Question

2. Reaction time, operation speed, and roadway longitudinal grade are the three primary factors impacting stopping sight distance.

(1) Fill in the numbers in the table below.

(2) Plot the relationships between ds and the three influencing factors.

(3) Provide your observations by comparing how the three factors impact on stopping sight distance, and discuss how the results may influence roadway design.

l Verizon 9:07 AM 84% Back Hwk01_F19-3.docx 2. Reaction time, operation speed, and roadway longitudinal grade are the three primary factors impacting stopping sight distance. Fill in the numbers in the table below. Plot the relationships between d, and the three influencing factors Provide your observations by comparing how the three factors impact on stopping sight distance, and discuss how the results may influence roadway design. Please provide details of an example calculation for full credit, Distance Speed Reaction Time -6% -3 % 0 % 3 % 6 % () (mph) (sec) d. 45 2.5 4.5 60 2.5 45 75 2.5 45 d 4 2.5 45 60 2.5 4.5 75 2.5 4.5 d 45 2.5 4.5 60 2.5 4.5 75 2.5 45 Dashboard Calendar To Do Notifications Inbox

Answer 1

gium (Speed = mph), Reaction Time, gradeint SSD = utt v = (beg distance & Breakling 2gf distance) = vrtu? 29 (ft.) uz velocity (speed)=mples t= Reaction Time in due te grain = 9.81 m/s f = logo longitudinal frictions. emph < 30 40 50 60 780 f 0.40 0.38 0.37 0.36 0.35 Speed (kmph) hot wat -31. 1037 1621 distance dictable 5 (125) us lamela Scoz ar to v 120.00) f = 0.38 4.51 to = ot (0.38 & 0.06) (-6, t6) = of 7.9638 (1) - 78.36 +0.66) (+6)= SSD = 18.09m (66) => SSD = 24.880m (-3, +3) = of 7.9638 (0.38 +0.02) (+3) =) SSD = 19.423m (-3)=) SSD = 22.7537m (o) - ot 7.9638 = 20.95 m. 0.38

At 45 amph t=25see, here change the leg distance only SSD Vt + g (fto.on) t=25se ute. 12.5X2.5 = 31.25m was totoon) is fora as (tro) 4 t = 4.55ee ut 12.5x4.5 = 56.25m dr 45 (koment to -6 1-3 oT3 6 ] +=0.38 lo 2498 22.75 20.95) 19.423 | 18.09. 12.5 | 56.13 54 522 50.673/ 49.34 14581.13 79 17.2 15.673 74.34 at bokmph (16.67 mps) f = 0.36. to see SSD = x2 29 (f 10.09) 17-0,90 1, 257 ,14.163 – 39.34m 2x9.87(0.36) (0.36) (+3) 16.163 36.315 (+3) T0.36 10.03) 42.918(-3) 14.16 33.721(+6) n(+6)= 10.36 10.067 47.21(-6) t=2.5 t- 4.5 - ut= 16.67%2.5 = = vt = 16.67 x 4.5 = 41.675 m 75.015 m

at Speed (60 emph) = (16.67 mps) f = 0.36 J .o 41.21 | 4292 39.34 | 36.315 33.121 25 88.668457 81.015 7979 75.391 45 | 122.23 11.93 | 114.35 11:33 I 108.14. at 15 umph (20.83 mps) f= 0.35 to, n=0 y = 20.882 22.11 29(f:£oon) 2x9.81 (0:35) 0.35 = 63.18 m n(+3)= 22.0 - 58.18 (+3) 10.35€ 0.03) = 69.09 (-3) n(16), 22.11 - 53.92 (76) (6.31 20.06) = 76,24 (-6) t-25= vt = 20.83825= 52.075 m t- 4.5=) Vt = 20.83 84.8 93.73 m. Trumps (0.35=f) to 10T - 3 o 1 + 1 +6 o 176.24 169.09 | 63.18 | 18.18 53.9 T128.32 121.11 | 115.255 | 110.25 T105.94 9. R8.5 4.5 l 189.97|162.83 156.91 152.44 | 142.65

SSD= 350 - Ut + g (f toon) utt O speed will increase SSD always Increases time (reaction increases the day distance Increass - So SSD will increases 3 Saul will increases f ( longitudinal is slight to decreases so, SSD increases automatically gradient will incrane (o to tve) Ir increases the (Finvalu So, it indicants the SSD value slightly reduces gradeens will decreese (ote -ve) Ir Increases the Envalue, so it indicate the SSD value is slightly increase -ve to tue (3) SSD will be the high to low (always) decrans

SSD US velosir t Whitle. SSD= vt + v2 C 28 (fin) graphical Nesprestation of at t=o (raketon times no (gradunt ) SSD m) 5 CO 2 as same lilee differ et = vales and no Valele also SSD will increased the velous will intrases, always at v=45umph, no SCO 145 C 2.5 t Creation The at diffunt vulerity and gradens the reaction increase The lag disteme always Incrare so, the SSD always increase the reaction for Increases N = 45 konph to... SSD SOD grodenin hesative gradient alway gives moressa (ml. Compared to postive prodent , will all velocity time is like that only. it will affut te roul design.