Question 40903: I need help...here is my question and the start to my answer...please help
(Q) 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?
(A) The formula for circumference is 2 * pi * radius ( or pi * diameter)
So the circumference for the 16 in tire would be 16*pi = 50.26
The circumference of the 16 in. tire is : 16 * pi = 50.26
In 2000 miles, the 16 in. tire has rotated 2000/50.26 = 39.79 times
So, the 16 in. tire has done more work.
Found 2 solutions by Nate, checkley71:
Answer by Nate(3500)   (Show Source):
You can put this solution on YOUR website! 2000 miles = x inches
126,720,000 inches
First car's tire:
 is about 47.12
126,720,000/47.12 = about 2,689,303 rotations
Second car's tire:
 is about 50.27
126,720,000/50.27 = about 2,520,787 rotations
The first car's tire rotates more, so there is probably more 'ware and tare'.
Answer by checkley71(8403) (Show Source):
You can put this solution on YOUR website! THE 16 IN TIRE WILL ROTATE ONLY 15/16 TIMES THE 15 INCH TIRE THEREFORE THE WEAR OF THE 16 INCH TIRE IS ONLY 15/16 TIMES THE 15 INCH TIRE.
ANOTHER WAY OF EVALUATING THIS PROBLEM THE 16 INCH TIRE WILL HAS 16/15 OR 1.0667 MORE TREAD AND TRAVELS 1.0667 TIMES THE 15 INCH TIRE THUS THE 16 INCH TIRE ROTATES ONLY 15/16 OR .9375 TIMES THE 15 INH TIRE.
SUGGESTION ALWAYS BUY/USE THE BIGGER TIRE FOR LONGER WEAR.
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