A level test track has a coefficient of road adhesion of 0.80, and a car being tested has a coefficient of rolling friction that is approximated as 0.018 for all speeds. The vehicle is tested unloaded and achieves the theoretical minimum stop in 180 ft (from brake application). The initial speed was 60 mi/h. Ignoring aerodynamic resistance, what is the unloaded braking efficiency

the unloaded braking efficiency is 84.6 %

Explanation:

Given the data in the question;

by Ignoring aerodynamic resistance; we can find the theoretical stopping distance using the following formula

S = (Y( V₁² - V₂²)) / ( 2g( ηbμ + ± sin∅))

now given that the tracked is levelled, ∅ = 0, also Y = 1.04 for level or flat road

Speed V₁ = 60mil/hr = (60×5280)/(1×60×60) = 316800ft/3600s = 88ft/s

now, we substitute in our values to get the braking efficiency;

180ft = (1.04( (88ft/s)² - 0²)) / ( 2(32.2( (ηb/100)(0.80) + (0.018) ± sin(0°)))

180ft = 8053.76 / ( 64.4)(0.008ηb + 0.018)

180ft = 8053.76 / ( 0.5152ηb + 1.1592)

180( 0.5152ηb + 1.1592)  = 8053.76

( 0.5152ηb + 1.1592) = 8053.76 /180

0.5152ηb + 1.1592 = 44.7431

0.5152ηb = 44.7431 - 1.1592

0.5152ηb = 43.5839

ηb = 43.5839 / 0.5152

ηb = 84.596 ≈ 84.6 %

Therefore,  the unloaded braking efficiency is 84.6 %

there are no correct statementd on the list of choices you provided.

double the distance = divide strength by 4