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.
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)
Krushna Akotkar said:
4 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)
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)
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)
Nisar said:
6 years ago
Stopping dist.= (0.2 *vb+6).
= (0.2*30+6) = 12.
= (0.2*30+6) = 12.
(1)
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)
Arsh said:
9 years ago
@Arpyit.
For speed 30km/hr, the Coeff = 0.4.
For speed 30km/hr, the Coeff = 0.4.
(1)
Starlin said:
8 years ago
V^2/(254f) =30* 30* / (254 *.35)= 10.12 next to 10 is 12 or may be 10.
Naina said:
7 years ago
It's V coefficient.
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