\(\def \u#1{\,\mathrm{#1}}\) \(\def \abs#1{\left|#1\right|}\) \(\def \ast{*}\) \(\def \deg{^{\circ}}\) \(\def \tau{\uptau}\) \(\def \ten#1{\times 10^{#1}}\) \(\def \redcancel#1{{\color{red}\cancel{#1}}}\) \(\def \BLUE#1{{\color{blue} #1}}\) \(\def \RED#1{{\color{red} #1}}\) \(\def \PURPLE#1{{\color{purple} #1}}\) \(\def \th#1,#2{#1,\!#2}\) \(\def \lshift#1#2{\underset{\Leftarrow\atop{#2}}#1}}\) \(\def \rshift#1#2{\underset{\Rightarrow\atop{#2}}#1}}\) \(\def \dotspot{{\color{lightgray}{\circ}}}\) \(\def \ccw{\circlearrowleft}\) \(\def \cw{\circlearrowright}\)
Chapter 3: Linear Motion
17.

The Five Equations

The set of two equations on the previous page are enough to solve any one-dimensional constant-acceleration kinematics problem. However, we can take those two equations and, by recombining them in different ways, end up with this set of five equations:
$$\begin{align}
v_f&=v_i+a\Delta t &\color{blue}{\hbox{no}\,\Delta x}\\
\Delta x&=\frac{1}{2}(v_i+v_f)\Delta t &\color{blue}{\hbox{no}\,a}\\
\Delta x&=v_i\Delta t+\frac12a\Delta t^2 &\color{blue}{\hbox{no}\,v_f}\\
\Delta x&=v_f\Delta t-\frac12a\Delta t^2 &\color{blue}{\hbox{no}\,v_i}\\
v_f^2&=v_i^2+2a\Delta x &\color{blue}{\hbox{no}\,\Delta t}\\
\end{align}$$

A warning: these are not five independent equations; we can still only solve for two of the five variables.

So why complicate things? Notice that each of the equations has exactly one of the five variables missing. To solve a kinematics problem for one of the variables, you need to be given three: that leaves one variable that you don't know and don't care about. If you choose the equation that does not include that "don't-know-don't-care" (DKDC) variable, then you can solve that equation directly for the variable you want to know.

Example

For example, if I know a car accelerates from 10m/s to 15m/s over a distance of 500m, and I want to know its acceleration, then my DKDC variable is time. The last equation above doesn't include time, and so if I use it I can solve directly for acceleration: $$15^2=10^2+2a(500) \implies a=0.125\u{m/s^2}$$