Question

The radius of curvature of a road at a certain turn is $50m$. The width of the road is $10m$ and its outer edge is $1.5m$ higher than the inner edge. The safe speed for such an inclination will be

• A. $6.5\mspace{6mu} m/s$
• B. $8.6\mspace{6mu} m/s$
• C. $8\mspace{6mu} m/s$
• D. $10\mspace{6mu} m/s$

Answer 1

Answer: (B)

$8.6\mspace{6mu} m/s$

Answer 2

$h = 1.5m,r = 50m,l = 10m,g = 10m/s^{2}$ (given)

$\frac{v^{2}}{rg} = \frac{h}{l}$ ⇒ $v = \sqrt{\frac{hrg}{l}} = \sqrt{\frac{1.5 \times 50 \times 10}{10}} = 8.6m/s$