Question

Traffic on a three-lane freeway observes a triangular flow-density relationship with free-flow speed vf=100km/h, capacity c=2000veh/h/lane, jam density kj=150veh/km/lane. Flow in the freeway from upstream is 5000veh/h, while the ramp flow is negligible (almost zero).At 9:45am, an accident occurs at point A. The vehicles involved close one lane of traffic (total capacity reduces to 4000veh/h) for 15min and afterwards the scene is clear (capacity returns to 6000veh/h). Lucy enters the freeway via the ramp at point B at 10:00am. Distance AB is equal to 6.5km.What time [h] does Lucy hit the queue?

Hint: Draw all the shockwaves during the lane closure and the opening of the lane at 10am, and consider the trajectory of Lucy in the space-time diagram.

Answer 1

Given Free blow Speed (VF) - 2 100 km br Jam Capacity (C) = 2000 Venlhollane density (Kj) = 150 Veh lhyllano no. of lanes = 3 time bebore the accident Situation o Saturation For Each of the lane 9 = 5000 Veh|bv * → q = kivi Kivi = 5000 Vehlhy Since traffic observes triangular flow-density - Y=Vf (1 - 4 / 4 - 4a= K=kj ☆ vfik [ - ] = 5000 lop Ki Chhot- Sog (0) Looki - 5000 → K, ?- 1503, +2500 =0 cs Scanned with CamScanner

② On solving K2_150k, 42500 to we get I k, = .130.9 Veblicos From o Vik, = 5000 → Vi = 5000 3x130.9 V = 12.73 Venlor Saturation after closing down one - Situation ③ q upstream flow > Capacity Since flow flow will become Zero is at max. Capacity , V₂ = 0 , Ky = kj = 150 Veh Em => KxV2 = 9, 20 From Situation 0 & 2 Speed of Vw = shock ware 92-91 kg-t = = (0-5000) Veblings (150-130.9) vehlim -87.26 km hy Vw = -87026 km hr | Scained with CamScanner

Situation - 3 : Lane in open again → 93 = 45000 = 5000 5000 5000 venlhr From K3 V2 = 5000 3 V3 = V4 [1-4] – 100 k3 (1-) = 5000 » 150 ks - kg - 2500 to on solving K3 = 130.9 Vehlby Speed of Shock wave 1. Vw = 93-92 kg-k2 = 5000 - - 30.9-150 - - B 7.26 km hy | Vw = -87.26 km . -87 26 .br ] 1.2 Scanned with CamScanner

e Time for lucy to hit queue 12-13 og 87.26 km - 6.5 km - Let Let the time be t 12:73 + + 87.26€ - 6.5 = 100t = 6.5 km > t= 65 km 100km lb te 65 km 65 x 60 min STT TOYOTIRONET PTPTN 100 I t = 309 min | Lucy hit the Queue at = 9.45 +3.9 min = 9.45Am + 3 min su sec = 9:48:54 Surmany Summary 9:45 am 54 sec 9:48 AM 54 sec Scanned with - At 9:48:54 Am CamScanner Lucy hit the queue