Answer:
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[tex]_{b}[/tex]( Vâ² - Vâ²)) / ( 2g( ΡbÎź + [tex]f_{rl}[/tex] Âą sinâ [tex]_{g}[/tex]))
now given that the tracked is levelled, â [tex]_{g}[/tex] = 0, also Y[tex]_{b}[/tex] = 1.04 for level or flat road
Speed Vâ = 60mil/hr = (60Ă5280)/(1Ă60Ă60) = 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( (Ρb/100)(0.80) + (0.018) ¹ sin(0°)))
180ft = 8053.76 / ( 64.4)(0.008Ρb + 0.018)
180ft = 8053.76 / ( 0.5152Ρb + 1.1592)
180( 0.5152Ρb + 1.1592)  = 8053.76
( 0.5152Ρb + 1.1592) = 8053.76 /180
0.5152Ρb + 1.1592 = 44.7431
0.5152Ρb = 44.7431 - 1.1592
0.5152Ρb = 43.5839
Ρb = 43.5839 / 0.5152
Ρb = 84.596 â 84.6 %
Therefore, Â the unloaded braking efficiency is 84.6 %