Question

A road is 8m wide average radius of curvature 40m the outer edge is above the lower edge by distance of 1.28m find velocity of vechicle for which road is most suited.

Answer 1

Answer:

the velocity will be 7.91 m/s

Explanation:

According to the problem the width of the road is 8m

the curvature radius,r  is 40 m and the distance between lower to upper limit is 1.28 m

let the velocity be v

now ,

tanθ= v^2/rg [ g is the gravitational force]

sinθ= v^2/rg

=> 1.28/8= v^2/40 x  9.8

=> v= 7.91 m/s

Answer 2

Answer:

we take the velocity to be v as we have to find out the value of v.

radius of curvature is given 40m

the road is 8m wide

outer edge is above the lower edge by 1.28m

= tan¤=v^2/rg

perpendicular/base=v^2/rg

1.28/8=v^2/40*10

$v = \sqrt{128\%2}$

so we get 8