document.write( "Question 133136: The formula d = 0.05s^2 + 1.1s estimates the minimum stopping distance d in feet for a car traveling s miles per hour.\r
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document.write( "a. a car traveling at 60 miles per hour, how many feet does this car need to safely stop?\r
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document.write( "b. What is the maximum speed a car can be traveling if it has only 50 feet to stop?\r
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document.write( "Thank you for your help. \n" );
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Algebra.Com's Answer #97355 by checkley71(8403)![]() ![]() ![]() You can put this solution on YOUR website! d=.05s^2+1.1s \n" ); document.write( "d=.05*60^2+1.1*60 \n" ); document.write( "d=.05*3600+66 \n" ); document.write( "d=180+66 \n" ); document.write( "d=246 feet to stop when traveling 60 mph. \n" ); document.write( "----------------------------------------------------- \n" ); document.write( "50=.05s^2+1.1s \n" ); document.write( ".05s^2+1.1s-50=0 \n" ); document.write( "using the quadratic equation: \n" ); document.write( "s=(-1.1+-sqrt[1.1^2-4*.05*-50])/2*.05 \n" ); document.write( "s=(-1.1+-sqrt[1.21+10])/.1 \n" ); document.write( "s=(-1.1+-sqrt11.21)/.1 \n" ); document.write( "s=(-1.1+-3.348)/.1 \n" ); document.write( "s=(-1.1+3.348)/.1 \n" ); document.write( "s=(-1.1+3.348)/.1 \n" ); document.write( "s=2.248/.1 \n" ); document.write( "s=22.48 mph is the max speed for a 50 foot stopping distance. \n" ); document.write( " |