Solutions to Homework QuestionsProblem 5-3 Discussed in the previous classes. See your class notes.Problem 5-4Problem 6-2Problem 6-4Problem 6-6P(0) = (2.8)0 * e-2.8/ 0! = 0.061Problem 6-7P(wait) = 1-0.37 = 0.63Problem 6-10CE222 Highway EngineeringSolutions to Homework QuestionsProblem 5-3 Discussed in the previous classes. See your class notes.Problem 5-4Speed = 50 mph * 5280 /3600 = 73.5 ft/secMedian brake reaction time for unexpected situations = 1.35 * 0.66 = 0.89 secDistance traveled during brake-reaction time = 0.89 * 73.5 ft/sec = 65 ftTotal distance = 65 + 160 = 225 ftProblem 6-2By Equation 6-1,T1 = 200/35 = 5.71 secT2 = 200/40 = 5.00 secT3 = 200/45 = 4.44 secBy Equation 6-2,Problem 6-4By Equation 6-3,40.00 = 39.60 + S2/39.60S2 = 15.84Problem 6-5Future ADT = 8,000 veh/dayK30th = 0.15K10th = 0.18K50th = 0.14DHV30th = 0.15 * 8,000 = 1200 Veh/DayDHV10th = 0.18 * 8,000 = 1440 Veh/DayDHV50th = 0.14 * 8,000 = 1120 Veh/Daysec/403/)454035( ftutsec/60.39)44.400.571.5/(200*3 ftusProblem 6-6By Equation 6-4,P(0) = (2.8)0 * e-2.8/ 0! = 0.061% time no cars will arrive = P(0) * 100% = 6.1%Problem 6-7By Equation 6-5,P(g> 10) = e –360*10/3600 = 0.37P(wait) = 1-0.37 = 0.63Problem 6-10By equation 6-9,Fhv = 1/[1 + Pt(Et –1) + Pr (Er –1)] = 1 /[ 1+ 0.12 (2.8-1) + 0.02 (2.0 –1)]= 0.809By equation 6-8,SFb = MSFb * N* fw*fhv * fpSFb = 1120 *3 * 1.0 * 0.809 * 0.95 = 2582 Veh /hr** Use data for a 70 mph design speedProblem 6-13By Table 6-14, the level of service is