A hiker walking at a constant rate of 4 miles per hour is passed by a cyclist traveling in the...

1. Translate the problem requirements: A cyclist passes a hiker, travels for 5 more minutes, then stops and waits. We need to find how long the cyclist waits from the stopping point until the hiker reaches that same location.
2. Calculate the gap created during cyclist's additional travel: Determine how far ahead the cyclist gets during the 5 minutes of continued travel after passing the hiker.
3. Determine hiker's catch-up time: Calculate how long it takes the hiker, walking at her constant speed, to cover this gap distance.

Execution of Strategic Approach

1. Translate the problem requirements

Let's break down what's happening in this problem step by step:

We have a hiker walking at 4 miles per hour. A cyclist passes her while traveling at 20 miles per hour in the same direction. After passing the hiker, the cyclist continues traveling for exactly 5 minutes, then stops and waits.

The question asks: How long must the cyclist wait until the hiker catches up to where the cyclist stopped?

Process Skill: TRANSLATE - Converting the word problem into a clear timeline of events

Think of it this way: At the moment the cyclist passes the hiker, they're at the same location. Then the cyclist pulls ahead for 5 minutes while the hiker keeps walking. We need to find how long it takes the hiker to walk to the cyclist's stopping point.

2. Calculate the gap created during cyclist's additional travel

During those 5 minutes after passing the hiker, both people are moving, but the cyclist is moving much faster.

Let's see how far each person travels in those 5 minutes:

The cyclist travels at 20 miles per hour. In 5 minutes (which is \(\frac{5}{60} = \frac{1}{12}\) of an hour):
Distance cyclist travels = \(20 \times \frac{1}{12} = \frac{20}{12} = \frac{5}{3}\) miles

The hiker travels at 4 miles per hour. In the same 5 minutes:
Distance hiker travels = \(4 \times \frac{1}{12} = \frac{4}{12} = \frac{1}{3}\) miles

The gap between them when the cyclist stops is:
Gap = \(\frac{5}{3} - \frac{1}{3} = \frac{4}{3}\) miles

So when the cyclist stops, the hiker is \(\frac{4}{3}\) miles behind the cyclist's position.

3. Determine hiker's catch-up time

Now we have a simpler situation: the cyclist is stopped, and the hiker needs to cover \(\frac{4}{3}\) miles at her walking speed of 4 miles per hour.

Time = Distance ÷ Speed
Time = \(\frac{4}{3} \text{ miles} ÷ 4 \text{ miles per hour}\)
Time = \(\frac{4}{3} \times \frac{1}{4} = \frac{1}{3}\) hours

Converting to minutes: \(\frac{1}{3} \text{ hours} \times 60 \text{ minutes per hour} = 20\) minutes

Process Skill: MANIPULATE - Converting between time units to match answer choices

4. Final Answer

The cyclist must wait 20 minutes for the hiker to catch up.

Looking at our answer choices, this matches choice C: 20.

Let's verify this makes sense: The cyclist had a 5-minute head start at much higher speed, creating a significant gap. It's reasonable that it would take the slower hiker 20 minutes to close that gap.

Common Faltering Points

Errors while devising the approach

1. Misunderstanding the timeline of events: Students often assume the cyclist stops immediately after passing the hiker, missing the crucial detail that the cyclist continues traveling for 5 more minutes before stopping. This leads to incorrectly calculating that the wait time equals the time for the hiker to cover the distance the cyclist traveled in 5 minutes.

2. Confusing what needs to be calculated: Some students calculate the total time from when the cyclist passes the hiker until they meet again, rather than specifically calculating the waiting time (time from when cyclist stops until hiker arrives). This misinterpretation of the question leads to including the 5-minute travel period in the final answer.

3. Setting up relative motion incorrectly: Students may try to solve this as a relative motion problem where both parties are moving, rather than recognizing it should be broken into two phases: (1) both moving for 5 minutes creating a gap, and (2) only the hiker moving to close that gap.

Errors while executing the approach

1. Time unit conversion errors: Students frequently make mistakes converting 5 minutes to hours (using \(\frac{5}{100}\) instead of \(\frac{5}{60}\)) when calculating distances, or converting the final answer from hours back to minutes (using \(\frac{1}{3} \times 100\) instead of \(\frac{1}{3} \times 60\)).

2. Fraction arithmetic mistakes: When calculating the gap distance \(\left(\frac{5}{3} - \frac{1}{3} = \frac{4}{3}\right)\), students may make computational errors with fractions, or when dividing fractions in the final step: \(\frac{4}{3} ÷ 4 = \frac{4}{3} \times \frac{1}{4} = \frac{1}{3}\).

3. Forgetting to account for both people moving: During the 5-minute period, students may calculate only how far the cyclist travels but forget that the hiker is also moving during this time, leading to an incorrect gap calculation.

4. Leaving answer in wrong units: Students may correctly calculate \(\frac{1}{3}\) hour but forget to convert to minutes, selecting a non-existent answer choice or becoming confused about their solution.

2. Adding the 5-minute travel time: Even with correct calculations, students might add the initial 5 minutes to their answer (getting 25 minutes total) because they misunderstand that the question asks specifically for waiting time, not total elapsed time.