document.write( "Question 303155: A typical car's stopping distance on dry pavement \"d\" in feet can be approximated by the function
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document.write( "d= 0.034s2(squared) + 0.56s - 17.11. Where \"s\" is the speed in miles per hour, of the car before braking.\r
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document.write( "A. How fast is the car going if it requires 100 feet for the car to stop after the brakes are applied?
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Algebra.Com's Answer #217311 by nerdybill(7384)![]() ![]() You can put this solution on YOUR website! A typical car's stopping distance on dry pavement \"d\" in feet can be approximated by the function\r \n" ); document.write( "\n" ); document.write( "d= 0.034s2(squared) + 0.56s - 17.11. Where \"s\" is the speed in miles per hour, of the car before braking.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "A. How fast is the car going if it requires 100 feet for the car to stop after the brakes are applied? \n" ); document.write( ". \n" ); document.write( "Replace d with 100 and solve for s: \n" ); document.write( "d= 0.034s^2 + 0.56s - 17.11 \n" ); document.write( "100= 0.034s^2 + 0.56s - 17.11 \n" ); document.write( "0= 0.034s^2 + 0.56s - 117.11 \n" ); document.write( "Solve using the quadratic formula. Doing so yields: \n" ); document.write( "s = {51.0, -67.5} \n" ); document.write( "Toss out the negative solution -- doesn't make sense. \n" ); document.write( "so, the car was moving at 51 mph. \n" ); document.write( ". \n" ); document.write( "Details of quadratic formula: \n" ); document.write( ". \n" ); document.write( "
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