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:
13 comments Page 1 of 2.
Abhay said:
10 years ago
SSD = LAG DISTANCE + BRAKING DISTANCE.
Design speed = 80 Km/hr => v = 80x1000/(60x60) m/sec = 22.22 m/sec.
Reaction time = 2.5 sec.
Coefficient of friction f = 0.35.
= vt + v^2/(2gf).
= 22.22*2.5 + 22.22^2/(2*9.81*0.35).
SSD = 127.45 m.
Design speed = 80 Km/hr => v = 80x1000/(60x60) m/sec = 22.22 m/sec.
Reaction time = 2.5 sec.
Coefficient of friction f = 0.35.
= vt + v^2/(2gf).
= 22.22*2.5 + 22.22^2/(2*9.81*0.35).
SSD = 127.45 m.
Adel abu ghazaleh said:
10 years 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.
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)
Bhosale Vrushali Ramesh said:
9 years ago
If speed is in Kmph then formula for SSD is;
SSD = (0.278 * v * t ) + (v^2/254f).
SSD = (0.278 * 80 * 205) + ((80^2) / (254 * 0.35)) = 127.59.
SSD = (0.278 * v * t ) + (v^2/254f).
SSD = (0.278 * 80 * 205) + ((80^2) / (254 * 0.35)) = 127.59.
(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)
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.
Shahid hamid said:
3 years ago
SD= vt+(v^2÷2gf).
V=80, v=80÷3.6=22.22{assume 3.6 by IRC),
SD=22.22+{(22.22)*(22.22)}÷2*9.8*0.35,
SD=127.52.
V=80, v=80÷3.6=22.22{assume 3.6 by IRC),
SD=22.22+{(22.22)*(22.22)}÷2*9.8*0.35,
SD=127.52.
(1)
Akhilesh said:
10 years 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.
Mani Bhushan said:
9 years ago
V * t + v^2/(2gf).
V = 80/3.6 = 22.2.
T = 2.5 sec.
G = 9.8.
F = 3.5.
SSD = 127m.
V = 80/3.6 = 22.2.
T = 2.5 sec.
G = 9.8.
F = 3.5.
SSD = 127m.
(1)
Dolly said:
7 years ago
Here, Stopping distance =d1+d2.
d1=0.7*v ,d2=v^2/(254f),
v=80 f=0.35.
d1=0.7*v ,d2=v^2/(254f),
v=80 f=0.35.
(1)
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