Civil Engineering - Highway Engineering - Discussion
Discussion Forum : Highway Engineering - Section 5 (Q.No. 36)
36.
In a braking test, a vehicle travelling at 36 km ph was stopped at a braking distance of 8.0 m. The average value of the vehicle's skid resistance (friction coefficient) is
Discussion:
16 comments Page 2 of 2.
Raj said:
8 years ago
Yes, absolutely correct @Kshitiz.
The Limiting value is 0.4. The correct option A is correct.
The Limiting value is 0.4. The correct option A is correct.
Sourish said:
8 years ago
What about none of this? 16 also more than .15.
Akhila said:
6 years ago
Yes, in sight distance its always Longitudinal friction not lateral. So, Option D as it exceeds the range (0.35-0.40).
Koushal said:
6 years ago
Here f= a/g.
Swain lipan said:
5 years ago
Option c is correct. Because it asking in skid resistance coefficient.
Hence.
S=1/2*at^2,
Hence a= g*f.
S is given 8.
t = 2.5 sec. As per IRC.
g = 9.81,
f = ?.
If we put this formula answer comes 0.26 which is nearly equal to option C.
Hence.
S=1/2*at^2,
Hence a= g*f.
S is given 8.
t = 2.5 sec. As per IRC.
g = 9.81,
f = ?.
If we put this formula answer comes 0.26 which is nearly equal to option C.
Shubham gour said:
5 years ago
Option c is correct.
Given:- SSD :- 8 M,
V=36kmph then convert.
V= 36 * 5/18= 10 t=2.5 sec by IRC.
SSD = (V*T)+(V^2/2gf),
8 = (10*2.5)+(10^2/2*9.18*f),
8 = 25+100/18.36fm
8 = 30.44f,
f = 8/30.44,
f = 0.26.
It is the nearest value options C.
So, options C is correct.
Given:- SSD :- 8 M,
V=36kmph then convert.
V= 36 * 5/18= 10 t=2.5 sec by IRC.
SSD = (V*T)+(V^2/2gf),
8 = (10*2.5)+(10^2/2*9.18*f),
8 = 25+100/18.36fm
8 = 30.44f,
f = 8/30.44,
f = 0.26.
It is the nearest value options C.
So, options C is correct.
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