Question

The radius of an exit taxiway with design exit speed of 80 kmph and coefficient of friction 0.15 is ________m.

• A. 341.33

Answer 1

The correct option is A 341.33
Radius of taxiway is given by
$R=\frac{V^2}{125f}=\frac{{80}^2}{125\times 0.15}=341.33m$