Civil Engineering - Highway Engineering - Discussion
Discussion Forum : Highway Engineering - Section 5 (Q.No. 36)
36.
In a braking test, a vehicle travelling at 36 km ph was stopped at a braking distance of 8.0 m. The average value of the vehicle's skid resistance (friction coefficient) is
Discussion:
16 comments Page 2 of 2.
Raj said:
8 years ago
Yes, absolutely correct @Kshitiz.
The Limiting value is 0.4. The correct option A is correct.
The Limiting value is 0.4. The correct option A is correct.
Kshitiz said:
8 years ago
Friction coefficient value never exceed 0.4
Longitudinal friction coefficient range 0.3 to o.4
Lateral friction coefficient value generally 0.15
You can use the formula for finding friction coefficient but it exceeds limiting value. So Ans is 0.16 correct ie it took lateral friction value.
Longitudinal friction coefficient range 0.3 to o.4
Lateral friction coefficient value generally 0.15
You can use the formula for finding friction coefficient but it exceeds limiting value. So Ans is 0.16 correct ie it took lateral friction value.
Gubendhiran said:
8 years ago
SSD=lag distance + braking distance.
SSD=Vt +[(V^2)÷(2g(f+-g))],
Breaking distance=(v^2)÷(254f+-g),
We don't have gradients so leave it.
8=(36^2)÷(254f),
f=0.63779=0.64,
So option A is correct.
SSD=Vt +[(V^2)÷(2g(f+-g))],
Breaking distance=(v^2)÷(254f+-g),
We don't have gradients so leave it.
8=(36^2)÷(254f),
f=0.63779=0.64,
So option A is correct.
(1)
Roy said:
8 years ago
The correct Answer is A.
ARUN KUMAR said:
9 years ago
f' = V*2 / 2gL
= (36/3.6)**2/ 2 * 9.81 * 8.0 v = V /3.6
= 36/3.6
= 10m/sec
= 0.64m (say)
= (36/3.6)**2/ 2 * 9.81 * 8.0 v = V /3.6
= 36/3.6
= 10m/sec
= 0.64m (say)
(1)
Nabin Adhikari said:
1 decade ago
F = V^2/(254S) = 0.64.
So correct answer might be A.
So correct answer might be A.
(1)
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