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GATE Exam Questions

Q1:

For a highway with design speed of 100 kmph, the safe overtaking sight distance is (assume accelerations as 0.53 m/sec2)

A 300 m

B 750 m

C 320 m

D 470 m

ANS:B - 750 m

(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.



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