Question

A road is 8 m wide. Its average radius of curvature is 40 m. The outer edge is above the lower edge by a distance of 1.28 m. Find the velocity of vehicle for which the road is most suited? $(g=10m/s^2$

Answer 1

The figure shows banking of road.

$\sin \theta =\frac{1.28}{8}=0.16$

$⟹\tan \theta =\frac{0.16}{\sqrt{1-{0.16}^2}}=0.162$

Now, $\tan \theta =\frac{v^2}{rg}$

$⟹v=\sqrt{\tan \theta rg}=\sqrt{0.162\times 40\times 10}$

$⟹v=8.05m/s$