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The rear wheels of a car crossed a certain line 0.5 second after the front wheels crossed the same line. If the centers of the front and rear wheels are 20 feet apart and the car traveled in a straight line at a constant speed, which of the following gives the speed of the car in miles per hour? (5280 feet = 1 mile)
Let's start by understanding what's happening in everyday terms. Imagine a car driving down the road at a steady speed. The car has front wheels and rear wheels that are 20 feet apart from each other. When this car crosses a line on the road, the front wheels cross first, and then exactly 0.5 seconds later, the rear wheels cross the same line.
What we need to find is the car's speed, but not just any speed - we need it in miles per hour specifically. This is our target.
The key insight here is that both the front and rear wheels are moving at exactly the same speed (they're part of the same car!), but they cross the line at different times because they're physically separated by 20 feet.
Process Skill: TRANSLATE - Converting the physical scenario into mathematical relationships
Here's the core logic: Since both wheels travel at the same speed, we can think of this problem as asking "How fast does something need to move to travel 20 feet in 0.5 seconds?"
Why 20 feet in 0.5 seconds? Because that's exactly what happened - in the time it took for the rear wheels to "catch up" to where the front wheels were (0.5 seconds), the entire car moved forward by the distance between the wheels (20 feet).
Using the most basic relationship in motion:
\(\mathrm{Speed} = \mathrm{Distance} ÷ \mathrm{Time}\)
So our speed in feet per second = 20 feet ÷ 0.5 seconds = \(\frac{20}{0.5}\) feet per second
Process Skill: INFER - Recognizing that the wheel separation distance equals the distance traveled in the time difference
Now we have the speed as \(\frac{20}{0.5}\) feet per second, but we need it in miles per hour. Let's convert step by step:
First, let's convert feet to miles:
Next, let's convert seconds to hours:
Rearranging: \(\left(\frac{20}{5280}\right) × \left(\frac{60^2}{0.5}\right)\) miles per hour
Our calculated expression is: \(\left(\frac{20}{5280}\right) × \left(\frac{60^2}{0.5}\right)\)
Looking at the answer choices, this matches exactly with choice A: \(\left(\frac{20}{5280}\right)\left(\frac{60^2}{0.5}\right)\)
Let's verify this makes sense:
The answer is A: \(\left(\frac{20}{5280}\right)\left(\frac{60^2}{0.5}\right)\)
This expression correctly captures the speed conversion from 20 feet per 0.5 seconds to miles per hour by converting distance units (feet to miles) and time units (0.5 seconds to hours).
1. Misunderstanding what distance to use in the speed calculation
Students often get confused about which distance matters here. They might think they need to find how far the entire car traveled during the 0.5 seconds, rather than recognizing that the key insight is simpler: since both wheels move at the same speed, the car travels exactly 20 feet (the wheel separation distance) in 0.5 seconds. This is because when the rear wheel reaches the line, it has traveled exactly the distance that separated it from the front wheel initially.
2. Incorrectly setting up the relative motion relationship
Some students may try to create complex equations involving the positions of both wheels over time, rather than recognizing this is fundamentally just a basic speed = distance/time problem. They might think the front and rear wheels have different speeds, or try to account for some kind of rotational motion, when in reality both wheels move at identical speeds since they're part of the same rigid car.
1. Unit conversion errors with time conversion
The trickiest part of the conversion is going from "per 0.5 seconds" to "per hour". Students often make errors here by either using 60 instead of \(60^2\) (forgetting that 1 hour = 3600 seconds, not 60 seconds), or by incorrectly handling the 0.5 in the denominator. They might write \(\frac{60}{0.5}\) instead of \(\frac{60^2}{0.5}\), not realizing they need to convert all the way from seconds to hours.
2. Incorrect order of operations in unit conversion
When converting from feet per 0.5 seconds to miles per hour, students might multiply or divide by conversion factors in the wrong order. For example, they might write \((20 × 5280)\) instead of \(\left(\frac{20}{5280}\right)\) when converting feet to miles, or they might place the time conversion factors in the wrong part of the fraction.
1. Choosing answer choices with incorrect time conversion factors
Even if students set up the problem correctly, they might select answer choice B with \(\left(\frac{60}{0.5}\right)\) instead of the correct choice A with \(\left(\frac{60^2}{0.5}\right)\). This happens because they recognize that 60 is related to time conversion but forget that they need \(60^2\) to convert from seconds to hours. Answer choice B would give speed in miles per minute, not miles per hour.