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 2 of 2.
Dipak said:
10 years ago
Without calculator can't be calculated.
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.
Virender singh said:
7 years ago
The formula is e * f +v^2/2gf.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers