document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=371997>Question 371997</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I own a 2006 Honda CRV. Last spring, I was still driving on the original tires. I looked at them and decided that I had worn down enough tread, so it was time to buy a whole new set of tires. Let us say that I began with 16-inch tires (11/16 inch of tire tread depth). If I drove at 60mph, how many revolutions per hour would one new tire make? Let us say that I drove 40,000 miles on those tires so that the tread depth wore down to 3/16 inch. If I drove at 60mph, how many revolutions per hour would one worn tire make? Calculate the difference in revolutions per hour between the new tire and the worn tire. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #265006 by </B><A HREF=/tutors/aboutme.mpl?userid=CharlesG2><B>CharlesG2(834)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=CharlesG2><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=265006\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> \"I own a 2006 Honda CRV. Last spring, I was still driving on the original tires. I looked at them and decided that I had worn down enough tread, so it was time to buy a whole new set of tires. Let us say that I began with 16-inch tires (11/16 inch of tire tread depth). If I drove at 60mph, how many revolutions per hour would one new tire make? Let us say that I drove 40,000 miles on those tires so that the tread depth wore down to 3/16 inch. If I drove at 60mph, how many revolutions per hour would one worn tire make? Calculate the difference in revolutions per hour between the new tire and the worn tire.\"\r<BR>\n" );
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document.write( "circumference = diameter * pi<BR>\n" );
document.write( "circumference new tire = 16 inches * pi = 16pi inches<BR>\n" );
document.write( "11/16 inch tire tread depth - 3/16 inch tire tread depth = 8/16 inch tire tread depth<BR>\n" );
document.write( "8/16 inch = 1/2 inch --> diameter old tire = 15.5 inches<BR>\n" );
document.write( "circumference old tire = 15.5pi inches<BR>\n" );
document.write( "60 mph = 60 miles in 1 hour<BR>\n" );
document.write( "new = 60 miles * 5280 ft/mile * 12 inches/ft * 1/16pi rev/inches<BR>\n" );
document.write( "new = 3801600/16pi revolutions<BR>\n" );
document.write( "new = 75630.428957 rounded to 6 places revolutions in 60 miles<BR>\n" );
document.write( "old = 60 * 5280 * 12 * 1/15.5pi revolutions<BR>\n" );
document.write( "old = 3801600/15.5pi revolutions<BR>\n" );
document.write( "old = 78070.120214 rounded to 6 places revolutions in 60 miles<BR>\n" );
document.write( "new - old = 75630.428957 - 78070.120214 = -2439.691257 revolutions<BR>\n" );
document.write( "75630 - 78070 = -2440 revolutions<BR>\n" );
document.write( "old tires make 2440 revolutions more in 60 miles at 60 mph\r<BR>\n" );
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document.write( "if i made a mistake, someone let me know, and i apologize if i did<BR>\n" );
document.write( " </SPAN>\n" );
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