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Braking Distance
Task
The braking distance, in feet, of a car traveling at $v$ miles per hour is given by $$ d= 2.2v+\frac{v^2}{20}. $$
 What is the braking distance, in feet, if the car is going 30 mph? 60 mph? 90 mph?
 Suppose that the car took 500 feet to brake. Use your computations in part (a) to make a prediction about how fast it was going when the brakes were applied.
 Use a graph of the distance equation to determine more precisely how fast it was going when the brakes were applied, and check your answer using the quadratic formula.
IM Commentary
The purpose of this task is to give an application arising from a realworld situation in which a quadratic equation arises, and where it is natural to use a graphical method to find an approximate solution and the quadratic formula to find an exact solution. Notice that although the graphical method can give a good approximation for the particular value $d=500$ chosen here, the quadratic formula is necessary for expressing $v$ as a function of $d$ in general.
Solution
 \begin{eqnarray*} \mbox{At 30 mph, braking distance} &=& d = 2.2(30) + \frac{30^2}{20} = 111 \mbox{ ft.}\\ \mbox{At 60 mph, braking distance} &=& d = 2.2(60) + \frac{60^2}{20} = 312 \mbox{ ft.}\\ \mbox{At 90 mph, braking distance} &=& d = 2.2(90) + \frac{90^2}{20} = 603 \mbox{ ft.} \end{eqnarray*}

From our answers in part (a), we see that the speed for a braking distance of 500 ft is between 60 mph and 90 mph. Estimates within this range can vary, though based on the computations, it is reasonable to predict a velocity closer to 90 than to 60.

The figure shows a graph of $d = 2.2v + v^2/20$. We see that the speed corresponding to a braking distance of 500 ft is approximately 80 mph.
To find the speed more accurately, we use the quadratic formula to solve the quadratic equation: $$ 500 = 2.2v + \frac{v^2}{20} \mbox{ which is } \frac{1}{20}v^2+2.2v500=0, $$ giving $v=80.39$ and $v=124.39$. On physical grounds we want $v>0$, so $v=80.39$, which is in good agreement with our graphical estimate of 80 mph.
Braking Distance
The braking distance, in feet, of a car traveling at $v$ miles per hour is given by $$ d= 2.2v+\frac{v^2}{20}. $$
 What is the braking distance, in feet, if the car is going 30 mph? 60 mph? 90 mph?
 Suppose that the car took 500 feet to brake. Use your computations in part (a) to make a prediction about how fast it was going when the brakes were applied.
 Use a graph of the distance equation to determine more precisely how fast it was going when the brakes were applied, and check your answer using the quadratic formula.
Comments
Log in to commentHeidi says:
over 2 yearsI made a little TInspire document to use with this activity in class: https://www.dropbox.com/s/u17hoamt0ks25p2/Braking%20distance.tns?dl=0
with additional information from http://www.harristechnical.com/articles/skidmarks.pdf
kassymov says:
over 4 yearsI've been trying to better understand what differentiates algebra and functions in the treatment of several topics that appear in both. In addition to the part in the Algebra progression, these topics from commoncore.me were helpful: "the function f where f(x)=3x+4" and "ACED.2, FBF.1, or FLE.2". Based on that, I can see how this task relates to IF.2, IF.4, IF.7a, IF.8a and F.LE (no standard there deals directly with quadratic models, but, I guess, they are there in spirit). There is clearly a dependent and independent variable, in intent as well as in the presentation. Given the nature of the problem, increasing the precision through the quadratic formula does not add much to the approximate solution. So, the last part of the solution, which essentially connects the task to REI.4b is not needed. Among algebra standards, it seems like REI.10 and CED.2 show more direct relationships with this task.
I suspect that what belongs to algebra and what belongs to functions already causes a lot of confusion. I'm afraid, this task may contribute to the confusion rather than clarify the issue.
mprice says:
almost 5 yearsThe example ought to specify that braking distances (really, this seems to be stopping distance in fact) are in feet. Otherwise a natural assumption would be miles, given the units on speed.
Also, my assumption is that the stopping distance is a combination of reaction distance (linear term) and braking distance (quadratic term)?
Cam says:
almost 5 yearsThanks, good catch on units.
There was some discussion about stopping distance vs. braking distance in the review process for this problem. The decision we ended up making was that the braking distance was more appropriate since it is the one that is the attribute of the car itself; in contrast, the stopping distance involves the breaking distance, but also road conditions, visibility, reaction times, etc.