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      <text>5.2.2.5 Modelling with parabolas</text>
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      <text><![CDATA[<p></p><p>Consider the situation in which position is measured&nbsp;<b>downwards</b>&nbsp;from the cliff-top again but the villain is lying down on the cliff-top and throws the rock&nbsp;<b>upwards</b>&nbsp;with speed \(5\ \text{m s}^{-1}\). The distance it would travel in time \(t\) seconds if gravity were not acting would be \(-5t\) metres (distance is speed multiplied by time but in the negative \(s\) direction in this case). To obtain the resulting distance in the presence of gravity we add this to the distance function \(s=4.9t^2\) that applies when the rock is simply dropped. The appropriate quadratic function for \(s\) is now&nbsp;</p><p>\(\qquad s = 4.9t^2 - 5t\qquad (0\le t \le T)\)&nbsp;</p><p>The nature of this quadratic function means that for any given value of \(s\) there are&nbsp;<b>two</b>&nbsp;possible values of \(t\). If we write the function in a slightly different way, taking out a common factor of \(t\),&nbsp;</p><p>\(\qquad s = t(4.9t - 5)\qquad (0\le t \le T)\)&nbsp;</p><p>it is possible to see that \(s = 0\) at two different times. These are when \(t = 0\) and when \(4.9t-5=0\). The first possibility is consistent with the initial position of the rock. The second possibility gives \(t = \dfrac {5}{4.9}\) which is a little more than 1. The rock will be at the cliff-top level at two different times. It is there at the instant when it is thrown. It rises until its speed is zero and then descends, passing cliff-top level again on its way to impact with the ground below or with the vehicle roof. Since the initial motion of the rock is upwards and position is defined as positive downwards, the initial part of the rock’s path corresponds to negative \(s\). The parabola associated with the appropriate function crosses the \(s=0\) axis twice and has a vertex at which \(s\) is negative. A sketch of \(s\) against \(t\) for this case is shown in the figure below.<br></p><p><img src="https://stack2.maths.ed.ac.uk/demo2018/brokenfile.php#/13204/user/draft/71353977/5-2-fig-2-rock_position.svg" alt="" width="317" height="137" role="presentation"><img src="@@PLUGINFILE@@/5-2-fig-2-rock_position.svg" alt="" width="317" height="137" role="presentation" class="img-responsive atto_image_button_text-bottom"><br></p><p>Graph of&nbsp;rock's position&nbsp;(measured downwards) when rock is thrown upwards</p><p><p>Note that the form of the parabola makes it inevitable that, as long as it is plotted over a sufficiently wide range, and apart from its vertex, there will&nbsp;<b>always be two values</b>&nbsp;on the curve for each value of one of the variables. Which of these values makes sense in a mathematical model will depend on the modelling context. In each of the contexts mentioned so far in this Section each context has determined the part of the parabola that is of interest.</p><p>Note also that there is a connection between the vertex on a parabola and the point where the gradient of that parabola is zero. In fact these points are the same!</p></p><p></p>]]></text>
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