectory of an object moving uniformly in addition to the knowledge about its velocity vector.

Let vector P0 represent the initial position {x0, y0, z0}. Then the equations of motion can be expressed in vector form as

P(t) = t·V + P0

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), y(t) and z(t).

Let's now prove that the motion described by equations

x(t) = a·t + x0

y(t) = b·t + y0

z(t) = c·t + z0

goes along a straight line.

To prove this, we will prove that any two vectors from the point where the motion started P0{x0, y0, z0} to any two points along its trajectory P1{x(t1), y(t1), z(t1)} and P2{x(t2), y(t2), z(t2)} are collinear.

In other words, we will prove that for some real number k the following is true:

P0P1 = P0P2

Vector P0P1 has coordinates equal to a difference between coordinates of P1{x(t1), y(t1), z(t1)} and P0{x0, y0, z0}, that is

P0P1 =

={x(t1)−x0, y(t1)−y0, z(t1)−z0}=

= {a·t1, b·t1, c·t1}

Vector P0P2 has coordinates equal to a difference between coordinates of P2{x(t2), y(t2), z(t2)} and P0{x0, y0, z0}, that is

P0P2 =

={x(t2)−x0, y(t2)−y0, z(t2)−z0}=

= {a·t2, b·t2, c·t2}

Obviously, vectors

P0P1 = {a·t1, b·t1, c·t1} =

= t1·{a, b, c} = t1·V


P0P2 = {a·t2, b·t2, c·t2} =

= t2·{a, b, c} = t2·V

are collinear to each other:

P0P1 = (t1/t2P0P2

In addition, both these vectors are collinear to velocity vector V = {a, b, c}, which proves that the direction of a trajectory in uniform motion, defined by vector P0P1, is the same as the direction of the velocity vector {a, b, c}. So, an object in uniform motion always moves in a direction of the constant 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 y(t)=z(t)=0 and y'(t)=z'(t)=0 for all moments of time t. In particular, velocity vector looks now as {a, 0, 0}.

For these cases traditionally the letter v is used for the X-component of the velocity vector.

The only non-trivial equation of motion then will be about x(t):

x(t) = v·t + x0

If the origin of the system of coordinates coincides with the initial position of the object, x0=0 and the equation of motion looks really simple:

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 v in the above equation of motion.


An object moving from the initial position P0 = {x0, y0, z0} with constant velocity vector V = {a, b, c} moves along a straight line collinear to a velocity vector and has the following equations of motion:

x(t) = a·t + x0

y(t) = b·t + y0

z(t) = c·t + z0

Or in vector form

P(t) = t·V + P0

If the X-axis coincides with the line of motion and a vector of velocity has X-component v (Y- and Z-comonents are, of course, equal to zero), the equations are simplified:

x(t) = v·t + x0

y(t) = 0

z(t) = 0

If, in addition, the origin of coordinates coincides with the point of beginning of motion, that is x0=0, the equations are simplified even further:

x(t) = v·t

y(t) = 0

z(t) = 0

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