Question

a road is 10m wide. It's radius of curvature is 50m. The outer edge is above the lower edge by a distance of 1.5 m. This road is most suited for the velocity

Answer 1

let,
angle of inclination = x
then,

tan(x) = v^2/(r×g)

here,
for very small angle x,
we can take
sin(x) = tan(x) ........ (almost equal)
=> 1.5/10 = v^2/ (50×9.81)
=> v^2 = 0.15×50×9.81
=> v =√(73.575) = 8.575 m/sec
______________________
Hope it helps....
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@realsujaykumar

Answer 2

Answer:

This road is most suited for the velocity of 8.57m/s ≈ 8.6m/s

Explanation:

Let the length of the road is L=10m

The road's radius of curvature is 50m. So assume that the radius of curvature is R=50m

As in the question the outer edge is above the lower edge by a distance of 1.5m. So the distance can be denoted by d=1.5m.

Let the velocity is V

Now,

V²/RG=d/L          ----(We know gravitational acceleration G= 9.8 m/s²)

=> V= $\sqrt{RGd/L}$

=> V= $\sqrt{(50*1.5*9.8)/10}$

∴ V= 8.57m/s ≈ 8.6m/s