Obtain the formula for the maximum safe speed $(v_{max})$ of a vehicle on a level curved road.

(N/A) Consider a vehicle of mass $m$ moving on a level curved road of radius $R$. The forces acting on the vehicle are:
$(1)$ The gravitational force $(mg)$ acting downwards.
$(2)$ The normal reaction force $(N)$ acting upwards from the road surface. Since there is no vertical motion,$N = mg$.
$(3)$ The static frictional force $(f_s)$ acting towards the center of the circular path,which provides the necessary centripetal force.
For safe turning,the required centripetal force must be provided by the static friction:
$\frac{mv^2}{R} \leq f_s$
Since the maximum value of static friction is $f_{s,max} = \mu_s N = \mu_s mg$,we have:
$\frac{mv_{max}^2}{R} = \mu_s mg$
Solving for $v_{max}$:
$v_{max}^2 = \mu_s Rg$
$v_{max} = \sqrt{\mu_s Rg}$
Where $\mu_s$ is the coefficient of static friction between the tyres and the road surface.