Kinematic equations
Today we will talk about equations that involve motion, or what we call kinematic equations. These are used mostly in physics when trying to determine things involved in the motion of an object. Let's start with a very basic example, like that of a falling object.
The Freefall
Suppose we threw an object from a high point, like a tower or some tall building, and we'd like to know the height of such a structure assuming we're aware of the other variables. Now obviously we're talking about the distance formula:
d = r * t
but in our case the rate is of a specific type we call velocity which include changes over a period of time:
Δd = Va * Δt
but since we are expecting a fixed distance we can cancel out the delta sign (Δ) meaning we no longer concern ourselves with the change in distance and time:
d = Va * t
For now we will ignore other environmental factors but assume we don't have uniform velocity throughout the fall, so we start with our initial velocity:
Vi = 0 meters / second
which makes sense because we're starting on a resting position. Now, if we have an initial velocity then we have a final velocity, or the velocity of the object as it hits the ground. We'll asign an arbitrary value to it for our example:
Vf = - 100 m / s
The reason for a negative number is to emphasize that we are falling down (negative on the y-axis). This could have easily been a positive number but we'd like to be precise on our direction in this example.
Another arbitrary value that we will consider is the acceleration, or, as in our case, gravity itself:
a = gravity
a = -10 m/(s**2)
So, going back to our distance formula, which of these should we use for our velocity? Well we can't favor one or the other, so we'll use an average velocity:
d = Va * t
and to compute for the average velocity we'll just take the average of the initial and the final velocity:
Va = ( Vi + Vf ) / 2
Va = ( 0 - 100 ) / 2
Va = -100 / 2
Va = -50 m/s
In order to get the time value we have to consider the change in velocity:
ΔV = Vf - Vi
which is the difference between the initial and the final velocity. Incidentally, the change in velocity can also be the acceleration (or gravity) multiplied with the time value.
ΔV = a * t
The change in time will suffice for our time value needed in the distance formula above.
If we combine these two formulae we can isolate the time variable as an unknown, giving us the time value:
ΔV = Vf - Vi = a * t
ΔV = -100 m/s - 0 = -10 m/s(**2) * t
ΔV = -100 m/s = -10 m/s(**2) * t
ΔV = -100 m/s = -10 m/s(**2) * t
ΔV = 10 s = t
There we could see the m/s rate gets cancelled, leaving the squared seconds with just seconds. Then we divide -100 with -10 cancelling out the negative sign.
So now that we have the average velocity and the time value we can then proceed with our distance calculation:
d = Va * t
d = -50 m/s * 10 s
d = -500 m
Which says that the distance the falling object travelled from the top of the structure to the ground is -500 meters.
La formule de Distance
Since we are dealing with a couple of related equations we can probably simplify the whole distance formula for falling objects. At the start we are given:
Initial velocity or Vi
Final velocity or Vf
Acceleration or a
and we need to find the distance d.
So we start with the formula that is meant for the distance travelled by accelarating objects:
Δd = Va * t
The average distance is straightforward:
Va = ( Vf - Vi ) / 2
The time formula needs a bit work since it is dereived the the change in velocity equation:
ΔV = Vf - Vi = a * t
Vf - Vi = a * t
1/a * ( Vf - Vi ) = a * t * 1/a
t = ( Vf - Vi ) / a
so, combining the average velocity formula with the time formula we get a new distance formula:
Δd = ( ( Vf - Vi ) / 2 ) * ( ( Vf - Vi ) / a )
Since the two terms are multiplied we can Use some algebraic love to shorten it:
Δd = ( Vf^2 - Vi^2 ) / 2a
Other useful variations of this formula are:
2a Δd = Vf^2 - Vi^2
Vf^2 = Vi^2 + 2aΔd
The Throw
Let's do an example making use of a variation of our distance formula. Suppose we are observing the same object fall but instead of just having it fall we throw it up first, and we'd like to know how fast it is as it hits the ground.
The obvious formula variation choice would be the one involving the final velocity:
Vf^2 = Vi^2 + 2aΔd
Suppose that our initial throw propelled the object into the air at 30 m/s:
Vi = 30 m/s
Δd = 500 m
a = -10 m/s
So now we let's see the difference if we incorporate an initial velocity by jumping (and so a positve value).
Vf^2 = Vi^2 + 2aΔd
Vf^2 = ( 30 m/s ) ^2 + ( 2 ( 10 m/s^2 ) ( 500 m ) )
Vf^2 = 900 + 10,000
Vf^2 = 10,900
sqrt (Vf^2) = sqrt( 10, 900 )
Vf = 104 m/s
And so the final velocity of the object as it hits the ground is 104 m/s. We notice a difference of 4 m/s as compared to our first final velocity should we throw the object in the air first at 30 m/s.
Lost in Time
What if we need to find time? With these as given:
Vi
a
Δd
Let's start with the simple yet equivalent original distance formula instead:
d = Va * t
d = ( ( Vi + Vf ) / 2 ) * t
As mentioned we don't know Vf, so we can use the equation of ΔV:
ΔV = a * t
Vf - Vi = a * t
Vf = Vi + ( a * t )
Then apply the formula for time:
T = D / V
T = D / Va
T = D / ( Vf + Vi ) / 2
T = 2D / ( Vf + Vi )
T = 2D / ( Vi + Vi + aT )
T = 2D / ( 2Vi + aT )
( 2Vi + aT ) T = 2D
2ViT + aT^2 = 2D
2ViT + aT^2 - 2D = 0
( 2ViT + aT^2 - 2D ) / 2 = 0
ViT + (a/2) T^2 - D = 0
Now this should look more familiar as a quadratic equation:
(a/2) T^2 + ViT - D = 0
with the root being T. If we remember the quadratic formula:
x = ( -n (+/-) sqrt( n^2 - 4mo ) ) 2m
with x being the root variable, m as the coefficient of the 1st term, n as the coefficient of the 2nd term and o as the coefficient of the 3rd term. Using this we finally get our time equation:
aT^2 + 2ViT - 2D = 0
T = ( -n (+/-) sqrt( n^2 - 4mo ) ) 2m
T = ( -Vi (+/-) sqrt( (Vi)^2 - (4aD)/2 ) ) ) / 2 (a/2)
T = ( -Vi +/- sqrt( Vi^2 - 2aD ) ) / a
So, basically, the time equation when only D, Vi and a is known and Vf is unknown is actually a variation of the quadratic equation.
Wrapping up
So to summarize kinematic equations aka motion formulas help us determine things about a moving object. This might not sit well on more realistic environments or in outer space but for the simplest of applications these will do. In closing I'll leave you with the list of formulas we discovered plus a bit more of their other forms.
d = ( vf^2 - vi^2 ) / ( 2*a )
d = ( ( vi + vf ) * t ) / 2
d = ( vf * t ) - ( ( a * t**2 ) / 2 )
d = ( vi * t ) - ( ( a * t**2 ) / 2 )
t = ( vf - vi ) / a
t = ( vi +/- sqrt( -vi**2 - 2*a*d ) ) / a
t = ( -vf +/- sqrt( vf**2 - 2*a*d ) ) / -a
t = 2d / ( vf + vi )
va = ( vf - vi ) / 2
a = ( 2dt - 2vi ) / t
a = ( vf**2 - vi**2 ) / ( 2*d )
a = ( vf - vi ) / t
vi = sqrt ( vf**2 - 2*a*d )
vi = ( 2d / t ) - vf
vi = vf - ( a*t )
vi = ( d / t ) - ( (a*t) / 2 )
vi = ( d - ( ( 1/2 )*a*( t**2 ) ) ) / t
vf = sqrt ( vi**2 + 2*a*d )
vf = ( 2d / t ) - vi
vf = ( a*t ) + vi
vi = ( d + ( ( 1/2 )*a*( t**2 ) ) ) / t
