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.
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.
Nisar said:
6 years ago
Stopping dist.= (0.2 *vb+6).
= (0.2*30+6) = 12.
= (0.2*30+6) = 12.
(1)
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)
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.
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)
Krushna Akotkar said:
5 years ago
F = u^2/29*L
L = 8.33*2/19*0.2 = 11.96~12m.
Assume f = .20.
L = 8.33*2/19*0.2 = 11.96~12m.
Assume f = .20.
(2)
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)
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