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 1 of 2.
Sudhanshu said:
4 years ago
@Shubham.
It's given breaking distance, not SSD.
Braking dist:V^2/254f.
It's given breaking distance, not SSD.
Braking dist:V^2/254f.
(5)
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.
Shubham gour said:
5 years ago
Options C is correct.
We know that;
SSD= (V*T)+(V^2/2*g*f).
First off ..V= 36*5/18=10 and t=2.5 by IRC.
8 = (10*2.5) +(10^2/2*9.81*f)
8 = 25+100/19.62*f
8 = 25+30.09*f
f = 8/30.09
f = 0.26.
So, take the value is nearest in options is 0.16.
We know that;
SSD= (V*T)+(V^2/2*g*f).
First off ..V= 36*5/18=10 and t=2.5 by IRC.
8 = (10*2.5) +(10^2/2*9.81*f)
8 = 25+100/19.62*f
8 = 25+30.09*f
f = 8/30.09
f = 0.26.
So, take the value is nearest in options is 0.16.
(1)
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.
Salim said:
5 years ago
f = v2/254L = 36*36/254*8 = 0.64 but the value of friction factor lies in between 0.35 to 0.40. So, the correct Ans will be (D) none of the above.
(4)
Binny bana said:
5 years ago
A is correct.
Breaking distance =v^2/2gf.
8=10^2/2*9.81*f,
so = 100/156.7.
=.64 answer.
Breaking distance =v^2/2gf.
8=10^2/2*9.81*f,
so = 100/156.7.
=.64 answer.
(6)
Koushal said:
5 years ago
Here f= a/g.
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).
Suman said:
7 years ago
It seems to be option D is correct as 'f' in the breaking distance formula signify longitudinal friction, which varies from 0.4 to 0.35 depending on the speed, while the lateral friction 'f' comes to design the superelevation instead. So option D be the correct answer.
(1)
Sourish said:
7 years ago
What about none of this? 16 also more than .15.
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers