Q.The road in the example 14 above is constructed as per the requirements. The coefficient of static friction between the tyres of a vehicle and this road is 0.8. Will there be any lower speed limit? By how much can the upper speed limit exceed in this case?

[!NOTE]
The book says “the example 14” — it refers to question 14 of this exercise (the racing-track design).

With $\tan\theta=5$ and $\mu_s=0.8$, since $\tan\theta>\mu_s$ there IS a nonzero lower speed limit, $v_{min}=\sqrt{rg\frac{\tan\theta-\mu_s}{1+\mu_s\tan\theta}}\approx24.6$ m/s $\approx88.5$ kmph; but since $\theta\approx78.69^\circ>45^\circ$, there is NO upper speed limit (the road can support any …

From question 14, this road has r = 72 m and $\tan\theta=5$; here $\mu_s=0.8$. Since $\tan\theta=5>\mu_s=0.8$, the lower speed limit is genuinely nonzero (a road banked steeply enough relative to its friction can still slide DOWN at too low a speed): $v_{min}=\sqrt{rg\,\frac{\tan\theta-\mu_s}{1+\mu_s\tan\theta}}=\sqrt{72\times10\times\frac{5-0.8}{1+0.8\times5}}=\sqrt{720\times\frac{4.2}{5}}=\sqrt{604.8}\approx24.59\text{ m/s}\approx88.5\text{ km/h}$ For the upper limit, $v_{max}=\sqrt{rg\,\frac{\tan\theta+\mu_s}{1-\mu_s\tan\theta}}$ requires the denominator $1-\mu_s\tan\theta=1-(0.8)(5)=1-4=-3$, which is NEGATIVE -- meaning there is no finite value of $v_{max}$ that satisfies the limiting-friction condition; physically, once $\mu_s\tan\theta>1$ (equivalently $\mu_s>\cot\theta$), friction acting down the slope is always more than sufficient to prevent outward skidding at ANY speed, however large. This condition is guaranteed here beca …

Substitute the given r, $\tan\theta$ and $\mu_s$ into the lower-limit formula (Eq. 1.3) to get $v_{min}$, and check the upper-limit formula's (Eq. 1.4) denominator to see it is negative wh …

Assuming every banked road automatically has both a finite lower AND upper speed limit, without checking whether $\mu_s\tan\theta>1$ (which re …