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
10 metres
12 metres
15 metres
18 metres
Answer: Option
Explanation:
No answer description is available. Let's discuss.
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.
(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.
(2)

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.
(3)

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.

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.
(1)

Nisar said:   6 years ago
Stopping dist.= (0.2 *vb+6).
= (0.2*30+6) = 12.
(1)

Thakur791 said:   7 years ago
8.33^2/2 * .30 * 9.81,
= 11.78~~12.

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.

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).
(1)

Naina said:   7 years ago
It's V coefficient.


Post your comments here:

Your comments will be displayed after verification.