Two curves on a highway have the same radii. However, one is unbanked and the other is banked at an angle ɵ. A car can safely travel along the unbanked curve at a maximum speed v under conditions when the coefficient of static friction between the tires and the road is μs = 0.787. The banked curve is frictionless, and the car can negotiate it at the same maximum speed v. Find the angle ɵ of the banked curve.

Explanation:

For unbanked curve , centripetal force is provided by friction and for banked curve , centripetal force is provided by component of reaction from the ground.

centripetal force = m v² / r , m is mass , v is velocity and r is radius of curve.

friction force ( maximum ) = μ mg , μ is coefficient of friction ,

for motion on unbanked curve

m v² / r  =  μ mg

v² = μ gr

= .787 x 9.8 x r

= 7.712 r

In case of motion on banked curve having angle of banking θ

If R be the ground reaction

R cosθ = mg

Rsinθ = centripetal force

= m v² / r

Dividing the two

Tanθ = v² / rg

=  7.712 r / 9.8 r

tanθ = .787

θ = 38 degree .

The angle ɵ of the banked curve of the given highway is 38.2⁰.

The net force on the car for the unbanked curve is calculated as follows;

The normal force on the car for the banked curve is calculated as follows;

The horizontal force on the car for the banked curve is calculated as follows;

Divide 2 by 1;

Thus, the angle ɵ of the banked curve of the given highway is 38.2⁰.

Learn more about banking angle here:

we're given that v₀ = 0 and a = 5.4 m/s². therefore, since v = at + v₀ for constant acceleration the equation for velocity is

after 28s,

m/s

which i would round to 150 m/s as my final answer