### Discussion :: Highway Engineering - Section 5 (Q.No.36)

Nabin Adhikari said: (Oct 15, 2014) | |

F = V^2/(254S) = 0.64. So correct answer might be A. |

Arun Kumar said: (Jun 18, 2016) | |

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) |

Roy said: (Aug 26, 2017) | |

The correct Answer is A. |

Gubendhiran said: (Dec 20, 2017) | |

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. |

Kshitiz said: (Jan 18, 2018) | |

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. |

Raj said: (Jan 22, 2018) | |

Yes, absolutely correct @Kshitiz. The Limiting value is 0.4. The correct option A is correct. |

Sourish said: (Mar 13, 2018) | |

What about none of this? 16 also more than .15. |

Suman said: (Nov 22, 2018) | |

It seems to be option D is correct as 'f' in the breaking distance formula signify longitudinal friction, which varies from 0.4 to 0.35 depending on the speed, while the lateral friction 'f' comes to design the superelevation instead. So option D be the correct answer. |

Akhila said: (Jun 20, 2019) | |

Yes, in sight distance its always Longitudinal friction not lateral. So, Option D as it exceeds the range (0.35-0.40). |

Koushal said: (Mar 30, 2020) | |

Here f= a/g. |

Binny Bana said: (Jun 10, 2020) | |

A is correct. Breaking distance =v^2/2gf. 8=10^2/2*9.81*f, so = 100/156.7. =.64 answer. |

Salim said: (Nov 30, 2020) | |

f = v2/254L = 36*36/254*8 = 0.64 but the value of friction factor lies in between 0.35 to 0.40. So, the correct Ans will be (D) none of the above. |

Swain Lipan said: (Jan 20, 2021) | |

Option c is correct. Because it asking in skid resistance coefficient. Hence. S=1/2*at^2, Hence a= g*f. S is given 8. t = 2.5 sec. As per IRC. g = 9.81, f = ?. If we put this formula answer comes 0.26 which is nearly equal to option C. |

Shubham Gour said: (Feb 5, 2021) | |

Options C is correct. We know that; SSD= (V*T)+(V^2/2*g*f). First off ..V= 36*5/18=10 and t=2.5 by IRC. 8 = (10*2.5) +(10^2/2*9.81*f) 8 = 25+100/19.62*f 8 = 25+30.09*f f = 8/30.09 f = 0.26. So, take the value is nearest in options is 0.16. |

Shubham Gour said: (Feb 6, 2021) | |

Option c is correct. Given:- SSD :- 8 M, V=36kmph then convert. V= 36 * 5/18= 10 t=2.5 sec by IRC. SSD = (V*T)+(V^2/2gf), 8 = (10*2.5)+(10^2/2*9.18*f), 8 = 25+100/18.36fm 8 = 30.44f, f = 8/30.44, f = 0.26. It is the nearest value options C. So, options C is correct. |

Sudhanshu said: (Jan 7, 2022) | |

@Shubham. It's given breaking distance, not SSD. Braking dist:V^2/254f. |

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