# Laws of motion

This resource will detail the 5 equations for motion under constant acceleration/deceleration. For the purposes of this resource, deceleration will be treated identically to acceleration in the opposite direction. Velocities are speeds with directions and displacements are distances with directions.

## Constant acceleration
### 1d motion

There are two 'intuitive' equations:
* [final velocity]=[initial velocity]+[acceleration][time elapsed] ($v_{f}=v_{i}+at$), or in other words, the final velocity is the sum of the initial and the product of the acceleration and elapsed time. This makes intuitive sense, for the final velocity is the sum of the initial velocity and the (signed) change in velocity ($v_{f}=v_{i}+\Delta v$). The (signed) change in velocity per unit time is the acceleration by definition ($a={\frac {\Delta v}{t}}$).
* [displacement]=(([initial velocity]+[final velocity])/2)[time elapsed] ($s={\frac {v_{f}+v_{i}}{2}}t$), or in other words, the displacement is the product of the time elapsed and the mean of the initial and final velocities. This makes intuitive sense, for the initial velocity would give a number too low (if the final velocity is higher) or too high (if the final velocity is lower), and the final velocity would give a number too low (if the initial velocity is higher) or too high (if the initial velocity is lower).
There are two, slightly less intuitive equations:
* [displacement]=[initial velocity][time elapsed]+([acceleration][time elapsed]^2)/2 ($s=v_{i}t+{\frac {1}{2}}at^{2}$), or in other words, the displacement is the sum of the product of the initial velocity and elapsed time and half the product of the acceleration and the square of the elapsed time. This is intuitive as the product of the final velocity and elapsed time is the sum of the product of the initial velocity and elapsed time and the product of the acceleration and the square of the elapsed time ($v_{f}t=v_{i}t+at^{2}$) (recall that the final velocity is the sum of the initial and the product of the acceleration and elapsed time ($v_{f}=v_{i}+at$)), so the halving serves as a middle ground.
* [displacement]=[final velocity][time elapsed]-([acceleration][time elapsed]^2)/2 ($s=v_{f}t-{\frac {1}{2}}at^{2}$), or in other words, the displacement is half the product of the acceleration and the square of the elapsed time subtracted from the product of the final velocity and elapsed time. This is intuitive as the product of the initial velocity and elapsed time is the product of the acceleration and the square of the elapsed time subtracted from the product of the final velocity and elapsed time ($v_{i}t=v_{f}t-at^{2}$) (recall that the final velocity is the sum of the initial and the product of the acceleration and elapsed time ($v_{f}=v_{i}+at$), therefore the initial velocity is the product of the acceleration and elapsed time subtracted from the final velocity ($v_{i}=v_{f}-at$)), so the halving serves as a middle ground.
There is one final equation that completely defies intuition:
[final velocity]^2=[intitial velocity]^2+2[acceleration][displacement] ($v_{f}^{2}=v_{i}^{2}+2as$), or in other words, the square of the final velocity is the sum of the square of the initial velocity and twice the product of the acceleration and displacement.