SSC JE Civil Engineering 2018 Set 3 Solved Paper

Stopping sight distance
(SSD) = $\mathit{v}\mathit{t}+\frac{{\mathit{v}}^2}{254\left(\mathrm{f}\pm \frac{\mathit{n}}{100}\right)}$
(t) → Ascending gradient
(–) → Descending gradient
v → velocity of vehicle = 80 km/hr.
f → coefficient of longitudinal friction = 0.35
n → Gradient = 3%
SSD $=0.278\times 80\times 2+\frac{{80}^2}{254\left(0.35+\frac{3}{100}\right)}$.
= 123.22 m.