Question:
When new, a rubber tire has an outside diameter of 24 inches. If the tire wears down 0.1 inch, about how many more times will the wheel go around a mile?
Please show work.
2007-05-06 04:30:39 · 4 answers · asked by chi_town_legnd 1 in Science & Mathematics ➔ Mathematics
If its diameter is 2 feet, its circumference is 2*pi feet, or about 6.28. In one mile (5280 feet), that tire would make (5280/6.28) = 840.3 rotations.
If the tire wears down 0.1 inches, its diameter is 0.2 inches less (because it wears down 0.1 inches on both ends of where the diameter is measured), leaving a new diameter of about 1.983 feet. (24 inches - 0.2 inches = 23.8 inches. 23.8 inches / 12 = 1.983 feet.)
The tire's circumference is now about 6.23 feet. In one mile, that tire would make (5280/6.23) or about 847.4 rotations.
It's about seven more (from 840 to 847) rotations per mile.
Note that the problem isn't entirely clear, in the way it is worded. One might take the wording to mean that the "diameter" has worn down 0.1 inches, instead of the "tire tread" wearing down 0.1 inches. That would change the diameter by 0.1 inches instead of 0.2 inches, which would change the answer.
2007-05-06 04:37:06 · answer #1 · answered by McFate 7 · 0⤊ 0⤋
1 mile = 5280 ft = 63360 in
63360[1/(2pi*23.9)-1/(2pi*24)] = 1.76 more times
2007-05-06 11:41:44 · answer #2 · answered by sahsjing 7 · 0⤊ 0⤋
There is nowhere near enough information given to solve this problem.
Doug
2007-05-06 11:40:07 · answer #3 · answered by doug_donaghue 7 · 0⤊ 0⤋
0.1 inches every mile, meter, km???
2007-05-06 11:36:07 · answer #4 · answered by alexandra 2 · 0⤊ 0⤋