Question

The radius of the turn of a road is 500 m. The
width of the road is 10 m. Its outer edge is 0.15 m
higher than the inner edge. For which speed of the
vehicle running over the road this banking is ideal?
(g = 10 m/s)
Ans. 8.66 m/s.​

Answer 1

Answer:

It is a problem of physics. Banking means the inclination which is given at the road curvatures so that the centrifugal force due to curvatures doesn’t make the moving car fall off the road.

Explanation:

Let the Angle of inclination of the road = x

Then,

$Tan(x) =\frac{v^{2}}{r \times g}$

Here we will take an approximation.  

If we suppose the angle x to be very small, we can take

Sin(x) = tan(x) almost equal.  

$\frac{15}{10}=\frac{v^{2}}{500 \times 9.81}$

$v^{2}=1.5 \times 500 \times 9.81$

$v=\frac{\sqrt{(73.575) m}}{s}=\frac{8.66 m}{s}$

So, the ideal speed is 8.6m/s.