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.
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)
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)
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
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)
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.
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)
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