What shall be the radius of an exit taxiway with design exit speed of 90 kmph and coefficient of friction 0.13?

Question

Question

This question was previously asked in

ESE Civil 2015 Paper 2: Official Paper

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Answer 1

Concept:

The radius of exit taxiway is given by,

\(R = \frac{{{V^2}}}{{125\;f}}\)

Where,

V = Design exit speed

f = Coefficient of friction

Calculation:

\(R = \frac{{{V^2}}}{{125\;f}}\)

\(\Rightarrow R = \frac{{{{90}^2}}}{{125 \times 0.13}}\)

⇒ R = 498.46 ≈ 500 m