Question

Part of a racing track is to be designed for a curvature of 72 m. We are not recommending the vehicles to drive faster than 216 kmph. With what angleshould the road be tilted? By what heightwill its outer edge be, with respect to theinner edge if the track is 10 m wide?​

Answer 1

Then Denominator=2x+1

Fraction =2x+1x

Reciprocal of the fraction =x2x+1

It is given that the sum fo the fraction and its reciprocal is 22116

∴2x+1x+x2x+1=22116

⇒x(2x+1)x2+(2x+1)2=2158⇒2x2+x5x2+4x+1=2158

⇒21(5x2+4x+1)=58(2x2+x)⇒105x2+84x+21=116x

⇒11x2−26x−21=0⇒11x2−33x+7x−21=0⇒11x(x−3)+7(x−3)=0

⇒(11x+7)(x−3)=0⇒x=3,−117

∵ x can not be a negative 

⇒x=3

Hence the fraction =2x+1x=73

Answer 2

Answer:

 Ө =  tan inverse( 5 )

 Ө = 78.69 degree

 h = 9.805 meter      

Explanation:

Given data :-  

radius of curvature r = 72m

vmax = 216 km/hr = 216 x (5/18 ) = 60 m/s

width of tack = w = 10 m

What we have to find out :-

 1) angle should the road be tilted i.e angle of banking = Ө= ?

 2) height of outer edge h = ?

 

Solution :-

1)The angle of banking is given by,  

 

Ө = tan inverse ( v square  / rg )

    = $tan^{-1}$ ( $60^{2}$  / 72 x10 )              g = 10 m/$s^{2}$  

     =  (3600 /720 )

 Ө =  ( 5 )

 Ө = 78.69 degree  

2) now,

 h = w (sin Ө)

 h = 10 (sin 78.69 degree  )  

 h = 10 ( 0.9805 )

 h = 9.805 meter