Question 133136
d=.05s^2+1.1s
d=.05*60^2+1.1*60
d=.05*3600+66
d=180+66
d=246 feet to stop when traveling 60 mph.
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50=.05s^2+1.1s 
.05s^2+1.1s-50=0
using the quadratic equation:{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} we get:
s=(-1.1+-sqrt[1.1^2-4*.05*-50])/2*.05
s=(-1.1+-sqrt[1.21+10])/.1
s=(-1.1+-sqrt11.21)/.1
s=(-1.1+-3.348)/.1
s=(-1.1+3.348)/.1
s=(-1.1+3.348)/.1
s=2.248/.1
s=22.48 mph is the max speed for a 50 foot stopping distance.