The Cadillac Escalade is a cross between a sports utility vehicle (SUV) and luxury car. Either way, it’s a big car. And it takes awhile to stop. One study showed that the 2010 Escalade traveling at 60 miles per hour takes about 144 feet to come to a complete stop from when the driver first hits the brakes. In fact, the braking distance of any car depends on how fast it is going. If someone is driving slowly they can stop in shorter distance than if they are driving fast. Which is why you should drive slowly on residential streets.
We would like to be able to calculate the braking distances at other speeds, so our two variables are
\begin{align*}
S \amp= \text{ speed (mph) } \sim \text{ indep} \\
B \amp= \text{ braking distance (feet) } \sim \text{ dep}
\end{align*}
Using the data and equations from physics, automobile analysts were able to determine that the equation relating these two variables is
\begin{equation*}
B=0.04S^2
\end{equation*}
Remember that the 0.04 written next to the \(S^2\) means they are multiplied. We might equally well have written
\begin{equation*}
B = 0.04 \ast S^2
\end{equation*}
You may be a little surprised to see the variable \(S\) squared (raised to the 2nd power) or wonder what the number 0.04 means. This equation is not something we can figure out because it relies both on the data and knowledge of the physics involved. But, we can still work with this equation to find the braking distances at any speed. (If you must know, this equation is only approximate since things like tire and road conditions are a factor, but for what we want it is good enough.)
Although in the last couple of sections we were able to find equations by generalizing examples, there are actually many different mathematical and statistical techniques for finding equations. A scientist might use lab experiments and some theory to figure it out. An economist might recognize that the equation fits a certain template because of the underlying economics. A store manager might know from years of experience that a certain equation works well to predict sales. It can be comforting to know where an equation comes from but whether we find an equation for ourselves or get it from an expert, we can use it to answer our questions and make predictions.
Now that we have an equation we should check the reported stopping distance at 60 mph. We have \(S=60\) so we substitute 60 in place of \(S\) in the equation to get
\begin{equation*}
B = 0.04 \ast 60^2 = 0.04 \times \underline{60} \wedge 2 = 144 \text{ feet} \quad \checkmark
\end{equation*}
Quick bit of terminology. When we know the independent variable, like \(S=60\) and we substitute into the equation to find the dependent variable, like \(B\text{,}\) we say we evaluate the function \(B\) at \(S=60\text{.}\) You might have noticed that the 60 was underlined in the calculation above. In this book we underline the value of the independent variable when we are evaluating. That way it’s easier to see which numbers come from the equation and which number we’re plugging in. Feel free to do the same. Or not.
Let’s use our equation to calculate the braking distance for a Cadillac Escalade traveling 30, 50, 70 or 90 miles per hour. For 30 miles per hour, we have \(S=30\text{.}\) So, we evaluate at \(S=30\) to get
\begin{equation*}
B = 0.04 \ast 30^2 = 0.04\times \underline{30} \wedge 2 = 36 \text{ feet}
\end{equation*}
At 30 mph, it takes the Cadillac Escalade 36 feet to stop. As we expected, it doesn’t take nearly as far to stop as it did at 60 mph. For the other speeds we do the same thing: evaluate at the appropriate value of \(S\text{.}\) When \(S=50\) mph we get
\begin{equation*}
B = 0.04 \ast 50^2 = 0.04\times \underline{50}\wedge2 = 100 \text{ feet}
\end{equation*}
When \(S=70\) mph we get
\begin{equation*}
B = 0.04 \ast 70^2 = 0.04\times \underline{70}\wedge2 = 196\text{ feet}
\end{equation*}
When \(S=90\) mph we get
\begin{equation*}
B = 0.04 \ast 90^2 = 0.04\times \underline{90}\wedge2 = 324 \text{ feet}
\end{equation*}
And, what does our equation tell us when the speed is 0 mph? We evaluate at \(S=0\) mph to get
\begin{equation*}
B = 0.04 \ast 0^2 = 0.04\times \underline{0} \wedge 2 = 0 \text{ feet}
\end{equation*}
Well, sure! If the car isn’t moving, then it won’t need any distance to stop. Here’s what we’ve found so far, displayed in a table.
\(S\) |
0 |
30 |
50 |
60 |
70 |
90 |
\(B\) |
0 |
36 |
100 |
144 |
196 |
324 |
My neighbor Jeff happens to drive a 2010 Cadillac Escalade. The other day he almost was in an accident on the highway. Luckily no one was hurt, but he had to slam on the brakes to stop. The police report mentioned they believe it took his car 183 feet to stop. Jeff says he was not driving over the posted speed limit of 65 mph. Should we believe him?
We can see from the table that braking distance of 183 feet falls in between the 144 and 196 on our table which leads us to believe that Jeff was traveling faster than 60 mph and slower than 70 mph. We can figure out if Jeff were driving at 65 mph, then his braking distance would have been
\begin{equation*}
B = 0.04 \ast 65^2 = 0.04\times \underline{65} \wedge 2 = 169 \text{ feet}
\end{equation*}
That’s less than the 183 feet Jeff took to stop. So, it appears that Jeff was driving faster than 65 mph.
But wait a minute. The braking distance is just the time it takes from when the driver’s foot hits the brake until the car stops. That distance doesn’t take into account the driver’s reaction time - how long between when the driver thinks to stop and when the driver’s foot actually hits the brake. We have a new dependent variable
\begin{align*}
D \amp= \text{ total stopping distance (feet) } \sim \text{ dep} \\
S \amp= \text{ speed (mph) } \sim \text{ indep}
\end{align*}
How can we include this reaction time into an equation? Suppose it takes 1 second to react. We would like to know how many feet that adds to the equation. This is something we can figure out. We know the speed and the time, so multiply them to get the distance, right? One small snag: the speed is in mph (miles per hour). We need to convert units.
\begin{align*}
1 \cancel{\text{ sec}} \ast
\frac{1 \cancel{\text{ min}}}{60 \cancel{\text{ sec}}} \ast
\frac{1 \text{ hour}}{60 \cancel{\text{ min}}} \ast
\frac{5{,}280\text{ feet}}{1 \text{ mile}} \amp =
1 \div 60 \div 60 \times 5{,}280\\
\amp= 1.4666\ldots \approx 1.47 \text{ feet per mph}
\end{align*}
added to the stopping distance. Notice the fancy fraction work with the units?
\begin{equation*}
\frac{\text{hour}\ast \text{feet}}{\text{mile}}=\frac{\text{feet}}{\quad \frac{\text{miles}}{\text{hour}}\quad~}=\frac{\text{feet}}{\text{mph}}
\end{equation*}
What all this mess means is that we should add 1.47 feet for every mph of speed. So \(S\) mph adds \(1.47\ast S\) feet to the stopping distance. Long story short, our new equation is
\begin{equation*}
D=0.04S^2+1.47S
\end{equation*}
Something interesting about this equation. The independent variable \(S\) appears twice; first for the braking distance and again because of the reaction time. When we evaluate the equation we need to plug in the value of \(S\) in two places. Check it out. When \(S=30\) mph we have
\begin{equation*}
D = 0.04 \ast 30^2 + 1.47 \ast 30 = 0.04\times \underline{30} \wedge 2 +1.47 \times \underline{30} = 80.10 \approx 80 \text{ feet}
\end{equation*}
That’s a lot further than just the braking distance of 36 feet. When \(S=50\) mph we have
\begin{equation*}
D = 0.04 \ast 50^2 +1.47\ast 50 = 0.04\times \underline{50} \wedge 2 +1.47\times \underline{50} = 173.50 \approx 174 \text{ feet}
\end{equation*}
again, much more than the braking distance of 100 feet. Here’s the revised table of values.
\(S\) |
0 |
30 |
50 |
60 |
70 |
90 |
\(D\) |
0 |
80 |
174 |
232 |
299 |
456 |
These numbers make us rethink Jeff’s assertion. Given that he stopped in 183 feet, which is much less than the 232 feet it takes to stop at 60 mph, it looks like Jeff was driving less than 60 mph. To be sure, calculate that at 65 mph, it would have taken Jeff
\begin{equation*}
D = 0.04 \ast 65^2 + 1.47 \ast 65 = 0.04\times \underline{65} \wedge 2+ 1.47\times \underline{65} =264.55 \approx 265 \text{ feet}
\end{equation*}
Again, we should believe Jeff. And, be glad nobody was hurt.
A quick glance at the graph confirms our findings.
The distance of 183 feet falls just above the unlabeled gridline for 175 and below the gridline for 200. Looking at the corresponding point on the braking distance curve, it looks like it falls around 67 mph, but looking at the corresponding point on the stopping distance curve, it looks like just over 50 mph.
By the way, our first equation
\begin{equation*}
B = 0.04 \ast S^2
\end{equation*}
is a power equation because the independent variable is being raised to a power, \(n=2\text{,}\) and then scaled by a proportionality constant, \(k=0.04\text{.}\) Any power equation fits this template.
Power Equation Template
\begin{equation*}
\text{dep} = k \ast \text{indep}^{n}
\end{equation*}
Our second equation
\begin{equation*}
D=0.04S^2 + 1.47S
\end{equation*}
is a polynomial equation because it includes both a linear term and powers. The exercises introduce more polynomial equations. Polynomials can have any powers, but in this equation the highest power happens to be 2. This type of polynomial equation has a special name. It is a quadratic equation. Any quadratic equation fits this template.
Quadratic Equation Template
\begin{equation*}
\text{dep} = a \ast \text{indep}^2 + b \ast\text{indep} + c
\end{equation*}
For our equation \(a=0.04\text{,}\) \(b=1.47\text{,}\) and the mysteriously missing \(c=0\text{.}\)