### Discussion :: Highway Engineering - Section 6 (Q.No.9)

Manish said: (Dec 5, 2015) | |

Square root of (4*s/a). S = 0.2*Speed of overtaken vehicle + 6 (Base length of vehicle). A = Acceleration. |

Pavan said: (Jun 20, 2016) | |

First change the overtaken vehicle speed into m/sec by multiply with .278 then apply S = .7*- speed of overtaken vehicle + wheel base ( assume 6 meter) Where a = acceleration And T = Square root of (4* s/a) |

Leo Basil George said: (Mar 24, 2017) | |

Overtaking time T = [2(s1+s2)/a]^1/2. Where s1 , s2 = .7v+l. v - velocity of overtaken vehicle. s1 -distance b/w overtaking and over taken vehicle. s2- distance covered after overtaking. l = wheel Base of vehicle ~ 6.1m. a -overtaking acceleration =1.25 m/s2. v = 30 * 5/18=8.33m/s s1 = s2= .7 * 8.33+6.1 = 11.93m. T = [2x(11.93+11.93)/1.25]^1/2. = 6.17 seconds. |

Anbarasan said: (Aug 29, 2017) | |

FORMULAS: OSD = 0.278Vbt + 0.278Vb T + 2(.2Vb +6) + VT WHERE T = (4S/a)^0.5 and S = .2vb+6 a= acceleration= 1.25 Vb=30kmph V=50 KMPH SOLUTION : S= .2 * 30 + 6 =12 T= (4*12/1.25)^.5= 6.197(ANSWER) OSD= .278*30*2.5 + .278 *30*6.197+2(12)+50*6.197. |

Gubendhiran said: (Dec 21, 2017) | |

T=√[(4S)÷a]. S=6+(0.7Vb). Vb=Design speed - 16 =50 - 16 =34 kmph, S=6 + (.7*0.278*34) =12.61 m/s, T=√[(4*12.61)÷1.25] = 6.35 seconds. So option B was correct. |

Umesh said: (Jul 24, 2019) | |

Thanks @Gubendhiran. |

Renis Gajipara said: (May 1, 2020) | |

V= 50 km/ph = 13.88 m/s ....vehicle with design speed Vb= 50 km/ph = 13.88 m/s....vehicle moving at a speed a = 1.25 m/s^2. Now, S =0.7*VB +6 = 11.831 sec. T = √(4S÷ a) = 6.15 sec. |

Jahangir Abbas Sahil said: (Aug 28, 2021) | |

V= 50 km/ph = 13.88 m/s -> vehicle with design speed. Vb= 30 km/ph = 13.88 m/s -> vehicle moving at a speed. a = 1.25 m/s^2. Now, S = 0.7*VB +6 = 11.831 sec. T = √(4S÷ a) = 6.14 sec. |

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