Finding the Equations of Motion

Extension: Finding the Equations of Motion

The following are derivations of the equations of motion.

Derivation of Equation 1

According to the definition of acceleration:

\[\vec{a}=\frac{\Delta \vec{v}}{t}\]

where \(\Delta \vec{v}\) is the change in velocity, i.e. \(\Delta v={\vec{v}}_{f} - {\vec{v}}_{i}\). Thus we have

\begin{align*} \vec{a} & = \frac{{\vec{v}}_{f} - {\vec{v}}_{i}}{t} \\ {\vec{v}}_{f} & = {\vec{v}}_{i} + \vec{a}t \end{align*}

Derivation of Equation 2

We have seen that displacement can be calculated from the area under a velocity vs. time graph. For uniformly accelerated motion the most complicated velocity vs. time graph we can have is a straight line. Look at the graph below - it represents an object with a starting velocity of \({\vec{v}}_{i}\), accelerating to a final velocity \({\vec{v}}_{f}\) over a total time t.

To calculate the final displacement we must calculate the area under the graph - this is just the area of the rectangle added to the area of the triangle. This portion of the graph has been shaded for clarity.

\begin{align*} {\text{Area}}_{△} & = \frac{1}{2} b\times h \\ & = \frac{1}{2} t \times \left({v}_{f} - {v}_{i}\right) \\ & = \frac{1}{2}{v}_{f}t - \frac{1}{2}{v}_{i}t \end{align*}\begin{align*} {\text{Area}}_{\square } & = l \times b \\ & = t\times {v}_{i} \\ & = {v}_{i}t \end{align*}\begin{align*} \text{Displacement } & = {\text{Area}}_{\square } + {\text{Area}}_{△} \\ \Delta \vec{x} & = {v}_{i}t + \frac{1}{2}{v}_{f}t - \frac{1}{2}{v}_{i}t \\ \Delta \vec{x} & = \frac{\left({v}_{i} + {v}_{f}\right)}{2}t \end{align*}

Derivation of Equation 3

This equation is simply derived by eliminating the final velocity \({v}_{f}\) in Equation 2. Remembering from Equation 1 that

\[{\vec{v}}_{f} = {\vec{v}}_{i} + \vec{a}t\]

then Equation 2 becomes

\begin{align*} \Delta \vec{x} & = \frac{{\vec{v}}_{i} + {\vec{v}}_{i} + \vec{a}t}{2}t \\ & = \frac{2{\vec{v}}_{i}t + \vec{a}{t}^{2}}{2} \\ \Delta \vec{x} & = {\vec{v}}_{i}t + \frac{1}{2}\vec{a}{t}^{2} \end{align*}

Derivation of Equation 4

This equation is just derived by eliminating the time variable in the above equation. From Equation 1 we know

\[t = \frac{{\vec{v}}_{f} - {\vec{v}}_{i}}{a}\]

Substituting this into Equation 3 gives

\begin{align*} \Delta \vec{x} & = {\vec{v}}_{i}\left(\frac{{\vec{v}}_{f} - {\vec{v}}_{i}}{a}\right) + \frac{1}{2}a{\left(\frac{{\vec{v}}_{f} - {\vec{v}}_{i}}{\vec{a}}\right)}^{2} \\ & = \frac{{\vec{v}}_{i}{\vec{v}}_{f}}{\vec{a}} - \frac{{\vec{v}}_{i}^{2}}{\vec{a}} + \frac{1}{2}\vec{a}\left(\frac{{\vec{v}}_{f}^{2} - 2{\vec{v}}_{i}{\vec{v}}_{f} + {\vec{v}}_{i}^{2}}{{\vec{a}}^{2}}\right) \\ & = \frac{{\vec{v}}_{i}{\vec{v}}_{f}}{\vec{a}} - \frac{{\vec{v}}_{i}^{2}}{\vec{a}} + \frac{{\vec{v}}_{f}^{2}}{2\vec{a}} - \frac{{\vec{v}}_{i}{\vec{v}}_{f}}{\vec{a}} + \frac{{\vec{v}}_{i}^{2}}{2\vec{a}} \\ 2\vec{a}\Delta \vec{x} & = -2{\vec{v}}_{i}^{2} + {\vec{v}}_{f}^{2} + {\vec{v}}_{i}^{2} \\ {\vec{v}}_{f}^{2} & = {\vec{v}}_{i}^{2} + 2\vec{a}\Delta \vec{x} \end{align*}

This gives us the final velocity in terms of the initial velocity, acceleration and displacement and is independent of the time variable.