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 1 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.
Arsh said:
9 years ago
@Arpyit.
For speed 30km/hr, the Coeff = 0.4.
For speed 30km/hr, the Coeff = 0.4.
(1)
Dhruba said:
8 years ago
What is the formula to calculate the answer?
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.
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.
Murali said:
8 years ago
Option A is correct. s=((v*v)/(2gf)).
S = 10.11.
S = 10.11.
Starlin said:
8 years ago
V^2/(254f) =30* 30* / (254 *.35)= 10.12 next to 10 is 12 or may be 10.
Kshitiz said:
8 years ago
All persons are wrong here. Ans is 12m.
rxn distn=3*3=9.
braking distance=3*0.4=3.6.
So, 9+3.6=12.6 closer to 12. So 12 m is the answer.
rxn distn=3*3=9.
braking distance=3*0.4=3.6.
So, 9+3.6=12.6 closer to 12. So 12 m is the answer.
(1)
Naina said:
7 years ago
It's V coefficient.
Harsh Shukla said:
7 years ago
For velocity > 60kmph, f = 0.35.For velocity <60kmph, f= 0.36.
While making a vehicle stop all the kinetic energy will be converted into frictional energy i.e.
0.5*m*v^2=f*m*g*L.
So, L = v^2/2*g*f.
L = 10.12m ~10m (Answer).
While making a vehicle stop all the kinetic energy will be converted into frictional energy i.e.
0.5*m*v^2=f*m*g*L.
So, L = v^2/2*g*f.
L = 10.12m ~10m (Answer).
(1)
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