What is the ideal speed to take a \(100{\rm{ m}}\) radius curve banked at \(20.0^\circ \) angle?

To supply the centre gravity force required for a vehicle to perform a safe turn, the outside edge of curved roadways is raised above the inner edge. This is referred to as road banking.

The ideal speed for a banked road is,

\(v = \sqrt {Rg\tan \theta } \)

Here, \(R\) is the radius of the turn, \(g\) is the acceleration due to gravity, and \(\theta \) is the ideal banking angle.

Substitute \(100{\rm{ m}}\) for \(R\), \(9.8{\rm{ m}}/{{\rm{s}}^2}\) for \(g\), and \(20.0^\circ \) for \(\theta \),

\(\begin{array}{c}v = \sqrt {\left( {100{\rm{ m}}} \right) \times \left( {9.8{\rm{ m}}/{{\rm{s}}^2}} \right) \times \tan \left( {20.0^\circ } \right)} \\ = 18.9{\rm{ m}}/{\rm{s}}\end{array}\)

Hence, the ideal speed is \(18.9{\rm{ m}}/{\rm{s}}\).