Car Tire Wear and Odometer Accuracy Calculation

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The discussion centers on a homework problem involving tire wear and odometer accuracy. A person measures a drive at 37.35 miles with worn tires and 36.68 miles with new tires, indicating the new tires are larger. Calculations reveal the worn tire's diameter is approximately 3.93 meters, derived from the number of revolutions based on the odometer readings. The second part of the problem raises questions about calculating the amount of tire lost per revolution, highlighting a lack of information regarding the initial tire wear during the first trip. Overall, the conversation emphasizes the importance of precise measurements and critical thinking in physics-related problems.
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Even numbered book problem

Homework Statement


A person goes for a drive, and after, measured by the odometer, reads 37.35 miles in length. This person then replaces the tires with new ones and goes for the same drive, however, now the odometer reads 36.68 miles. Obviously the new tires are larger in diameter. The odometer counts the number of revolutions of a tire and then, using a calibrated diameter for said tire, converts this into a distance of travel.

1) Assume a tire diameter of 4 meters is the calibration entered into the car's odometer and calculate the diameter of the worn tire.

2) Can you calculate an approximation for the amount of tire lost per revolution? If you can, calculate it. If not, why?


Homework Equations




Circumference = pi x Diameter
Distance = Circumference x Revolutions

The Attempt at a Solution



First I want to calculate the first question, that is diameter of the worn tire. Before I do that I want to calculate the number of revolutions.

Revolutions = (37.35 miles)/(4 meters)(pi) ... After come unit conversions, I came up with 4780 revolutions.

Using that number, and 36.68 miles, I can calculate the diameter of the worn tires.

Diameter = (36.68 miles)/(4780 revolutions)(pi) = 3.93 meters. To me this makes sense because the diameter of the worn tire should be smaller.

Second part of the question,
Im assuming the question means, from the worn tires to the new tires, how much tire did I lose per revolution. This seemingly would translate to (change in circumference)/(change in revolution) . Though I feel I need to know more, and the above translation would not give me how much tire I lost per revolution. Thus part of me feels there is not enough information because I don't know how much tire I lost during the first trip that took 37.35 miles.
 
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Isn't the circumference of a circle 2pi r?

For the second part, I agree with your "answer", but not your reasoning. This is honestly just more of a critical thinking than a physics or anything.

Whats a good amount of tread to have on a tire? How long does a set of tires last roughly? You don't have to answer me, just think about it.
 
BiGyElLoWhAt said:
Isn't the circumference of a circle 2pi r?
Yes, and one diameter is 2r :wink:
For the second part, I agree with your "answer", but not your reasoning. This is honestly just more of a critical thinking than a physics or anything.
I suspect you are using the same reasoning, but would have expressed it in different words. OP says:
I don't know how much tire I lost during the first trip that took 37.35 miles
Seems fair enough to me.
 
Tires 4 meters in diameter? Those are some big tires. Must be driving some kind of Earth mover. :biggrin:

titan.jpg
 
doh! for some reason i took 4m to be the radius. Stupid words...

As for the second part well... I think I see what op means, but I still feel there's something missing... It's along the lines of what I'm thinking, but the word 'trip' is what's throwing me off. It's amazing what an apostraphe s can do to the meaning of a sentence.
 
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