The design speed for a two-way traffic on a two way lane road is 50 kmph. What is the value of safe stopping sight distance, if co-efficient of friction is 0.37 and reaction time of driver as 2.5 sec?
- A34.8 m
- B61.4 mCorrect
- C122.8 m
- D193.5 m
Solution
Concept
SSD = lag distance (during reaction time) + braking distance.
Formula
SSD = v t + v2/(2 g f), where v is in m/s
Step-by-step solution
1. v = 50/3.6 = 13.89 m/s 2. Lag distance = v t = 13.89 × 2.5 = 34.72 m 3. Braking distance = v2/(2 g f) = 13.892/(2 × 9.81 × 0.37) = 192.9/7.26 = 26.57 m 4. SSD = 34.72 + 26.57 = 61.3 m, which is about 61.4 m
Explanation
On a two-lane road with two-way traffic, each lane carries one direction of traffic, so only one vehicle needs to stop. SSD is the lag distance plus the braking distance. This gives about 61.4 m.
Why the other options are wrong
A: 34.8 m is only the lag distance. C: 122.8 m is 2 × SSD, which is used only for a single-lane road with two-way traffic. D: 193.5 m is close to v2 in m²/s², not a distance.
Shortcut
SSD = 0.278 V t + V2/(254 f) = 34.75 + 26.6 = 61.4 m (V in kmph)
Common mistake
Doubling the SSD for a two-lane road. Doubling is done only on a single-lane road with two-way traffic.
Correct answer: B) 61.4 m