### Discussion :: Highway Engineering - Section 2 (Q.No.23)

Dinesh said: (Jun 12, 2014) | |

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. |

Sarang Mote said: (Sep 26, 2016) | |

What is the lag distance? |

Jitendra Singh said: (Oct 16, 2016) | |

Lag distance = vt. V = m/ s. T = sec. |

Salman Mohammed said: (Nov 19, 2016) | |

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. |

Navi said: (Jan 23, 2017) | |

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). |

Ankur said: (Feb 6, 2017) | |

How to predict the value of f in such cases? |

Asita said: (Feb 17, 2017) | |

Yes, right @Navi. |

Ankur said: (Jul 1, 2017) | |

We can predict because we have to calculate min SSD if take f= .15 we get more value but if we take f= .35 we get min value so we take f=.35. |

Pakkirappa said: (Sep 16, 2017) | |

V is given in kmph convert to m I.e 1000/60*60=0.278 Ssd=vt+{v2/2gf} Where t is time required for understanding the situation this is a constant value. Ssd=0.27*80*2.5+{(0.27*80)^2/(2*9.81*0.35)} Ssd=121.6m. |

Shubham said: (Aug 11, 2018) | |

Right @Navi. |

Dinesh said: (Jul 21, 2019) | |

0.28 * V.t + 0.01V^2/brake efficiency. V = 80 t = 2.5 And We get exact ans 120. |

Lucky Kumar said: (Mar 12, 2020) | |

SSD = 0.278Vt+V^2/254(f+S%). = 0.278*80*2.5+80*80/254(.35). = 127.6 mtr. Cofficent of friction (f) = 0.35. |

Mayu said: (May 6, 2020) | |

Longitudinal friction --------- 0.35 - 0.4. Lateral friction --------------- 0.15. |

Dipunku said: (Aug 28, 2020) | |

The correct answer is 127.6m. |

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