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Answer 1 · 2017-07-11T21:17:34

#t = 10.7# #"s"#

The function #d(t)# gives the value of how much a car will skid for some time #t#.

Let's set #d(t) = 190#:

#Rightarrow d(t) = 190#

#Rightarrow 11 t^(2) - 100 t = 190#

Subtracting #190# from both sides of the equation:

#Rightarrow 11 t^(2) - 100 t - 190 = 0#

#Rightarrow t = frac(- (- 100) pm sqrt((- 100)^(2) - 4(11)(- 190)))(2(11))#

#Rightarrow t = frac(100 pm sqrt(10,000 + 8360))(22)#

#Rightarrow t = frac(100 pm sqrt(18,360))(22)#

#Rightarrow t approx frac(100 pm 135.5)(22)#

#Rightarrow t approx frac(100 - 135.5)(22), frac(100 + 135.5)(22)#

#Rightarrow t approx - 1.6, 10.7#

However, #t ne - 1.6#

Therefore, the time #t# it would take the car to skid #190# #"ft"# is around #10.7# #"s"#.