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      <text>9.1.3.4 Resolving vectors applications</text>
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      <text><![CDATA[<p>The need to <span><strong>resolve</strong></span> a vector along a given direction occurs in other areas. For example, as a police car or ambulance with siren operating passes by, the pitch of the siren appears to increase as the vehicle approaches and decrease as it goes away. This change in pitch is known as the <span><strong>Doppler effect</strong></span>. This effect occurs in any situation where waves are reflected from a moving object.</p>
<p>A radar gun produces a signal which is bounced off the target moving vehicle so that when it returns to the gun, which also acts as a receiver, it has changed pitch. The speed of the vehicle can be calculated from the change in pitch. The speed indicated on the radar gun is the speed directly towards or away from the gun. However it is not usual to place oneself directly in front of moving vehicle when using the radar gun (panel (a) of the diagram below). Consequently the gun is used at an angle to the line of traffic (panel (b)).</p>
<figure class="HELM_figure">
<img src="@@PLUGINFILE@@/9-1-3RadarGun.svg" alt="Two panels.  The first shows a radar gun positioned in line with an oncoming vehicle.  The second shows it offset by an angle theta from the vehicle's direction of motion" class="img-responsive atto_image_button_text-bottom" width="390" height="238">
</figure>

<p>This means that it registers only the component of the velocity towards the gun. Suppose that the true speed along the road is \(v\). Then the component measured by the gun (\(v\cos\theta\)) is less than \(v\).</p>
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<h4 class="HELM_example">Example<br></h4>
<p>A safety inspector wishes to check the speed of a train along a straight piece of track. She stands \(10\ \mathrm{m}\) to the side of the track and uses a very accurate radar speed gun. If the reading on the gun is to be within 5% of the true speed of the train, how far away from the approaching train should the reading be taken?</p>
<h4 class="HELM_solution">Solution</h4>
<figure class="HELM_figure">
<img src="@@PLUGINFILE@@/9-1-3-2ExampleRadarGunSoln.svg" alt="A radar gun is located a distance, d, diagonally from a train moving horizontally to the left at speed,v.  The radar gun is a distance, s, vertically below the line of the train's movement..  " class="img-responsive atto_image_button_text-bottom" width="312" height="156">
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<p>For an error of 5%, the gun should read \(0.95|{\underline v}|\).  So \[|{\underline v}|\cos\theta = 0.95|{\underline v}|\] or \[\cos\theta = 0.95\]
If the distance to the side of the track at which the gun is used is \(s\) m and the distance between the radar gun and the train is \(d\) m, then \[\sin\theta = \frac{s}{d}\]
Here \(s=10\) so \[\sqrt{1-\cos^2\theta}=\frac{10}{d}\]
So if \(\cos\theta = 0.95\)
\[d=\frac{10}{\sqrt{1-(0.95)^2}}=32.03\]
This means that the reading should be taken when the train is over \(32\ \mathrm{m}\) from the radar gun to ensure an error of less than 5%.</p>]]></text>
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