Tuesday, October 19, 2010

Equation Three

There are five equations that can be derived from a velocity time graph. This applies to motion that has a constant acceleration

The first two equations are
1. V2 = V1 + a .Δt
2. d = 1/2 (V2 + V1) Δt
The following three equations can be derived from the first two by rearranging the variables. This is done by isolating each of the variables in the first equation and plugging them into the second equation.
3. d = V1 . Δt  + 1/2 . a . Δt
4. d = V2 . t - 1/2 . a .Δt
5. V2²  = V1 + 2 . a . d

Equation One
The velocity at a certain point in time is equal to the initial velocity + (the acceleration in seconds squared times the change of time in seconds)
V2 = V1 + a . t
Find the slope of the graph.
The rise is V2 - V1
The run is the change in time

Equation Two
The area (of the trapezoid) calculated in this equation is the change in distance.
the base = time
heights = V2 and V1
d = 1/2 (V2 + V1) Δt
Now that we have equation 1 and equation 2 we can find equation 3.

Equation 1
V2 = V1 + a . Δt
V2 is ready for substitution into Equation 2.

d = 1/2 (V2 + V1) Δt
d = 1/2 ( V1 + a .Δt + V1)
d = 1/2 (2V1 + a .Δt)Δt
d = V1 .Δt + 1/2 . a . Δt²           < Equation 3

This can be proven from the velocity time graph.
The graph can be split into a reactangle and a triangle on top of it.
Distance is equal to the area of the reactangle from where the slope starts, and the triangle on top of it.
the area of a triangle can be found using the formula
d = 1/2 (V2- V1) . Δt
The area of the rectangle can be found using the formula
d = V1 . Δt
Putting the two together gives equation 3

We can find three equations from equation 1 and 2, just like we found equation 3 by
isolating V1 in equation 1
isolating V2 in equation 1
isolating Δt in equation 1
This is how we can get the five fundamental equations in kinematics!

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