Answer:
1. 
2. 163.3 m
Explanation:
Static friction between road and rubber, μs =0.06
The maximum speed of the car, v = 50 km/h
                                               = (50)(1000/3600) m/s
                                                = 13.89 m/s
The acceleration due to gravity, 
The frictional force, f = μsN     ...... (1)
 The component mg cosθ which balance the normal reaction N
The component mg sinθ acts in an opposite direction to the frictional force f.
         ΣF = mg sinθ-f = 0      ...... (2)
Substitute the equation (1) in equation (2), we get
 ΣF = mgsinθ-μsN = 0
 mgsinθ-μsmgcosθ =0
 μs = sinθ/cosθ
    tanθ = μs
     θ = tan-1( μs) = tan-1(0.06) = 
(b)The vertical component of the force is
N cosθ = fsinθ+mg
 N cosθ = μsNsinθ+mg
 N[cosθ- μs sinθ] = mg     ...... (3)
The horizontal component of the force along the motion of the car is
 Nsinθ+fcosθ = ma  (Centripetal acceleration, 
   Nsinθ+fcosθ = 
    Nsinθ+μsNcosθ = 
 N[sinθ+μs cosθ] =  ...... (4)
     ...... (4)    
Dividing the equation (4) with equation (3),
 [sinθ+μscosθ]/[cosθ- μs sinθ] = 
 cosθ[sinθ/cosθ+μs]/cosθ[1- μs sinθ/cosθ] =
 [tanθ+μs]/[1-μs tanθ] =  
      
  From part (1), tanθ = μs
 Then the above equation becomes
 ![\frac {(\mu_s+\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%28%5Cmu_s%2B%5Cmu_s%5D%7D%7B%5B1-%5Cmu_s%5E%7B2%7D%5D%7D%20%3D%5Cfrac%20%7Bv%5E%7B2%7D%7D%7Brg%7D)
![\frac {(2\mu_s]}{[1-\mu_s^{2}]} =\frac {v^{2}}{rg}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%282%5Cmu_s%5D%7D%7B%5B1-%5Cmu_s%5E%7B2%7D%5D%7D%20%3D%5Cfrac%20%7Bv%5E%7B2%7D%7D%7Brg%7D)
Therefore, the minimum radius of the curvature of the curve is
                ![r = \frac {v^{2}}{{2 \mu_s/[1-\mu_s^{2}]}g}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%20%7Bv%5E%7B2%7D%7D%7B%7B2%20%5Cmu_s%2F%5B1-%5Cmu_s%5E%7B2%7D%5D%7Dg%7D) 
 
                    = ![\frac {v^{2}[1-\mu_s^{2}]}{2\mu_s g}](https://tex.z-dn.net/?f=%5Cfrac%20%7Bv%5E%7B2%7D%5B1-%5Cmu_s%5E%7B2%7D%5D%7D%7B2%5Cmu_s%20g%7D)
                    = ![\frac {(13.89 m/s)^{2}[1-(0.06)^{2}]}{(2)(0.06)(9.81)}](https://tex.z-dn.net/?f=%5Cfrac%20%7B%2813.89%20m%2Fs%29%5E%7B2%7D%5B1-%280.06%29%5E%7B2%7D%5D%7D%7B%282%29%280.06%29%289.81%29%7D)
                  = 163.3 m