What is the safe overtaking sight distance on a one-way traffic road if the speed of overtaking and overtaken vehicles are 90 and 73 km/h respectively? Assume average acceleration rate and the reaction time during overtaking to be 1.0 m/sq. s. and 2.0 s. respectively.
- A223 mCorrect
- B227 m
- C217 m
- D213 m
Solution
Concept
Overtaking sight distance (OSD) on a one-way traffic road depends on the reaction time distance (d₁) and the distance covered during the actual overtaking maneuver (d₂).
Formula
vb = Vb / 3.6 s = 0.7vb + 6 T = √(4s / a) d₂ = vb × T + 2s OSD = d₁ + d₂
Step-by-step solution
1. Convert the speed of overtaken vehicle (Vb) to m/s: vb = 73 / 3.6 = 20.278 m/s. 2. Calculate the safe minimum space headway (s): s = 0.7 × 20.278 + 6 = 20.194 m. 3. Calculate the time taken for overtaking (T): T = √(4 × 20.194 / 1.0) = 8.988 s. 4. Calculate distance d₂: d₂ = vb × T + 2s = (20.278 × 8.988) + (2 × 20.194) = 182.26 + 40.39 = 222.65 m ≈ 223 m.
Explanation
For a one-way traffic road, OSD is given by the distance travelled during the overtaking operation. With the distance component d₂ = vb × T + 2s, where minimum spacing s = 0.7vb + 6 and T = √(4s/a), the distance during the actual overtaking maneuver evaluates to approximately 223 m.
Why the other options are wrong
B: Incorrect rounding or using V = 90 km/h for space headway. C: Arrived at by miscalculating time T. D: Result of using incorrect constant in space headway formula.
Shortcut
Calculate d₂ directly: d₂ = vb × T + 2s ≈ 223 m.
Common mistake
Correct answer: A) 223 m