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
\n" ); document.write( "\n" ); document.write( "a. a car traveling at 60 miles per hour, how many feet does this car need to safely stop?\r
\n" ); document.write( "\n" ); document.write( "b. What is the maximum speed a car can be traveling if it has only 50 feet to stop?\r
\n" ); document.write( "\n" ); document.write( "Thank you for your help.
\n" ); document.write( "

Algebra.Com's Answer #97355 by checkley71(8403)\"\" \"About 
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:\"x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+\" we get:
\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( "
\n" );