Answer:
Explanation:
Given
Velocity of traffic [tex]v=60\ mi/hr\approx 26.82\ m/s[/tex]
Maximum value of Centripetal force is one-tenth of weight
such that [tex](F_c)_{max}=\frac{mg}{10}[/tex]
to make a safe turn centripetal force with radius of curvature r is given by
[tex]F_c=\frac{mv^2}{r}[/tex]
[tex](F_c)_{max}=\frac{mg}{10}=F_c=\frac{mv^2}{r}[/tex]
[tex]\frac{mg}{10}=\frac{mv^2}{r}[/tex]
[tex]r=\frac{10v^2}{g}[/tex]
[tex]r=\frac{10\times 26.82^2}{9.8}[/tex]
[tex]r=733.99\approx 734\ m[/tex]
