# 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 ?
127 m
132 m
76 m
56 m
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Shahid hamid said:   10 months ago
SD= vt+(v^2Ã·2gf).
V=80, v=80Ã·3.6=22.22{assume 3.6 by IRC),
SD=22.22+{(22.22)*(22.22)}Ã·2*9.8*0.35,
SD=127.52.

Wagh Akash Kailas said:   4 years ago
Formula : v*t+ v*v/2gf.
Convert all in m & s as per value.
(1)

Virender singh said:   5 years ago
The formula is e * f +v^2/2gf.

Dolly said:   6 years ago
Here, Stopping distance =d1+d2.
d1=0.7*v ,d2=v^2/(254f),
v=80 f=0.35.

Bhosale Vrushali Ramesh said:   7 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.
(2)

Mani Bhushan said:   7 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.

Abhay said:   8 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.

Dipak said:   8 years ago
Without calculator can't be calculated.

Adel abu ghazaleh said:   8 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.

Akhilesh said:   8 years ago
SSD = Lag Distance+Braking Distance.

= V*t+V^2/2gf.

= 22.22*2.5+{22.22/2*9.81*0.35}.

= 127.