Acurve of radius 78m is banked for design speed of 85km/h. if coeficient of static friction is 0.3[wet pavement]at what range of speeds can a car safelymake the curve?

The speed of the car safely make the curve, V = 65.23 Km/h

Explanation:

Given,

The radius of the curve, r = 78 m

The designed speed of the curve, v = 85 km/h

                                                            = 23.61 m/s

The coefficient of static friction, μₓ = 0.3

The static friction of the curve is given by the relation,

                                       Fₓ = μₓ η

The acceleration responsible for the static friction

                                      aₓ  = μₓ x g

Substituting the values in the equation

                                      aₓ = 0.3 x 9.8

                                           = 2.94 m/s²

The designed acceleration of the curve

                                        a₀ = v²/r

                                           = 23.61²/78

                                          = 7.15  m/s²

The acceleration supported by the static friction, aₓ can be subtracted from the designed acceleration of the curve.

Therefore the net acceleration,

                                         a  = a₀ - aₓ

                                             = 7.15 - 2.94

                                              = 4.21 m/s²

The centripetal velocity associated with this acceleration is

                                           a = V²/r

∴                                         V² = a x r

                                          V = √(a x r)

                                              =√ (4.21 x 78)

                                               = 18.12 m/s      

                                               = 65.23 Km/h

Hence, the speed required by the car to safely make the curve is, V = 65.23 Km/h                                

the answer is a. at the equator