Origin of the Time Independent Acceleration Equation

Back Kinematics Equations Kinematics Mechanics Physics Math Contents Index Home

Here we will take a look at the derivation of the following kinematics equation:

velocity final squared minus velocity original squared equals two times acceleration times displacement


This equation in its above form is not solved for any particular variable.


Notice that it relates four quantities: final velocity, vf, original velocity, vo, constant acceleration, a, and displacement, d. Using algebra you can rearrange this equation to solve for any one of those quantities. That algebraic rearrangement is covered here.


What is important to notice is that the quantity of time is not present in this equation. We say, therefore, that the equation is time independent.

This equation is often useful in kinematics problems where you do not know the time period of the acceleration but still have to work with the velocities, acceleration, and displacement.


We derive this equation by combining the other two kinematics equations in this section, and, through substitutions, eliminating the time variable.

Here are the other two equations; first, the displacement equation:

displacement equals velocity original times time plus one half times acceleration times time squared

And second, the velocity equation:

velocity final equals velocity original plus acceleration times time


For the substitution this velocity equation is rearranged so that it is solved for t:

time equals the quantity velocity final minus velocity original divided by acceleration

The above expression for t can be substituted into the displacement equation, and the resultant equation can be simplified and arranged till our time independent equation takes form.


The algebra for this is a bit more complicated than the algebra in our other examples. We will step through it as follows.

Let's start here:

displacement equation

In the above equation we will substitute the following expression for t:

equation solved for time

This gives us:

displacement equation with substitution for time

And on the right side of the equation in the second term, since the square of a fraction equals the square of the numerator over the square of the denominator, we get:

algebra for displacement equation

Multiplying each side by a,

Carefully canceling where a/a=1 and a2/a2=1,

We get:

equation algebra

Now, on the right side, first term we distribute vo into the parentheses and get:

equation algebra

On the right side, third term, by expanding (vf - vo)2 we get:

equation algebra

For reasons that are really only cosmetic, inside of that expansion we will note that vfvo=vovf:

equation algebra

On the right side, third term we will distribute the 1/2 into the parentheses:

equation algebra

On the right side noting that vovf cancels with -vovf we get:

equation algebra

Now, multiplying each side by 2:

equation algebra

On the right side, since -2vo2+vo2=-vo2:

equation algebra

Commuting terms on the right side we get:

equation algebra

And, lastly, we rearrange by switching the left and the right sides of the equation:

equation algebra

Back Kinematics Equations Kinematics Mechanics Physics Math Contents Index Home

Custom Search