Civil Engineering - Highway Engineering - Discussion
Discussion Forum : Highway Engineering - Section 6 (Q.No. 39)
39.
If brakes of vehicles are effective, the vehicle-running at 30 km/hour comes to a stop in
Discussion:
18 comments Page 2 of 2.
Arpyit Gupta said:
9 years ago
What will be the value of Coeff of long friction here? Will it not be 0.35?
By taking this value and by applying.
S= (V2) / (2gf).
I am getting 10. Please tell me how (B) is the correct option.
By taking this value and by applying.
S= (V2) / (2gf).
I am getting 10. Please tell me how (B) is the correct option.
Thakur said:
7 years ago
Value of friction coefficient will be 30.
So v2/2 * 30 * 9.81.
= 8.33^2/2 * 30 * 9.81.
So v2/2 * 30 * 9.81.
= 8.33^2/2 * 30 * 9.81.
Thakur791 said:
7 years ago
8.33^2/2 * .30 * 9.81,
= 11.78~~12.
= 11.78~~12.
Murali said:
8 years ago
Option A is correct. s=((v*v)/(2gf)).
S = 10.11.
S = 10.11.
Naren said:
8 years ago
For 30kmph the value of 'f' is 0.4. The answer is 8.9 and round of to the higher value. The answer is option A.
Mhathung Kikon said:
5 years ago
For speed (kmph) and longitudinal coefficient of friction (f).
20-30 -> 0.4.
40. -> 0.38.
50 ->0.37.
60 -> 0.36.
65 -> 0.36.
80 -> 0.35.
100 -> 0.35.
20-30 -> 0.4.
40. -> 0.38.
50 ->0.37.
60 -> 0.36.
65 -> 0.36.
80 -> 0.35.
100 -> 0.35.
Chhaya said:
8 years ago
Here, s = V2/2gf.
Where V in m/s and F = 0.35,
Answer is 10.11.
Where V in m/s and F = 0.35,
Answer is 10.11.
Dhruba said:
8 years ago
What is the formula to calculate the answer?
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