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UPSC Civil Service Exam Questions

Q1: For total reaction time of 2.5 seconds, coefficient of friction 0.35, design speed 80 km/h, what is the stopping sight distance on a highway ?

A 127 m

B 132 m

C 76 m

D 56 m

ANS:A - 127 m

SD= vt+(v^2÷2gf).
V=80, v=80÷3.6=22.22{assume 3.6 by IRC),
SD=22.22+{(22.22)*(22.22)}÷2*9.8*0.35,
SD=127.52.



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