Civil Engineering - Highway Engineering - Discussion
Discussion Forum : Highway Engineering - Section 2 (Q.No. 23)
23.
The absolute minimum sight distance required for stopping a vehicle moving with a speed of 80 km ph, is
Discussion:
15 comments Page 2 of 2.
Navi said:
9 years ago
Actually its from the direct values given in the books (for 20kmph = 20).
So as (25=25) (30=30) (40=45) (50=60) (60=80) (80=120) & (100=180).
So as (25=25) (30=30) (40=45) (50=60) (60=80) (80=120) & (100=180).
(1)
Salman mohammed said:
9 years ago
SSD = 0.27vt + v.v/254(f +-g/100).
Where t = 2.5.
f = b/n 0.43 and 0.4 interpolate.
So to answer the question in meter we must use this equation.
Where t = 2.5.
f = b/n 0.43 and 0.4 interpolate.
So to answer the question in meter we must use this equation.
Jitendra singh said:
9 years ago
Lag distance = vt.
V = m/ s.
T = sec.
V = m/ s.
T = sec.
Sarang mote said:
9 years ago
What is the lag distance?
Dinesh said:
1 decade ago
SSD = lag distance+braking distance.
= v.t + v^2/2gf.
= 22.22*2.5 + (22.2)^2/(2*9.81*0.35).
= 126.5 (near to 120) So answer is (a)120m.
Where v is in m/s and t = reaction time(2.5sec acc. to IRC).
f is longitudinal coefficient of friction.
= v.t + v^2/2gf.
= 22.22*2.5 + (22.2)^2/(2*9.81*0.35).
= 126.5 (near to 120) So answer is (a)120m.
Where v is in m/s and t = reaction time(2.5sec acc. to IRC).
f is longitudinal coefficient of friction.
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