<h2>Motion at Constant Velocity</h2>
<p>Motion at a constant velocity or <em>uniform motion</em> means that the position of the object is changing at the same rate.</p>
<p>Assume that Vivian takes $\text{100}$ $\text{s}$ to walk the $\text{100}$ $\text{m}$ to the taxi-stop every morning. If we assume that Vivian's house is the origin and the direction to the taxi is positive, then Vivian's velocity is:</p>
<p>\begin{align*} v &amp; = \frac{\Delta \vec{x}}{\Delta t} \\ &amp; = \frac{{x}_{f} - {x}_{i}}{{t}_{f} - {t}_{i}} \\ &amp; = \frac{\text{100}\text{ m} - \text{0}\text{ m}}{\text{100}\text{ s} - \text{0}\text{ s}} \\ &amp; = \text{1}\text{ m·s$^{-1}$} \end{align*}</p>
<p>Vivian's velocity is $\text{1}$ $\text{m·s$^{-1}$}$. This means that she walked $\text{1}$ $\text{m}$ in the first second, another metre in the second second, and another in the third second, and so on. For example, after $\text{50}$ $\text{s}$ she will be $\text{50}$ $\text{m}$ from home. Her position increases by $\text{1}$ $\text{m}$ every $\text{1}$ $\text{s}$. A diagram of Vivian's position is shown below:</p>
<p><img alt="1fe55c2738eba0dcd5775f2b3d50fbae.png" src="https://data.ulearngo.com/assets/fab94786-500e-4e0b-bff9-a6184bb3c098"/></p>
<p>We can now draw graphs of position vs.time ($\vec{x}$ vs. $t$), velocity vs. time ($\vec{v}$ vs. $t$) and acceleration vs. time ($\vec{a}$ vs. $t$) for Vivian moving at constant velocity. The graphs are shown here:</p>
<p><img alt="13d558164b8e45289b6ed0b968403a62.png" src="https://data.ulearngo.com/assets/554fe553-93e9-48fa-8f5c-c3f415dc8125"/></p>
<p><img alt="cbed3687c6c79629d9cc5108c6edefc5.png" src="https://data.ulearngo.com/assets/e8d629ac-fd16-4689-98cc-04ef9bc050c5"/></p>
<div class="wp-caption alignnone" style="width: 459px"><img alt="52164b5a2d089648e797c4da0f4b8b29.png" height="400" src="https://data.ulearngo.com/assets/53b1c7cc-c75a-4525-b296-f2933bef9a28" width="459"/><p class="wp-caption-text">Graphs for motion at constant velocity (a) position vs. time (b) velocity vs. time (c) acceleration vs. time. The area of the shaded portion in the $v$ vs. $t$ graph corresponds to the object's displacement.</p></div>
<p>In the evening Vivian walks $\text{100}$ $\text{m}$ from the bus stop to her house in $\text{100}$ $\text{s}$. Assume that Vivian's house is the origin. The following graphs can be drawn to describe the motion.</p>
<p><img alt="49a8f6b3cc76d310b7a84b538562e099.png" src="https://data.ulearngo.com/assets/5efcf493-3e20-4ca1-abc1-d77d0de29fce"/></p>
<p><img alt="26f15b954e6a653c3988c5ea082df5e9.png" src="https://data.ulearngo.com/assets/f25df480-11eb-407a-bc0d-b93eb2970441"/></p>
<div class="wp-caption alignnone" style="width: 459px"><img alt="52164b5a2d089648e797c4da0f4b8b29.png" height="400" src="https://data.ulearngo.com/assets/929a3391-7dd4-4e21-b6bb-e453914a063a" width="459"/><p class="wp-caption-text">Graphs for motion with a constant negative velocity. The area of the shaded portion in the $v$ vs. $t$ graph corresponds to the object's displacement.</p></div>
<p>We see that the $\vec{v}$ vs. $t$ graph is a horizontal line. If the velocity vs. time graph is a horizontal line, it means that the velocity is <em>constant</em> (not changing). Motion at a constant velocity is known as <em>uniform motion</em>. We can use the $\vec{x}$ vs. $t$ graph to calculate the velocity by finding the gradient of the line.</p>
<p>\begin{align*} v &amp; = \frac{\Delta \vec{x}}{\Delta t} \\ &amp; = \frac{{\vec{x}}_{f} - {\vec{x}}_{i}}{{t}_{f} - {t}_{i}} \\ &amp; = \frac{\text{0}\text{ m} - \text{100}\text{ m}}{\text{100}\text{ s} - \text{0}\text{ s}} \\ &amp; = -\text{1}\text{ m·s$^{-1}$} \end{align*}</p>
<p>Vivian has a velocity of $-\text{1}$ $\text{m·s$^{-1}$}$, or $\text{1}$ $\text{m·s$^{-1}$}$ towards her house. You will notice that the $\vec{v}$$t$$-\text{1}$ $\text{m·s$^{-1}$}$. The horizontal line means that the velocity stays the same (remains constant) during the motion. This is uniform velocity.</p>
<p>We can use the $\vec{v}$ vs. $t$ graph to calculate the acceleration by finding the gradient of the line.</p>
<p>\begin{align*} a &amp; = \frac{\Delta \vec{v}}{\Delta t} \\ &amp; = \frac{{\vec{v}}_{f} - {\vec{v}}_{i}}{{t}_{f} - {t}_{i}} \\ &amp; = \frac{\text{1}\text{ m·s$^{-1}$} - \text{1}\text{ m·s$^{-1}$}}{\text{100}\text{ s} - \text{0}\text{ s}} \\ &amp; = \text{0}\text{ m·s$^{-2}$} \end{align*}</p>
<p>Vivian has an acceleration of $\text{0}$ $\text{m·s$^{-2}$}$. You will notice that the graph of $\vec{a}$$t$$\text{0}$ $\text{m·s$^{-2}$}$. There is no acceleration during the motion because her velocity does not change.</p>
<p>We can use the $\vec{v}$ vs. $t$ graph to calculate the displacement by finding the area under the graph.</p>
<p>\begin{align*} \Delta \vec{x} &amp; = \text{Area under graph} \\ &amp; = l \times b \\ &amp; = 100(-1) \\ &amp; = -\text{100}\text{ m} \end{align*}</p>
<p>This means that Vivian has a displacement of $\text{100}$ $\text{m}$ towards her house.</p>
<div class="ns-school-emp">
<h2>Experiment: Motion at Constant Velocity</h2>
<div class="ns-school-defn">
<section class="activity">
<h3>Aim</h3>
<p>To measure the position and time during motion at constant velocity and determine the average velocity as the gradient of a “Position vs. Time” graph.</p>
</section>
<section class="activity">
<h3>Apparatus</h3>
<p>A battery operated toy car, stopwatch, meter stick or measuring tape.</p>
</section>
<section class="activity">
<h3>Method</h3>
<ol>
<li>
<p>Work with a friend. Copy the table below into your workbook.</p>
</li>
<li>
<p>Complete the table by timing the car as it travels each distance.</p>
</li>
<li>
<p>Time the car twice for each distance and take the average value as your accepted time.</p>
</li>
<li>
<p>Use the distance and average time values to plot a graph of “Distance vs. Time” <em>onto graph paper</em>. Stick the graph paper into your workbook. (Remember that “A vs. B” always means “$y$ vs. $x$”).</p>
</li>
<li>
<p>Insert all axis labels and units onto your graph.</p>
</li>
<li>
<p>Draw the best straight line through your data points.</p>
</li>
<li>
<p>Find the gradient of the straight line. This is the average velocity.</p>
</li>
</ol>
</section>
<section class="activity">
<h3>Results</h3>
<table>
<tbody>
<tr>
<td rowspan="2">
<p>Distance (m)</p>
</td>
<td colspan="3">
<p>Time (s)</p>
</td>
</tr>
<tr>
<td>
<p>1</p>
</td>
<td>
<p>2</p>
</td>
<td>
<p>Ave.</p>
</td>
</tr>
<tr>
<td>
<p>$\text{0}$</p>
</td>
<td> </td>
<td> </td>
<td> </td>
</tr>
<tr>
<td>
<p>$\text{0.5}$</p>
</td>
<td> </td>
<td> </td>
<td> </td>
</tr>
<tr>
<td>
<p>$\text{1.0}$</p>
</td>
<td> </td>
<td> </td>
<td> </td>
</tr>
<tr>
<td>
<p>$\text{1.5}$</p>
</td>
<td> </td>
<td> </td>
<td> </td>
</tr>
<tr>
<td>
<p>$\text{2.0}$</p>
</td>
<td> </td>
<td> </td>
<td> </td>
</tr>
<tr>
<td>
<p>$\text{2.5}$</p>
</td>
<td> </td>
<td> </td>
<td> </td>
</tr>
<tr>
<td>
<p>$\text{3.0}$</p>
</td>
<td> </td>
<td> </td>
<td> </td>
</tr>
</tbody>
</table>
</section>
<section class="activity">
<h3>Conclusions</h3>
<p>Answer the following questions in your workbook:</p>
<ol>
<li>
<p>Did the car travel with a constant velocity?</p>
</li>
<li>
<p>How can you tell by looking at the “Distance vs. Time” graph if the velocity is constant?</p>
</li>
<li>
<p>How would the “Distance vs. Time” graph look for a car with a faster velocity?</p>
</li>
<li>
<p>How would the “Distance vs. Time” graph look for a car with a slower velocity?</p>
</li>
</ol>
</section>
</div>
</div>
<section></section>

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Motion at Constant Velocity

Physics is a science discipline concerned with the nature and properties of matter and energy. Physics topics include mechanics, heat, light and radiation, sound, electricity, magnetism, the structure of atoms, and so on.

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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.


Motion at Constant Velocity

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