Civil Engineering - UPSC Civil Service Exam Questions - Discussion
Discussion Forum : UPSC Civil Service Exam Questions - Section 1 (Q.No. 9)
9.
For total reaction time of 2.5 seconds, coefficient of friction 0.35, design speed 80 km/h, what is the stopping sight distance on a highway ?
Discussion:
15 comments Page 2 of 2.
Adel abu ghazaleh said:
1 decade ago
SSD = Recognizing distance + Breaking distance.
= V(in m/s)*t(in sec)+V^2(in m/s)/(2g (f-G)).
Where G is the difference in slope.
= 80*1000/3600*2.5+(80*1000/3600)^2/(2*9.81*0.35).
= 127.468 roughly 127.
= V(in m/s)*t(in sec)+V^2(in m/s)/(2g (f-G)).
Where G is the difference in slope.
= 80*1000/3600*2.5+(80*1000/3600)^2/(2*9.81*0.35).
= 127.468 roughly 127.
Akhilesh said:
1 decade ago
SSD = Lag Distance+Braking Distance.
= V*t+V^2/2gf.
= 22.22*2.5+{22.22/2*9.81*0.35}.
= 127.
= V*t+V^2/2gf.
= 22.22*2.5+{22.22/2*9.81*0.35}.
= 127.
Jaspreet singh said:
1 decade ago
Ssd = 0.278vt+0.039v^2/a.
v = Design speed, km/h.
t = Brake reaction tim, 2.5 sec.
a = Deceleration rate, 3.4m/s^2.
v = Design speed, km/h.
t = Brake reaction tim, 2.5 sec.
a = Deceleration rate, 3.4m/s^2.
ASHISH said:
1 decade ago
SSD = lag dist + breaking dist.
= (0.278*v*t) + (v*v/254*f).
= (0.278*80*2.5) + (80*80/254*0.35).
= 127.59.
v in km/hr.
= (0.278*v*t) + (v*v/254*f).
= (0.278*80*2.5) + (80*80/254*0.35).
= 127.59.
v in km/hr.
(1)
Danesh said:
1 decade ago
SSD = LAG DISTANCE + BRAKING DISTANCE.
= v*t + v^2/(2gf).
= 22.22*2.5 + 22.22^2/(2*9.81*0.35).
= 127.45m.
Where "v" in m/s.
= v*t + v^2/(2gf).
= 22.22*2.5 + 22.22^2/(2*9.81*0.35).
= 127.45m.
Where "v" in m/s.
(1)
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