Equations of motion: velocity-position
The first two equations of motion each describe one kinematic variable as a function of time. In essence…
- Velocity is directly proportional to time when acceleration is constant (v ∝ t).
- Displacement is proportional to time squared when acceleration is constant (∆s ∝ t2).
Combining these two statements gives rise to a third — one that is independent of time. By substitution, it should be apparent that…
- Displacement is proportional to velocity squared when acceleration is constant (∆s ∝ v2).
This statement is particularly relevant to driving safety. When you double the speed of a car, it takes four times more distance to stop it. Triple the speed and you'll need nine times more distance. This is a good rule of thumb to remember.
Combine the first two equations together in a manner that will eliminate time as a variable. The easiest way to do this is to start with the first equation of mortion and solve it for t.
\[v = v_{0} + at \]
\[t = \frac{v-v_{0}}{a} \]
and then substitute it into the second equation of motion
\[s = s_{0} + v_{0}t + \frac{1}{2}at^{2} \]
\[s -s_{0}= \left ( \frac{vv_{0}-v_{0}^{2}}{a} \right ) + \left ( \frac{v^{2} - 2vv_{0} + v_{0} + v_{0}^{2}}{2a}\right ) \]
\[s =s_{0}+v_{0}\left ( \frac{v-v_{0}}{a} \right )+\frac{1}{2}a\left ( \frac{v-v_{0}}{a} \right )^{2} \]
\[2a\left ( s-s_{0} \right )=2(vv_{0}-v_{0}^{2})+\left ( v^{2}-2vv_{0} + v_{0}^{2}\right ) \]
\[2a\left ( s-s_{0} \right )=v^{2}-v_{0}^{2} \]
\[v^{2}= v_{0}^{2}+ 2a\left ( s - s_{0} \right ) \]
the symbol s0 is the initial position and s is the position some time t
