<h2>Extension: Finding the Equations of Motion</h2>
<p>The following are derivations of the equations of motion.</p>
<section>
<h3 class="title">Derivation of Equation 1</h3>
<p>According to the definition of acceleration:</p>

\[\vec{a}=\frac{\Delta \vec{v}}{t}\]

<p>where \(\Delta \vec{v}\) is the change in velocity, i.e. \(\Delta v={\vec{v}}_{f} - {\vec{v}}_{i}\). Thus we have</p>

\begin{align*} \vec{a} &amp; = \frac{{\vec{v}}_{f} - {\vec{v}}_{i}}{t} \\ {\vec{v}}_{f} &amp; = {\vec{v}}_{i} + \vec{a}t \end{align*}</section>
<section>
<h3 class="title">Derivation of Equation 2</h3>
<p>We have seen that displacement can be calculated from the area under a velocity vs. time graph. For <em>uniformly accelerated motion</em> 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 <em>\({\vec{v}}_{i}\)</em>, accelerating to a final velocity <em>\({\vec{v}}_{f}\)</em> over a total time <em>t</em>.</p>
<img alt="35b0c9fa58a427ed07938b053b96a206.png" src="https://data.ulearngo.com/assets/7af39a2e-9f26-4a34-a9af-334efe75dc44"/>
<p>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.</p>

\begin{align*} {\text{Area}}_{△} &amp; = \frac{1}{2} b\times h \\ &amp; = \frac{1}{2} t \times \left({v}_{f} - {v}_{i}\right) \\ &amp; = \frac{1}{2}{v}_{f}t - \frac{1}{2}{v}_{i}t \end{align*}\begin{align*} {\text{Area}}_{\square } &amp; = l \times b \\ &amp; = t\times {v}_{i} \\ &amp; = {v}_{i}t \end{align*}\begin{align*} \text{Displacement } &amp; = {\text{Area}}_{\square } + {\text{Area}}_{△} \\ \Delta \vec{x} &amp; = {v}_{i}t + \frac{1}{2}{v}_{f}t - \frac{1}{2}{v}_{i}t \\ \Delta \vec{x} &amp; = \frac{\left({v}_{i} + {v}_{f}\right)}{2}t \end{align*}</section>
<section>
<h3 class="title">Derivation of Equation 3</h3>
<p>This equation is simply derived by eliminating the final velocity \({v}_{f}\) in Equation 2. Remembering from Equation 1 that</p>

\[{\vec{v}}_{f} = {\vec{v}}_{i} + \vec{a}t\]

<p>then Equation 2 becomes</p>

\begin{align*} \Delta \vec{x} &amp; = \frac{{\vec{v}}_{i} + {\vec{v}}_{i} + \vec{a}t}{2}t \\ &amp; = \frac{2{\vec{v}}_{i}t + \vec{a}{t}^{2}}{2} \\ \Delta \vec{x} &amp; = {\vec{v}}_{i}t + \frac{1}{2}\vec{a}{t}^{2} \end{align*}</section>
<section>
<h3 class="title">Derivation of Equation 4</h3>
<p>This equation is just derived by eliminating the time variable in the above equation. From Equation 1 we know</p>

\[t = \frac{{\vec{v}}_{f} - {\vec{v}}_{i}}{a}\]

<p>Substituting this into Equation 3 gives</p>

\begin{align*} \Delta \vec{x} &amp; = {\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} \\ &amp; = \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) \\ &amp; = \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} &amp; = -2{\vec{v}}_{i}^{2} + {\vec{v}}_{f}^{2} + {\vec{v}}_{i}^{2} \\ {\vec{v}}_{f}^{2} &amp; = {\vec{v}}_{i}^{2} + 2\vec{a}\Delta \vec{x} \end{align*}

<p>This gives us the final velocity in terms of the initial velocity, acceleration and displacement and is independent of the time variable.</p>
</section>
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Finding the Equations of Motion

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Learn about the basic principles of motion including reference frames, position, displacement, distance, speed, velocity, acceleration, stationary objects, graphs of motion, projectile motion, and equations of motion with practical examples and applications.


Finding the Equations of Motion

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