Kinematics 1:
Velocity and Acceleration
in One Dimension

by George E. Hrabovsky

Position Vector

We begin our discussion of the geometry of motion with a presentation of the concept of a position vector.  The position vector locates an object relative to the origin of the coordiante system.  In Mathematica we will denote the position vector in the following way,

r = Table[x[i], {i, 3}]				(1)

which produces the following output,

where x[1], x[2], and x[3] are the distances from the origin in each of the three coordinate axes, where the command Table[] is used to build the vector.
	A second important concept is that of displacement, which is the change in the position vector.  In Mathematica we denote displacement as follows,

delta_r = r1 - r0				(2)

where r1 is the final value of r, and r0 is the intiial-value of r.  Redefining (1) for r0,

r0 = Table[x0[i], {i, 3}];

and for r1,

r1 = Table[x1[i], {i, 3}];

thus we get the following output.

delta_r = r1 - r0

Putting this into practice, we look at an example of motion in one dimension, the principle works the same for an arbitrary number of dimensions, it is simpler to see how it works when using one.  Let us assume that some motion can be represented by a position vector of x(t) = 52mm Sin[(0.44 rad/s) t].  We can write this in Mathematica as,

x[t_] := 52 Sin[0.44 t]

We can plot a graph of this function quite easily, and we will choose an interval of time, 0 <= t <= 15.  In Mathematica we perform this task by using the Plot[] command, with t being the horizontal axis and x being the vertical.

Plot[x[t], {t, 0, 15}, 
	PlotLabel -> "Position Vector",
	AxesLabel -> {"t", "x[t]"}];

In order to determine the distance travelled by the object we note from the graph that x(t) increased until approximately 50mm.  Beginning at t=0 x(t) increased until approximately t=4, which turns out to be,

x[4]

which gives us a distance of approximately 51mm, and then it began decreasing until approximately t=11, or,

x[11]

which gives us a total distance travelled of about 103mm, where the objects motion again until t=14, or

x[15]

which gives a distance travelled of about 68mm.  The total distance travelled is the total of all three intervals, or approximately 222mm.
	Finding the displacement we need to find the final position, which is 16.2002 mm, and the initial position which is 0.  The displacement is x1 - x0, or 16.2002 - 0 = 16.2002 mm.

Velocity and Speed

Now that we know how to use the concept of the position vector, we can define the average velocity as the rate of change of the position vector with respect to time.  In Mathematica terms we define the interval of time as,

delta_t = t1 - t0

v_bar = delta_r/delta_t

This is nothing more than the displacement over the time interval.  As will be seen in the next example, this is not too useful.
	Returning to our example above, we have a displacement of 16.2002mm, and the time interval is t1 - t0 = 15 - 0 = 15 s.  So the average velocity, which is also a vector, is,

16.2002/15

in units of mm/sec.  But we know that the object has crossed more distance than indicated by the average velocity, so we need a better measure of velocity.  We need to find the instantaneous velocity for any point along the path of the particle.
	To accomplish this we need to turn to the technique of differential calculus.  The nature of the problem is this: we must find the rate of change of position with respect to time.  The way we do this is rather involved, so follow carefully.
1) Since we will be looking for the velocity over a very small time interval we need to express the position interval as x[t + delta_t] where delta_t is very small.  In our example this becomes,

x[t + delta_t]

2) We now apply the definition of the derivative in the following way, 
dx/dt =Limit[ (x[t + delta_t] - x[t])/delta_t, delta_t -> 0].  For our example we adopt a dummy variable which we will call rr and we will change our delta_t to simply dt and get,

rr = (x[t + dt] - x[t])/dt

we now take the limit of this,

Limit[rr, dt -> 0]

this takes nearly 30 seconds on a 386DX33, so we will try the built-in differentiation operator,

v = D[x[t], t]

Which is not only the correct answer, but only took 0.88 seconds.  Using this result we can determine the velocity at any point in time for the position vector.  Here is a plot of the velocity over the interval given in our example,

Plot[v, {t, 0, 15}, PlotLabel -> "Velocity",
	AxesLabel -> {"t", "v"}];

We can also produce a table,

TableForm[Table[v, {t, 0, 15}]]

We now know how to find the velocity for any position vector which is defined.  It is important to remember that velocity is also a vector even though here it is portrayed in only one dimension.  A velocity vector for Euclidean 3-Space in Mathematica would look like this,

vel = Table[v[i], {i, 3}]

where each component is the velocity in a particular direction.  Another way to represent this is,

vel = Table[D[x[i], t], {i, 3}]		(3)

where the components in this case are the derivatives of the components of the position vector.

Acceleration

Acceleration is the vector quantity which represents the time rate of change of velocity, that is the derivative of velocity with respect to time. It is also the second derivative of the position vector.  Here we take the derivative of the velocity from our example,

a = D[v, t]

Compare this with the second derivative of the position vector,

a = D[x[t], {t, 2}]

as you can see they are the same.  Here is a plot of the acceleration for our example,

Plot[a, {t, 0, 15}, PlotLabel -> "Acceleration",
	AxesLabel -> {"t", "a"}];

Now let's compare each of the plots so that we can see the relationship between them visually,

Notice that the acceleration vector is exactly the reverse of the original position vector.  Also note that velocity is half-way beteen the other two.  As you study physics you will find few types of motion with this property.  Any kind of motion which repeats itself continuously (as this does) is called PERIODIC MOTION.  This will be discussed at greater length in another notebook.