Civil Engineering - GATE Exam Questions - Discussion
Discussion Forum : GATE Exam Questions - Section 6 (Q.No. 46)
46.
For a highway with design speed of 100 kmph, the safe overtaking sight distance is (assume accelerations as 0.53 m/sec2)
Discussion:
4 comments Page 1 of 1.
NISHANT SINGH said:
3 years ago
Vb = V-16 how?
Explain me.
Explain me.
(2)
Suraj kadam said:
5 years ago
I got
d1 = 46.67m
d2 314m
d3 319.42m
Total OSD = 680.12.
d1 = 46.67m
d2 314m
d3 319.42m
Total OSD = 680.12.
Amar said:
6 years ago
Thank you @Lee.
Lee said:
8 years ago
(b) 750m.
V = 100kmph, Vb = V - 16 = 100 - 16 = 84kmph.
Take, t = reaction time = 2secs.
A = 0.53 m/s2 = 0.53 * 36 = 1.92 kN/s2.
Now, d1 = 0.28 * Vb * t = 0.28 * 84 * 2 = 47.04m.
d2 = 0.28Vb * T; T = ; s = 0.2Vb + 6 = 0.28 * 84 + 6 = 22.8m.
then T = 13.11secs.
d2 = 0.28 * Vb * T = 0.28*84*13.11 = 354.13m.
d3 = 0.28V * T = 0.28 * 100 * 13.11 = 367.08m.
d1 +d2 + d3 = 768.25m = 750m.
V = 100kmph, Vb = V - 16 = 100 - 16 = 84kmph.
Take, t = reaction time = 2secs.
A = 0.53 m/s2 = 0.53 * 36 = 1.92 kN/s2.
Now, d1 = 0.28 * Vb * t = 0.28 * 84 * 2 = 47.04m.
d2 = 0.28Vb * T; T = ; s = 0.2Vb + 6 = 0.28 * 84 + 6 = 22.8m.
then T = 13.11secs.
d2 = 0.28 * Vb * T = 0.28*84*13.11 = 354.13m.
d3 = 0.28V * T = 0.28 * 100 * 13.11 = 367.08m.
d1 +d2 + d3 = 768.25m = 750m.
(1)
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