For a car that is traveling on a dry road at a speed of r miles per hour, the stopping...

1. Translate the problem requirements: We need to find how much greater the stopping distance is when speed doubles from r to 2r mph. This means calculating \(\mathrm{d(2r) - d(r)}\) where \(\mathrm{d = 0.045r^2 + 1.1r}\).
2. Calculate stopping distance at speed 2r: Substitute 2r into the formula \(\mathrm{d = 0.045r^2 + 1.1r}\) to find the stopping distance at the higher speed.
3. Set up the difference expression: Create the expression \(\mathrm{d(2r) - d(r)}\) to represent how much greater the stopping distance is at speed 2r compared to speed r.
4. Simplify to find the final answer: Combine like terms and simplify the algebraic expression to match one of the answer choices.

Execution of Strategic Approach

1. Translate the problem requirements

Let's start by understanding what we're being asked to find. We have a car traveling at speed r miles per hour, and we want to know how much further it will travel before stopping if we double its speed to 2r miles per hour.

Think of it this way: if you're driving at 30 mph versus 60 mph, how much more distance do you need to stop at the higher speed? We're looking for that difference.

The stopping distance formula is given as \(\mathrm{d = 0.045r^2 + 1.1r}\), where d is in feet and r is the speed in mph.

We need to find: (stopping distance at speed 2r) - (stopping distance at speed r)
Mathematically, this means we need to calculate \(\mathrm{d(2r) - d(r)}\).

Process Skill: TRANSLATE - Converting the word problem into a clear mathematical comparison

2. Calculate stopping distance at speed 2r

Now let's find the stopping distance when the car is traveling at 2r miles per hour. We substitute 2r into our formula wherever we see r:

Starting with: \(\mathrm{d = 0.045r^2 + 1.1r}\)
Substituting 2r for r: \(\mathrm{d(2r) = 0.045(2r)^2 + 1.1(2r)}\)

Let's work this out step by step:

• \(\mathrm{(2r)^2 = 4r^2}\)
• So \(\mathrm{0.045(2r)^2 = 0.045 \times 4r^2 = 0.18r^2}\)
• And \(\mathrm{1.1(2r) = 2.2r}\)

Therefore: \(\mathrm{d(2r) = 0.18r^2 + 2.2r}\)

3. Set up the difference expression

Now we need to find how much greater the stopping distance is at speed 2r compared to speed r. This means we subtract the original stopping distance from the new stopping distance:

Difference = \(\mathrm{d(2r) - d(r)}\)
Difference = \(\mathrm{(0.18r^2 + 2.2r) - (0.045r^2 + 1.1r)}\)

When we subtract, we need to distribute the negative sign to both terms in the second expression:
Difference = \(\mathrm{0.18r^2 + 2.2r - 0.045r^2 - 1.1r}\)

4. Simplify to find the final answer

Now we combine like terms to get our final answer:

Grouping the \(\mathrm{r^2}\) terms: \(\mathrm{0.18r^2 - 0.045r^2 = 0.135r^2}\)
Grouping the r terms: \(\mathrm{2.2r - 1.1r = 1.1r}\)

Therefore: Difference = \(\mathrm{0.135r^2 + 1.1r}\)

Process Skill: SIMPLIFY - Combining like terms to reach the cleanest form

Final Answer

The stopping distance for a car traveling at 2r mph is \(\mathrm{0.135r^2 + 1.1r}\) feet greater than for a car traveling at r mph.

Looking at our answer choices, this matches choice D: \(\mathrm{0.135r^2 + 1.1r}\)

To verify: At double the speed, a car needs significantly more distance to stop - not just twice as much, but even more due to the squared term in the formula. This makes intuitive sense because kinetic energy increases with the square of velocity.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding what "greater" means in context
Students may confuse "how many feet greater" with finding the total stopping distance at speed 2r, rather than finding the difference between the two stopping distances. They might think they only need to calculate \(\mathrm{d(2r)}\) instead of \(\mathrm{d(2r) - d(r)}\).

2. Incorrect interpretation of "2r miles per hour"
Some students might misread the problem and think they need to compare speeds of r and r+2 (adding 2 to the speed) rather than r and 2r (doubling the speed). This fundamental misreading would lead to entirely wrong calculations.

Errors while executing the approach

1. Algebraic errors when squaring (2r)
When substituting 2r into the formula, students often make the error \(\mathrm{(2r)^2 = 2r^2}\) instead of the correct \(\mathrm{(2r)^2 = 4r^2}\). This is a very common algebraic mistake that would give \(\mathrm{0.045(2r^2) = 0.09r^2}\) instead of the correct \(\mathrm{0.045(4r^2) = 0.18r^2}\).

2. Sign errors during subtraction
When calculating \(\mathrm{d(2r) - d(r) = (0.18r^2 + 2.2r) - (0.045r^2 + 1.1r)}\), students may forget to distribute the negative sign properly, writing it as \(\mathrm{0.18r^2 + 2.2r - 0.045r^2 + 1.1r}\) instead of \(\mathrm{0.18r^2 + 2.2r - 0.045r^2 - 1.1r}\).

3. Arithmetic errors when combining like terms
Students may make calculation mistakes when subtracting coefficients: \(\mathrm{0.18 - 0.045 = 0.135}\) or \(\mathrm{2.2 - 1.1 = 1.1}\). These seem simple but are common sources of errors under test pressure.

Errors while selecting the answer

1. Selecting the wrong expression due to calculation errors
If students made errors in the execution phase (particularly the \(\mathrm{(2r)^2}\) mistake), they might arrive at \(\mathrm{0.09r^2 + 1.1r}\) and look for that in the answer choices. Since this isn't available, they might select the closest-looking option like choice B (\(\mathrm{0.09r^2 + 2.2r}\)) without recognizing their error.

Alternate Solutions

Smart Numbers Approach

Step 1: Choose a smart number for r

Let's use \(\mathrm{r = 10}\) mph. This is a convenient number that will make our calculations clean and easy to verify.

Step 2: Calculate stopping distance at speed r = 10 mph

\(\mathrm{d(10) = 0.045(10)^2 + 1.1(10)}\)
\(\mathrm{d(10) = 0.045(100) + 11}\)
\(\mathrm{d(10) = 4.5 + 11 = 15.5}\) feet

Step 3: Calculate stopping distance at speed 2r = 20 mph

\(\mathrm{d(20) = 0.045(20)^2 + 1.1(20)}\)
\(\mathrm{d(20) = 0.045(400) + 22}\)
\(\mathrm{d(20) = 18 + 22 = 40}\) feet

Step 4: Find the difference

Difference = \(\mathrm{d(20) - d(10) = 40 - 15.5 = 24.5}\) feet

Step 5: Test answer choices with r = 10

We need to find which expression equals 24.5 when \(\mathrm{r = 10}\):

• \(\mathrm{0.045r^2 + 1.1r = 0.045(100) + 1.1(10) = 4.5 + 11 = 15.5}\) ❌
• \(\mathrm{0.09r^2 + 2.2r = 0.09(100) + 2.2(10) = 9 + 22 = 31}\) ❌
• \(\mathrm{0.135r^2 + 2.2r = 0.135(100) + 2.2(10) = 13.5 + 22 = 35.5}\) ❌
• \(\mathrm{0.135r^2 + 1.1r = 0.135(100) + 1.1(10) = 13.5 + 11 = 24.5}\) ✓
• \(\mathrm{3.3r^2 + 0.225r = 3.3(100) + 0.225(10) = 330 + 2.25 = 332.25}\) ❌

Answer: D

The smart numbers approach confirms that choice D gives us the correct difference in stopping distances.