<p>In the previous lesson, we started looking at the equations of motion. In this lesson, we will look at examples.</p>
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<h2>Example 1:</h2>
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<h3>Question</h3>
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<p>A racing car is travelling North. It accelerates uniformly covering a distance of $\text{725}$ $\text{m}$ in $\text{10}$ $\text{s}$. If it has an initial velocity of $\text{10}$ $\text{m·s$^{-1}$}$, find its acceleration.</p>
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<h3>Step 1: Identify what information is given and what is asked for</h3>
<p>We are given:</p>

\begin{align*} {\vec{v}}_{i} &amp; = \text{10}\text{ m·s$^{-1}$} \\ \Delta \vec{x} &amp; = \text{725}\text{ m} \\ t &amp; = \text{10}\text{ s} \\ \vec{a} &amp; = ? \end{align*}</div>
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<h3>Step 2: Find an equation of motion relating the given information to the acceleration</h3>
<p>If you struggle to find the correct equation, find the quantity that is not given and then look for an equation that has this quantity in it.</p>
<p>We can use</p>

\[\Delta \vec{x} = {\vec{v}}_{i}t + \frac{1}{2}\vec{a}{t}^{2}\]</div>
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<h3>Step 3: Substitute your values in and find the answer</h3>

\begin{align*} \Delta \vec{x} &amp; = {\vec{v}}_{i}t + \frac{1}{2}\vec{a}{t}^{2} \\ \text{725}\text{ m} &amp; = \left(\text{10}\text{ m·s$^{-1}$} \times \text{10}\text{ s} \right) + \frac{1}{2} \vec{a} \times \left(\text{10}\text{ s}\right)^{2} \\ \text{725}\text{ m} - \text{100}\text{ m} &amp; = \left(\text{50}\text{ s$^{2}$}\right)\vec{a} \\ \vec{a} &amp; = \text{12,5}\text{ m·s$^{-2}$} \end{align*}</div>
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<h3>Step 4: Quote the final answer</h3>
<p>The racing car is accelerating at $\text{12.5}$ $\text{m·s$^{-2}$}$ North.</p>
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<h2>Example 2:</h2>
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<h3>Question</h3>
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<p>A motorcycle, travelling East, starts from rest, moves in a straight line with a constant acceleration and covers a distance of $\text{64}$ $\text{m}$ in $\text{4}$ $\text{s}$. Calculate</p>
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<p>its acceleration</p>
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<p>its final velocity</p>
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<p>at what time the motorcycle had covered half the total distance</p>
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<p>what distance the motorcycle had covered in half the total time.</p>
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<h3>Step 1: Identify what information is given and what is asked for</h3>
<p>We are given:</p>

\begin{align*} \vec{v}_{i} &amp; = \text{0}\text{ m·s$^{-1}$} \text{ (because the object starts from rest)} \\ \Delta \vec{x} &amp; = \text{64}\text{ m} \\ t &amp; = \text{4}\text{ s} \\ \vec{a} &amp; = ? \\ \vec{v}_{f} &amp; = ? \\ t &amp; = ? \text{ at half the distance } \Delta \vec{x} = \text{32}\text{ m} \\ \Delta \vec{x} &amp; = ? \text{ at half the time } t \text{2}\text{ s} \end{align*}

<p>All quantities are in SI units.</p>
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<h3>Step 2: <u>Acceleration</u>: Find a suitable equation to calculate the acceleration</h3>
<p>We can use</p>

\[\Delta \vec{x}={\vec{v}}_{i}t+\frac{1}{2}\vec{a}{t}^{2}\]</div>
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<h3>Step 3: Substitute the values and calculate the acceleration</h3>

\begin{align*} \Delta \vec{x} &amp; = {\vec{v}}_{i}t+\frac{1}{2}\vec{a}{t}^{2} \\ \text{64}\text{ m} &amp; = \left(\text{0}\text{ m·s$^{-1}$} \times \text{4}\text{ s}\right) + \frac{1}{2}\vec{a} \times (\text{4}\text{ s})^{2} \\ \text{64}\text{ m} &amp; = \left(\text{8}\text{ s$^{2}$}\right) \vec{a} \\ \vec{a} &amp; = \text{8}\text{ m·s$^{-2}$} \text{ East} \end{align*}</div>
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<h3>Step 4: <u>Final velocity</u>: Find a suitable equation to calculate the final velocity</h3>
<p>We can use Equation 1 - remember we now also know the acceleration of the object.</p>

\[{\vec{v}}_{f}={\vec{v}}_{i}+\vec{a}t\]</div>
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<h3>Step 5: Substitute the values and calculate the final velocity</h3>

\begin{align*} {\vec{v}}_{f} &amp; = {\vec{v}}_{i} + at \\ {\vec{v}}_{f} &amp; = \text{0}\text{ m·s$^{-1}$} + \left(\text{8}\text{ m·s$^{-2}$}\right)\left(\text{4}\text{ s}\right) \\ &amp; = \text{32}\text{ m·s$^{-1}$} \text{ East} \end{align*}</div>
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<h3>Step 6: <u>Time at half the distance</u>: Find an equation to calculate the time</h3>
<p>We can use Equation 3:</p>

\begin{align*} \Delta \vec{x} &amp; = {\vec{v}}_{i} + \frac{1}{2}\vec{a}{t}^{2} \\ \text{32}\text{ m} &amp; = \left(\text{0}\text{ m·s$^{-1}$}\right)t + \frac{1}{2}\left(\text{8}\text{ m·s$^{-2}$}\right){\left(t\right)}^{2} \\ \text{32}\text{ m} &amp; = 0 + \left(\text{4}\text{ m·s$^{-2}$}\right){t}^{2} \\ \text{8}\text{ s$^{2}$} &amp; = {t}^{2} \\ t &amp; = \text{2,83}\text{ s} \end{align*}</div>
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<h3>Step 7: <u>Distance at half the time</u>: Find an equation to relate the distance and time</h3>
<p>Half the time is $\text{2}$ $\text{s}$, thus we have ${\vec{v}}_{i}$, $\vec{a}$ and $t$ all in the correct units. We can use Equation 3 to get the distance:</p>

\begin{align*} \Delta \vec{x} &amp; = {\vec{v}}_{i}t+\frac{1}{2}a{t}^{2} \\ &amp; = \left(\text{0}\text{ m·s$^{-1}$}\right)\left(\text{2}\text{ s}\right) + \frac{1}{2}\left(\text{8}\text{ m·s$^{-2}$}\right){\left(\text{2}\text{ s}\right)}^{2} \\ &amp; = \text{16}\text{ m} \text{ East} \end{align*}</div>
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Equations of Motion Examples

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Equations of Motion Examples

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