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      <text>5.2.2.1 Modelling with parabolas</text>
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      <text><![CDATA[<p></p><h4>Modelling with parabolas</h4>The function&nbsp;<p>\(\qquad s = \dfrac {t^2}{9}\qquad (0\le t \le 30)\)&nbsp;</p><p></p><p>is part of a parabola starting at the origin (\(s = 0\) and \(t = 0)\) and rising to \(s = 100\) at the end of its range of validity. \(s\) represents the distance of the object from the origin - N.B. Do not confuse this \(s\) with the symbol for seconds. ‘Negative’ time corresponds to time before the motion of the object is being considered. What would this parabolic function have predicted if it were valid up to 30 s before the ‘zero’ time? The answer to this can be deduced from the left-hand part of the graph of the function shown as a dashed curve, in the figure below, i.e. the part corresponding to \(-30\le t \le 0\).</p><p><img src="@@PLUGINFILE@@/5-2-fig-1-quadratic.svg" alt="" width="287" height="188" role="presentation" class="img-responsive atto_image_button_text-bottom"><br></p><p><span>Graph of&nbsp;\(s = \dfrac{t^{2}}{9}\) for&nbsp;\(-30\le t \le 30\)</span></p><p>The parabolic form predicts that at \(t = -30\), the object was 100 m away and for \((-30\le t\le 0)\) it was moving towards the point at which the original timing started. The rate of change of position, or instantaneous velocity, is given by the gradient of the position-time graph. Since the gradient of the parabola for \(s\) is steeper near \(t = -30\) than near \(t = 0\), the chosen function for \(s\) and new range of validity suggests that the object was moving quickly at the start of the motion, slows down on approaching the initial starting point, and then moves away again accelerating as it does so. Note that the velocity (i.e. the gradient) for \((-30\le t \le 0)\) is negative while for \((0\le t \le 30)\) it is positive. This is consistent with the change in direction at \(t = 0\).</p><p>We will consider falling objects again and return to the context of the thriller film and the villain on a cliff-tip dislodging a rock. Suppose that, as film director, you are considering a variation of the plot whereby, instead of the ground, the rock hits the roof of a vehicle carrying the hero and heroine. This means that you might be interested in the position as well as the velocity of the rock at any time. We can start from the linear function relating velocity and time for the dislodged rock,&nbsp;</p><p>\(\qquad v = 9.8t\qquad (0\le t \le T)\)&nbsp;</p><p>where \(T\) represents the time at which the rock hits the roof of the vehicle. The precise value of \(T\) will depend upon the height of the vehicle. If \(s\) is measured <i>from the cliff-top</i> and timing starts with release of the rock, so that \(s = 0\) when \(t = 0\), the resulting function is</p><p>\(\qquad s = 4.9t^2\qquad (0\le t \le T)\)&nbsp;</p><p>(Note that \(s = 4.9t^2\) is a particular case of a standard model for falling objects: \(s = \frac{1}{2} gt^2\).)</p><p></p><p>In this modelling context, negative time would correspond to time before the villain dislodges the rock. It seems likely that the rock was stationary before this instant. The parabolic function would not be appropriate for \(t\le 0\) since it would predict that the rock was moving. An appropriate function would have two parts to its domain:</p><p>For \(t\le 0, s\) would be constant (\(=0\)) and for \(0\le t \le T,\ s = 4.9t^2\).&nbsp;</p><p>The corresponding graph would also have two parts:</p><p>A flat line along the \(s = 0\) axis for \(t\le 0\) and part of a parabola for \(0&lt; t \le T\).</p><p>A different form of quadratic function for position is appropriate if position is measured&nbsp;<b>upwards</b>&nbsp;as height \((h)\) above the ground below the cliff-top. This is given as&nbsp;</p><p>\(\qquad h = 35 - 4.9t^2\qquad (0\le t \le 2.6)\)&nbsp;</p><p>Note that once \(t=2.6\) then \(h=0\) and the rock cannot fall any further. When position is measured upwards, velocities and accelerations, which are downwards for falling objects, will be negative.</p>]]></text>
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