Five Equations of Motion Derivations
Physics
The five equations of motion are the most basic equations for figuring out what is going on in a situation
where an object is moving. On this sheet, I will show you where the equations come from, so that you use them
with confidence.
The first two equations are definitions.
d
1. v̄ = t
This is the definition of what we mean by an object’s velocity – the rate at which it covers
distance. Basically, velocity tells us how far the object moves each second.
This equation can be rearranged into another form, which is useful for some calculations:
d = v̄t
2. a =
vf – vi
t
This is the definition of what we mean by an object’s acceleration – the rate at which its
velocity changes. Basically, acceleration tells us how much the velocity changes each second.
This equation can also be rearranged into another form, which is useful for some calculations:
vf = vi + at
vf + vi
2
Although we almost never use this equation for calculation, it is very important because it is
what lets us derive the final three equations. This equation (and the next three) only apply when
the acceleration is constant. In real life, acceleration is almost always fluctuating – your car’s
acceleration goes up and down as you adjust the gas while you are speeding up. Similarly, the
acceleration is continuously changing as you adjust the pressure on the brake while you are
slowing down. However, in many situations the acceleration is close enough to constant that we
can safely ignore the variation. In this class we will only focus on situations where the
acceleration is constant or where we can act as if it is.
If we assume constant acceleration, then v̄ =
To prove that this equation works, let’s look at a velocity-time graph for a constantly
accelerating object. On this graph, the velocity increases in a straight line (same slope means
same acceleration!):
Notice that speed increase from some starting speed to some
final speed over t seconds. We would like to know what the
average speed of the car is during those t seconds. It may seem
intuitive that the average velocity will be halfway between the
lowest speed and the highest speed, but we want to prove it
definitively.
So we go back to our original definition of average speed –
total distance divided by total time. To find the total distance,
we need to find the area under the line on the velocity-time
graph. In this case we divide the area into a triangle resting on
a rectangle and find the area for both, then add them together to
get the total area.
So d = Arectangle + Atriangle
(see picture to right)
1
2b·h
d = l·w +
(general formulas for area)
Then we plug in specific values from the graph.
1
d = t·vi + 2·t·(vf – vi)
Rearrange the order of the variables:
1
d = vi·t + 2(vf – vi)·t
Use the distributive property on the term in parentheses:
1
1
d = vi·t + 2vf·t – 2vi·t
Combine like terms:
1
1
d = 2vi·t + 2vf·t
Use the distributive property to pull out similar terms:
1
d = 2(vi + vf)·t
This means that the average velocity is:
1
d 2(vi + vf)·t 1
vi + vf
v̄ = t =
=
(v
i + vf) =
2
t
2
QED (Which is what we set out to prove!)
We can also make sense of this outcome by looking at the
graphs and seeing what happens if we plot the average
velocity on the graph.
Since the average velocity is the constant velocity that
would cover the same distance as increasing speed over the
time shown. In graph terms, this should mean that they
cover the same amount of area.
You can observe that the areas under the two lines are
the same when v̄ is right between vi and vf by noticing that
the difference in distance (area) covered between the two
graphs is two triangles. There is a little triangle on the right
side of the graph that is under the sloped line, but not under
the flat line. There is also a little triangle on the left hand
side of the graph that is under the flat line, but not under the
sloped line.
For the two lines to cover the same distance, those two
triangles must be the same size so the total area under each line is the same. That only happens
when the flat line is right in between the top and bottom of the sloped line!
vf + vi
3. d =  2 ·t


This is simply combining our average speed equation with our new expression for the average
speed when the acceleration remains constant. Take the rearranged version of equation 1:
d = v̄t
vf + vi
and substitute in for v̄ based on our new expression v̄ = 2
vf + vi
d =  2 ·t


1
d = vi·t + 2at2 (for constant acceleration)
To prove that this equations holds, we will use the three equations we already know and some
substitution to arrive at the equation we want:
d
In the first three equations, the only v̄ = t includes d, so that is where we will start.
I will rearrange it to solve for d:
d = v̄·t
vf + vi
But our third equation tells us that v̄ = 2 , so we will substitute for v̄:
vf + vi
d =  2 ·t


One version of our second equation tells us that vf = vi + at, so we can substitute for vf:
vi + at + vi
d = 
·t
2


Simplify to get:
2vi + at
d = 
·t
 2 
Continue to simplify:
(
1
)
d = vi + 2at ·t
And apply the distributive property to bring the t to every term:
1
d = vi·t + 2at2
QED
5. vf2 = vi2 + 2a·d (for constant acceleration)
vf – vi
t , and get rid of the time.
d
d
We can substitute in for time from the first equation v̄ = t by rearranging to get: t = v̄
vf – vi
This gives us: a =
d
v̄
Since dividing by a fraction is the same as multiplying by its inverse:
v̄
a = (vf – vi)
d
vf + vi
Substitute in for v̄ by using our third equation: v̄ = 2 , and you get:
vf + vi
2
vf + vi
a = (vf – vi)
= (vf – vi)
d
2d
We will simplify things some more:
(vf – vi)(vf + vi)
a=
2d
Get rid of the denominator by multiplying both sides by 2d:
2a·d = (vf – vi)(vf + vi)
FOIL to get the product on the right hand side:
2a·d = vf2 – vivf + vfvi + vi2 = vf2 + vi2 or 2a·d = vf2 + vi2
Rearrange to get the final answer:
vf2 = vi2 + 2a·d
QED
To get this equation we start from the second equation, a =