a highway curve of radius 100 m, banked at an angle of 45 degrees, may be negotiated without friction at a speed of:

A car making this turn is pulled downward by its own weight, and pushed up by the road at an angle of 45°, so by Newton's second law,

• the net horizontal force on the car is

∑ F = N cos(45°) = m a = m v ² / R

• the net vertical force on the car is

∑ F = N sin(45°) - m g = 0

where

• N = magnitude of the normal force

• m = mass of the car

• a = v ² / R = centripetal acceleration of the car

• v = tangential speed of the car

• R = 100 m = radius of curvature

• g = 9.8 m/s² = acceleration due to gravity

From the net vertical force equation, we get

N = m g / sin(45°)

and substituting this into the net horizontal force equation and solving for v gives

(m g / sin(45°)) cos(45°) = m v ² / R

v = √(R g cos(45°) / sin(45°)) ≈ 31 m/s

the answer would be   dignified - apex

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