 # Kinematical Equations

In general, the equations of physics are relationships involving rates of change. This means that given the position, velocity, etc., at one time, we can predict the position and velocity at a later time. As a consequence, it is also necessary to consider the initial conditions of the motion: the initial position and velocity of the object. We will typically describe motion relative to an initial time t = 0, and use the subscript 0 in denoting the initial position x0 (or y0) and velocity v0.

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We can envision, then, using 11 position variables to describe the locations of the various parts of your body as you fell. In addition, although your fall is primarily along a vertical axis, your limbs are free to move in any direction in the plane perpendicular to that axis, as well as along that axis. Therefore each of the 11 position variables must have three parts: an x, a y and a z coordinate, corresponding to the three dimensional space in which you are falling: Since we can treat the x, y and z coordinates independently, this means we need 33 position variables to describe your fall. Since each part can move independently, we also need 33 velocity variables to describe your fall. Fortunately the acceleration is a constant, for reasons which we will discuss later. But this still leaves us with 66 degrees of freedom (free variables). So including the time, this means that 67 variables and 66 initial conditions are required for our so-called simplification of this problem.

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With this simplification to two degrees of freedom, we can describe the motion of your fall using four variables and two initial conditions:

y is your position at time t;
v is your velocity at time t;
t is of course the time;
a is the constant acceleration;

y0 is your position at time t = 0 and
v0 is your velocity at time t = 0.

We must also be careful to precisely define the origin of our coordinate system: y = 0 at ground level.

Since the problem specifically states that you fell and were not thrown, v0 = 0. Of course, your velocity changes constantly under the influence of gravity. Since we have defined the origin of the vertical axis to be at ground level, the value of your position decreases from y0 to 0, and therefore your velocity becomes more and more negative (as your speed becomes more and more positive). Since acceleration is the rate of change of velocity, the acceleration must be negative. As we will learn later, the acceleration due to gravity at the earth's surface can be assumed constant; that acceleration is denoted by the letter g and has a value of approximately 9.8 m/s2. So in this problem, as in any problem involving a freely falling object (and ignoring wind resistance), a = - g.

### Unit Analysis

In the problems we will work, the height from which you fell will be given. Since we will know two of the four variables (final height and acceleration) and both initial conditions, we only need two equations in order to completely describe your fall: one for y as a function of t, and one for v as a function of t. We can construct these equations using our knowledge of the units used to measure position, velocity, acceleration and time. The technique we are about to describe is one of the most important things you will learn in this text; you will be using it throughout your study of physics to construct the equations you need to solve problems. Relatively few equations will appear explicitly in this text: this technique will allow you to create them for yourself, using your knowledge of units.

The main idea is this: you cannot add or subtract quantities which do not have the same units. If you walk one half of a mile and run for three minutes, it makes no sense to describe your position with the number 3.5. You would first have to specify the speed at which you ran as, for instance, .1 miles per minute, and then multiply that speed times the three minutes to get .3 miles. Your final position is then .8 miles from where you started. You had to add two numbers which both had the same units before your final result made sense. We will use this idea to construct an equation for position as a function of initial position, initial velocity, acceleration and time.

The first step is to be clear about the goal of the equation you are trying to create: in this case, we want an equation for position, so the left hand side of the equation will be x. Since position is measured in units of distance, the right hand side of the equation will be a sum of terms, all of which are measured in the same units. It is usually helpful to be concrete about the units we are using, so we will use S.I. units of meters and seconds in order to construct our equation. Since initial position is also measured in meters, it is one of the terms on the right hand side of the equation. We would also expect the right hand side to include terms involving the initial velocity and the acceleration. Velocity is measured in units of meters per second, so it must be multiplied by a variable which has units of seconds so that the entire term has the desired units of meters:

v0 t.

Acceleration has units of meters per second squared, so we might expect to multiply acceleration by the square of time in order to get the necessary units of meters. This is almost, but not quite, correct: there is a numerical factor in the acceleration term which has no units, and therefore cannot be predicted by consideration of units alone. Calculus is required to properly derive this unitless numerical factor, although careful experimentation can lead us in the correct direction: if we carefully measure both the position and velocity of a falling object as a function of time, we would find that the constant of proportionality between position and time squared is precisely half the constant of proportionality between velocity and time. But we need calculus to know that this factor of 1/2 is always associated with a squared variable. So the final term is

a t2 / 2,
and our general equation for position as a function of time is:
x(t) = x0 + v0 t + a t2 / 2.
Note that every term in the equation has units of meters. If you have experience balancing chemical equations, you can think of this as a sort of "unit stoichiometry": instead of balancing the number of atoms of a given element on both sides of the reaction, you are balancing the units in every term of the equation by multiplying factors such that the product of their units gives the desired unit.
Δx = v0 t + a t2 / 2.
We will also refer to the change in velocity:
Δv = v - v0.

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We said that we need calculus to get the correct coefficient of one half in the term

a t2 / 2.
But we can borrow some ideas from calculus to see that it must be true for constant acceleration. In that case, the graph of your velocity as a function of time is a straight line. To be specific, we will assume that
a = 1 m/s2
and that you started from rest, so that the equation describing your velocity is
v = t.
Consider how you would figure out the total distance traveled as a function of time. You might split the time up into intervals, and use the average velocity in the middle of each interval to compute the distance traveled during that interval with the equation
Δx = v Δt.
Then you could add all of the distances up. But you notice that the smaller the interval, the more accurate the result:

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With a little imagination, you can see that if the intervals got so small that you couldn't see their width, the total distance traveled would just be the area under the velocity line. So let's use that area to compute the total distance as a function of time. Since v = t, at any given time the area under the graph is just half the area of a square of side t. But that gives us

x = t2 / 2.
So that's where the factor of one half comes from.

What we have done is to compute a Riemann Sum, taking the limit Δt -> 0. In effect, we have computed the integral of t*dt. In a calculus class, you would prove that the integral of

a xn dx
is
a xn + 1 / (n+1).
So we can see that the factor of one half only arises when n is one; that is, when the acceleration is assumed to be constant, and there are no higher powers of t in the computation of x.

The next section continues with an analysis of your fall in linear motion.