document.write( "Question 332220: The braking distance (in feet) of a car going V mph is given by v is greater or equal to 0.\r
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document.write( "how fast would the car have been traveling for a braking distance of 150feet? round to nearest mile per hour. \n" );
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Algebra.Com's Answer #238118 by ankor@dixie-net.com(22740)![]() ![]() You can put this solution on YOUR website! The braking distance (in feet) of a car going V mph is given by \n" ); document.write( " \n" ); document.write( "how fast would the car have been traveling for a braking distance of 150feet? \n" ); document.write( " round to nearest mile per hour. \n" ); document.write( ": \n" ); document.write( "Write it: \n" ); document.write( " \n" ); document.write( "multiply by 20, results \n" ); document.write( "v^2 + 20v = 20(150) \n" ); document.write( ": \n" ); document.write( "v^2 + 20v = 3000 \n" ); document.write( ": \n" ); document.write( "v^2 + 20v - 3000 = 0 \n" ); document.write( "Solve for v using the quadratic formula \n" ); document.write( " \n" ); document.write( "in this equation; x=v; a=1; b=20; c= -3000 \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( ": \n" ); document.write( " \n" ); document.write( "Two solutions, we only want the positive solution \n" ); document.write( " \n" ); document.write( "v = \n" ); document.write( "v = 45.68 mph for a stopping distance of 150 ft \n" ); document.write( ": \n" ); document.write( ": \n" ); document.write( "See if that flies in the original equation \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "d(v) = 150.00, confirms our solution \n" ); document.write( " |