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Fri 11 Nov, 2005 11:11 pm
Two cars with new tires are driven at an average speed of 60 mph for a test drive of 2000 miles. The diameter of the wheels of one car is 15 inches. The diameter of the wheels of the other car is 16 inches. If the tires are equally durable and differ only by diameter, which car will probably need new tires first? Why?
Both are driven 2000 miles? Neither will ever wear out.
How does this relate to trigonmetry, by the way?
May be the car with smaller diameter will wear out first
because it makes more contact with the road than the bigger one
Number of revolutions
A 15 inch tire will make more revolutions than a 16 inch tire over the same distance.
(2000 miles)(5280 feet/mile)(12 inches/foot) = 1.2672E8 inches
1.2672E8/(2*pi*15) = 1.34E6 revolutions
1.2672E8/(2*pi*16) = 1.26E6 revolutions
More revolutions on the smaller tire, but exactly the same amount of miles (highway contact). Possibly a slightly greater on the tires carrying the heaver load, which is not given. I'm guessing treadwear will be the same, given the same road, vehicle weight loading, driving style, and so forth.
In other words, I don't think we have a solveable problem here.
Since the tread thicknesses are the same, the larger tire has more tread volume. Its tread life will be longer than that of the smaller tire.
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