2 Modelling with parabolas

The function

$s=\frac{t^2}{9}\left(0\le t\le 30\right)$

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, for in Figure 4, i.e. the part corresponding to $-30\le t\le 0$ .

Figure 4 :

The parabolic form predicts that at $t=-30$ , the object was 100 m away and for $\left(-30\le t\le 0\right)$ 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 $\left(-30\le t\le 0\right)$ is negative while for $\left(0\le t\le 30\right)$ it is positive. This is consistent with the change in direction at $t=0$ .

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,

$v=9.8t\left(0\le t\le T\right)$

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 from the cliff-top and timing starts with release of the rock, so that $s=0$ when $t=0$ , the resulting function is

$s=4.9t^2\left(0\le t\le T\right)$

(Note that $s=4.9t^2$ is a particular case of a standard model for falling objects: $s=\frac{1}{2}g{t}^2$ .)

Task!

This Task refers to the model discussed above.

1. What kind of function is $s=4.9t^2$ ?

   Answer

   Quadratic, or parabolic

2. If the vehicle roof is 2 m above the ground and the cliff-top is 35 m above the ground, calculate a value for $T$ , the time when a rock falling from the cliff-top hits the car roof:

   Answer

   $t=T$ when $s=35-2=33$

   $\Rightarrow 33=4.9T^2\text{so}T=\sqrt{\frac{33}{4.9}}=2.5951\approx 2.6$ (only positive $T$ makes sense)

3. Given this value for $T$ sketch the function:

   Answer

   

   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:

   For $t\le 0,s$ would be constant ( $=0$ ) and for $0\le t\le T,s=4.9t^2$ .

   The corresponding graph would also have two parts:

   A flat line along the $s=0$ axis for $t\le 0$ and part of a parabola for $0<t\le T$ .

   A different form of quadratic function for position is appropriate if position is measured upwards as height $\left(h\right)$ above the ground below the cliff-top. This is given as

   $h=35-4.9t^2\left(0\le t\le 2.6\right)$

   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.

Task!

This Task refers to the model discussed above.

By comparing $h=35-4.9t^2$ with $y=a{x}^2+bx+c$ , deduce values for $a,b$ and $c$ and determine whether the parabola corresponding to this function has a highest or lowest point:

Answer

(b) Write down an appropriate function for the variation of $h$ with $t$ if height is measured upwards from the top of a 2 m high vehicle:

Answer

(c) Sketch this function:

Answer

Consider the situation in which position is measured downwards from the cliff-top again but the villain is lying down on the cliff-top and throws the rock upwards 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

$s=4.9t^2-5t\left(0\le t\le T\right)$

The nature of this quadratic function means that for any given value of $s$ there are two possible values of $t$ . If we write the function in a slightly different way, taking out a common factor of $t$ ,

$s=t\left(4.9t-5\right)\left(0\le t\le T\right)$

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 $t=\frac{5}{4.9}$ 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 Figure 5.

Figure 5 :

Task!

For the above modelling of falling rocks, calculate how high the rock rises after being thrown upwards at 5 m $s^{-1}$ . (Hint: use the previously determined value of the time when the rock reaches its highest point.)

Answer

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 always be two values 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.

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!