# Question #abf84

Oct 22, 2016

${24.6}^{\circ}$, rounded to one decimal place.

#### Explanation:

If banking angle is $\theta$ then we have the following expression connecting various variables
$\tan \theta = {v}^{2} / \left(r g\right)$
where $v$ is the velocity of the vehicle, $r$ is the radius of curvature and $g$ is the acceleration due to gravity equal to $9.81 m {s}^{-} 2$.

Inserting given values after converting speed to SI units, we get
$\tan \theta = {\left(\frac{108 \times {10}^{3}}{3600}\right)}^{2} / \left(200 \times 9.81\right) = {\left(30\right)}^{2} / \left(200 \times 9.81\right)$
$\implies \tan \theta = 0.4587$
$\implies \theta = {\tan}^{-} 1 0.4587$
$\implies \theta = {24.6}^{\circ}$, rounded to one decimal place.