A curve 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