Q.Part of a racing track is to be designed for a radius of curvature of 72 m. We are not recommending the vehicles to drive faster than 216 kmph. With what angle should the road be tilted? At what height will its outer edge be, with respect to the inner edge, if the track is 10 m wide?

With $v=216$ kmph $=60$ m/s and $r=72$ m, $\tan\theta=v^2/(rg)=5$, so $\theta=\tan^{-1}(5)\approx78.69^\circ$; for a 10 m wide (slant) track, the outer-edge height is $h=10\sin\theta\appro …

The 'most safe speed' relation (Eq. 1.1) is $\tan\theta=\dfrac{v^2}{rg}$. With v = 216 km/h $=\dfrac{216\times1000}{3600}=60$ m/s, r = 72 m and g = 10 m/s^2: $\tan\theta=\frac{(60)^2}{72\times10}=\frac{3600}{720}=5 \Rightarrow \theta=\tan^{-1}(5)\approx78.69^\circ$ For $\tan\theta=5$ (opposite 5, adjacent 1), the hypotenuse is $\sqrt{5^2+1^2}=\sqrt{26}$, so $\sin\theta=\dfrac{5}{\sqrt{26}}\approx0.9806$. Treating the given track width (10 m) as the width measured ALONG the now-tilted …

Use $\tan\theta=v^2/(rg)$ (the safe-speed banking relation) to find the angle, then use $h=(\text{track width})\sin\theta$ to find …

Using $h=(\text{width})\tan\theta$ instead of $\sin\theta$ (mixing up whether the given width is measured horizontally or along the actual slanted road surface), o …