Question

PART of a racing track is to design for a curvature of 72 m we are not recommending the vehicle to drive faster then 216 kmph with what angle should be the road by tilted ? By what height will its outer edge be with respect to the inner edge if the track am wide
[And theta = tan-1 (5)= 78.69,n=9.8m]

Answer 1

Answer:

Explanation:

v = $\sqrt{r g tanx}$

where r is radius of curvature and x is the angle of tilt

here max  velocity allowed = 216 kmph

= 60 m/s

tan x = v² / 72 *10

tan x = 3600 / 720

tan x = 5

x = tan⁻¹ ( 5 )

x = 78.69⁰

the outer edge will be 5 meters above inner edge

Answer 2

Answer:

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Explanation:

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