If the coefficient of kinetic friction between tires and drypavenment is .80, wh

Question

If the coefficient of kinetic friction between tires and drypavenment is .80, what is the shortest distance in which you canstop an automobile by locking the brakes when traveling at 29.1 m/s(about 65mi/h)? d=                       m Part two On wet pavement, the coefficient of kinetic friction may beonly .25. How fast should you drive on wet pavement in order to beable to stop in the same distance as in part one. v=        m/s If the coefficient of kinetic friction between tires and drypavenment is .80, what is the shortest distance in which you canstop an automobile by locking the brakes when traveling at 29.1 m/s(about 65mi/h)? d=                       m Part two On wet pavement, the coefficient of kinetic friction may beonly .25. How fast should you drive on wet pavement in order to beable to stop in the same distance as in part one. v=        m/s

Explanation / Answer

1) vi = 29.1 m/s k = .8 Notation - Fn(normal force)Fg(gravitational force) Ffr(Friction) Vertical component of force
Fy= Fn - Fg =ma.   Since a = 0 in the vertical Fn = Fg = mg Ffr = Fnk =mgk Fx = -Ffr or ma = -mgk a = -gk Now ... vf2 = vi2 + 2ad,where vf = 0 d = vi2/-2a =vi2/-2(-gk) d = (29.1m/s)2 / 2(9.8)(.8) = 54.006 m 2) vf2 = vi2 +2ad, where vf = 0 vi = -2ad =-2(-gk)d vi = 2(9.8)(.25)(54.006) = 16.27 m/s Now ... vf2 = vi2 + 2ad,where vf = 0 d = vi2/-2a =vi2/-2(-gk) d = (29.1m/s)2 / 2(9.8)(.8) = 54.006 m 2) vf2 = vi2 +2ad, where vf = 0 vi = -2ad =-2(-gk)d vi = 2(9.8)(.25)(54.006) = 16.27 m/s

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