Question

A car executes a turn of radius $22m$ on a banked road while travelling at a speed of $45km/h$. If the height of the outer edge above the inner edge of the road is $1.1m$, what is the breadth of the road? (in $m$)

• A. $1.745$
  
• B. $1.569$
  
• C. $1.875$
  
• D. $1.236$
  

Answer 1

Answer: (C)

$1.875$

given that $R=22m$, $v=45km/h=12.5m/s$

as the diagram shows, we have to find the value of $L$, we can see there is a right triangle $\Delta OQP$ so

$\Rightarrow L=\sqrt{(a^2+{1.1}^2)}...\left(1\right)$

and $\tan \theta =\frac{1.1}{a}\Rightarrow a=\frac{1.1}{\tan \theta }...\left(2\right)$

balancing the force in car we gets two equations

$\Rightarrow N\cos \theta =mg...\left(3\right)$

$\Rightarrow N\sin \theta =\frac{m{v}^2}{R}...\left(4\right)$

by dividing eqn(3) by eqn(4)

$\Rightarrow \tan \theta =\frac{v^2}{gR}$ now putting the value of $\tan \theta$ in eqn(2)

$\Rightarrow a=\frac{1.1\times gR}{v^2}=\frac{1.1\times 10\times 22}{{12.5}^2}=1.548m$

now again putting the value of $a$ in eqn(1)

$L=\sqrt{({1.5488}^2+{1.1}^2)}=1.875m$