First Midterm 1

First midterm 1 (October 8, 2003)

Problem 1: a car collided with a telephone pole and left 20-ft skid marks on the dry pavement. Based on the damages sustained, an engineer estimated that speed at collision was 15 miles/hour. If the roadway had a +3% grade, calculate the speed of the car at the onset of skidding.

Solutions: first convert different units into the same unit system. The final velocity v = 15 miles/hour = 22 ft/s. Thus, Db = x*cos(arctan(0.03)) = 0.99x  x. Since we have:

v 2  v 2 D  0 b 2g( f  G) 2 Thus, v0  Db  2 g  ( f  G)  v = 35.9 ft/s.

Problem 2: a study of the traffic using a tunnel showed that the following speed-density relationship applies: u  17.2ln(228/ k) miles / hour . Find a) the capacity of the tunnel; b). the value of the speed and concentration at capacity, and c) jam density.

Solution: since we know q = ku, we can multiple u  17.2ln(228/ k) with k and obtain the following: q  k17.2ln(228/ k) , by setting dq/dk = 0, we obtain km = 83.9 vehicles/mile. We can also obtain that um = 17.2 miles/hour by substituting km = 83.9 vehicles/miles into equation u  17.2ln(228/ k) . And then qm = 1443 vehicles/hour. At 228 jam density, u = 0, thus, rewriting u  17.2ln(228/ k) into k  and set u = 0, eu  e17.2 we obtain, kj = 228 vehicles/mile. Problem 3: an approach to a pre-timed signal has a maximum of 8 vehicles in a queue in a given cycle. If the saturation flow is 1440 vehicles/hour and the effective red time is 40 seconds, how much time it will take this queue to dissipate after the start of the effective green (assume that approach capacity exceeds arrivals and D/D/1 queuing applies).

Solution: since there is a maximum of 8 vehicles in a queue in a given cycle, we can write Q (number of vehicles during time t) = t into 8 = t, where t = 40 seconds, and we can obtain  = 0.2 vehicles/second. Once we obtain , we also have  = 1440 vehicles/hour = 0.4 vehicles/second. We can calculate  = / = 0.2/0.4 = 0.5. We then can calculate t0, which is the time it takes the queue to dissipate after the start of the green  r phase as t  = 40 seconds. 0 1 

Problem 4: empirical data showed that the average number of vehicles in 20 minutes period is equal to 0.15. Assume that traffic is poisson distributed and continues to arrive for the next 20 minutes period. What is the probability of exactly 8 vehicles arriving during the next 20 minutes? And what is the probability of having vehicle headways between 4 minutes and 8 minutes?

Solution: let X be the number of vehicles arriving during the 20 minute period. Then,

(t) n e t (0.15* 20)8 e (0.15*20) 6561*0.0498 Pr(X=8) =    0.0081 n! 8! 4032 let h be the headways between successive vehicles. Then,

(October 8, 2003)

Problem 1: a driver of car was traveling at a speed of 45 miles/hour on a roadway with (+3%) grade, he suddenly saw an obstacle on the highway. Suppose that the distance between the driver and the obstacle is 200 ft. Driver’s perception-reaction time is 1.5 s, and assuming that f = 0.6 for the coefficient of friction. Will the driver hit the obstacle? If yes, what was the speed when hitting the obstacle.

Solution: first convert different units into the same unit system. The initial velocity v0 = 45 miles/hour = 66 ft/s. There may be two scenarios: a) the car hit the object; and b). the car stopped before the object and so did not hit the object. If the car stopped before the object, the final velocity v = 0. Assuming scenario b) applies, we can calculate the stopping sight distance by SSD (stopping sight distance) = v0t + Db, where v 2  v 2 D  0 . SSD = 206 ft, which is greater than 200 ft. In other words, scenario b) b 2g( f  G) does not apply and scenario a) applies. Based on equation 2 v   Db  2 g  ( f  G)  v0 = 18.42 ft/s. Thus, the driver will hit the obstacle and the final velocity is 18.42 ft/s.

Problem 2: the u-k relationship for a particular freeway lane was found to be: u + 2.6 = 0.001(k – 240)2. find a) the free-flow speed, b) the jam density, and c) the lane capacity, and d) the speed at capacity.

Solution: u + 2.6 = 0.001(k – 240)2, we can write: u = 0.001(k – 240)2 – 2.6. Applying q = ku and set dq/dk = 0, we obtain km = 74. um = 24.9 (note that the other value of km is not possible as it leads to a negative value of um). qm= 1865. When u = 0, k = kj = 290. u = uf, when k = 0, thus, uf = 57.6.

Problem 3: an approach to a pretimed signal has 25 seconds of effective green in a 60- second cycle. The approach volume is 500 vehicles/hour and the saturation flow is 1400 vehicles/hour. Calculate the average vehicle delay using D/D/1 queuing.

r 2 Solution: r = 60 – 25 = 35 seconds. The average delay is equal to: d  = 15.9 2c(1 ) seconds. Problem 4: from the empirical data, we know that at a specific point of highway, the probability of having zero vehicles arriving during the 30 minute interval is equal to 0.05. Suppose that vehicles continue to arrive for the next 30 minute interval, what is the probability having more than 4 and less than 10 vehicle arrivals during the next 30 minute interval.

Solution: let X be the random variable representing the number of vehicle arrivals during the 30 minute period.

(t) n e t (0.05*30)5 e (0.15*30) Pr(X=5) =  = 0.014 n! 5! (t) n e t (0.05*30)6 e (0.15*30) Pr(X=6) =  = 0.021 n! 6! (t) n e t (0.05*30)7 e (0.15*30) Pr(X=7) =  = 0.032 n! 7! (t) n e t (0.05*30)8 e(0.15*30) Pr(X=8) =  = 0.048 n! 8! (t) n e t (0.05*30)9 e (0.15*30) Pr(X=9) =  = 0.071 n! 9!