Question

On a highway, the maximum speed of vehicles is $25\text{m/s}$. The material of the road provides a coefficient of friction of $0.2$. If the highway has a turn with a radius of curvature of $50\text{m}$, then what should be the banking angle for this turn to ensure the safety of the drivers?

• A. ${\sin }^{-1}\left(0.2\right)$
• B. ${\cos }^{-1}\left(0.5\right)$
• C. ${\tan }^{-1}\left(0.84\right)$
• D. ${\tan }^{-1}\left(1.2\right)$

Answer 1

The correct option is C ${\tan }^{-1}\left(0.84\right)$

Given $v=25\text{m/s},\mu =0.2,r=50\text{m}$

Banking angle is given by the formula,
$\tan \theta =\frac{v^2-\mu gr}{gr+\mu v^2}=\frac{{25}^2-0.2\times 10\times 50}{10\times 50+0.2\times {25}^2}=0.84$

$\therefore \theta ={\tan }^{-1}\left(0.84\right)$