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.
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)
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.
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.
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)
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.
Pavan said:
6 years ago
Braking distance is vt.
So it is 0.4 for brakes are effective in case. Now 0.4*30. And we will get 12.
So it is 0.4 for brakes are effective in case. Now 0.4*30. And we will get 12.
(1)
Ankush sherikar said:
3 years ago
S = (0.7* Vb) + 6.
Where Vb = 30*5/18.
=8.33m/s.
Therefore,
S = (0.7*8.33)+6.
=11.83~12m.
Where Vb = 30*5/18.
=8.33m/s.
Therefore,
S = (0.7*8.33)+6.
=11.83~12m.
(3)
Umesh said:
5 years ago
s = i+0.7xVb,
Where i = 6m( vehicle base length).
Vb = 8.33 in m/s.
s = 11.83 ----> 12m.
Where i = 6m( vehicle base length).
Vb = 8.33 in m/s.
s = 11.83 ----> 12m.
(3)
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.
Starlin said:
8 years ago
V^2/(254f) =30* 30* / (254 *.35)= 10.12 next to 10 is 12 or may be 10.
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