Question

 

A 14.6-kN vehicle (CD = 0.30, Af = 2.6 m2, ρ = 1.2356 kg/m3) is driven on a surface with a coefficient of adhesion of 0.4, and the coefficient of rolling friction is approximated as 0.018 for all speeds. Assuming minimum theoretical stopping distances, if the vehicle comes to a stop 82 m after brake application on a level surface and has a braking efficiency of 0.82, what was its initial speed (a) if aerodynamic resistance is considered and (b) if aerodynamic resistance is ignored?

Answer 1

nowo weight of the vehicle = w=14.6 KN coefficient of adhesion Up = 0.30, A = 216 m2, e = 1.2356 Vg/m3 tolal fracties for fe = for W t o GA anatuw due to friction where force due to rolling friction - he W ffy with fee = 0.018: Now is the where Ffx 0.82 ww is a un decceleration def . due to friction waking efficiency is 0.82 . fee W XO'82 "/g - la xq x0182 mass and Mag = .00144 m/s2 adhesive force and = = 0.018 x 19 870.82 had resistance due se FA = MW on due to = FAX adhesion 0.82 now.decceleration dad Mo uw x0 182 aug 40.82 - w/g - 0.4 x 9.8 x0.82... - 3.2144 m/st. now total dicceleration without avrodynamic drag = d= 0.144 +3.2144= 3.36 M2/32 : Let initial velocity here, and final relosely le & , then when va dt = 3.36 x 82860 m/ &=& dt o endt - where t= 82x60 s. = = 16.5 km/s.

(a) now, if aire drag is considered a then force due to air drag = Falue = vor CoA. % decceleration due to aire decag dair = Faire x0.82 Mo m/s2 = ;xlu ² G A x0.82x9 W 1 x (1605x103) X0.30 X 206 xo.82 x 9.8 12X.12356: 1406x103 103 and = 72 m/s2 - 144x10-5 now let the initial speed be a u final speed is r=0. Then → va o (dat dad + dain) xt. - u - (0.144 + 30 2144 + + 4x100x7 DX 82X 60 = 16.523 km/s.