if F(s) = 1.1s + .054s2 then df = 1.1 + .108s ds
at s = 25 dF = 1.1 + .108(25)(1) = 3.8 ft
for s = 65 dF = 1.1 + .108(65)(1) = 8.12 ft
Erik K.
asked 11/17/18The stopping distance for an automobile is F(s) = 1.1s + 0.054s2 ft, where s is the speed in mph.
Use the Linear Approximation to estimate the change in stopping distance per additional mph when s = 25.
Use the Linear Approximation to estimate the change in stopping distance per additional mph when s = 65.
if F(s) = 1.1s + .054s2 then df = 1.1 + .108s ds
at s = 25 dF = 1.1 + .108(25)(1) = 3.8 ft
for s = 65 dF = 1.1 + .108(65)(1) = 8.12 ft
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