--- title: "Linear speed-density V = 60 - 0.43K" subject: "Transportation Engineering II" chapter: 1 key: "semester_7/transportation_engineering_ii:linear-speed-density-v-60-minus" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/linear-speed-density-v-60-minus site: BCT Engineering --- # Linear speed-density V = 60 - 0.43K Transportation Engineering II · Chapter 1: Traffic Engineering - Asked in: 2074 Asoj, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/linear-speed-density-v-60-minus - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1 ## Question Assuming linear Speed-density relationship of $V = 60 - 0.43K$ a) Draw V-K, V-Q and Q-K diagram showing critical value b) Find the saturation flow? c) Find speed and density at flow of 1000 veh/hr ## Short answer Saturation flow Qmax = 2093 veh/h. At flow 1000 veh/h: uncongested speed 51.7 km/h at density 19.4 veh/km; congested speed 8.3 km/h at density 120.2 veh/km. ## Answer For the linear model $V = V_f - bK$ with $V_f = 60$ km/h and $b = 0.43$, the jam density is where $V = 0$: $$ k_j = \frac{60}{0.43} = 139.5\ \text{veh/km} $$ The flow is $Q = VK = 60K - 0.43K^2$, so the critical (maximum-flow) values are at $dQ/dK = 0$: $$ \begin{aligned} K_m &= \frac{k_j}{2} = 69.8\ \text{veh/km}\\ V_m &= \frac{V_f}{2} = 30\ \text{km/h}\\ Q_{max} &= \frac{V_f k_j}{4} = \frac{60 \times 139.5}{4} = 2093\ \text{veh/h} \end{aligned} $$ #### (a) Diagrams with critical values ```text v-k diagram (straight line) v (km/h) 60 |* | * 30 |- - - * <- at q max | * +-----------*--- k (veh/km) 0 69.8 139.5 q-k diagram (parabola) q (veh/h) 2093.0 | _ * _ <- q max | * * | * * +--*-------|-------*-- k 0 69.8 139.5 v-q diagram (half parabola) v (km/h) 60 |* | * 30 |- - - * <- q max (nose) | * | * (congested branch) +--------|--------- q (veh/h) 0 2093.0 ``` #### (b) Saturation flow The saturation flow is the maximum flow (capacity) at the critical density: **Saturation flow = $Q_{max}$ = 2093 veh/h** #### (c) Speed and density at a flow of 1000 veh/h $$ 0.43K^2 - 60K + 1000 = 0 \Rightarrow K = \frac{60 \pm \sqrt{60^2 - 4(0.43)(1000)}}{2(0.43)} = \frac{60 \pm 43.36}{0.86} $$ $$ K_1 = 19.35\ \text{veh/km},\quad K_2 = 120.18\ \text{veh/km} $$ - Uncongested: $K = 19.35$ veh/km, $V = 60 - 0.43(19.35) = 51.68$ km/h. - Congested: $K = 120.18$ veh/km, $V = 8.32$ km/h. **Answer: at 1000 veh/h the speed is 51.7 km/h at a density of 19.4 veh/km (stable flow); in congested flow it is 8.3 km/h at 120.2 veh/km.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For V = 60 - 0.43K, how is jam density found and what is it? A. Set K = 0 to get 60 veh/km B. Set V = 0 to get 139.5 veh/km C. Set dQ/dK = 0 to get 69.8 veh/km D. Set V = 30 to get 69.8 veh/km Answer: B. Jam density is where speed is zero: K = 60/0.43 = 139.5 veh/km. (Key point: Jam density) 2. At maximum flow for the linear model, what are Km and Vm? A. Km = kj, Vm = Vf B. Km = kj/2 = 69.8 veh/km, Vm = Vf/2 = 30 km/h C. Km = kj/4, Vm = Vf/4 D. Km = kj/2, Vm = Vf Answer: B. Q = 60K - 0.43K^2 is maximum at K = kj/2, where V = Vf/2. (Key point: Critical values) 3. What is the saturation flow (Qmax) here? A. 1000 veh/h B. 4186 veh/h C. 2093 veh/h D. 1046 veh/h Answer: C. Qmax = Vf x kj / 4 = 60 x 139.5 / 4 = 2093 veh/h. (Key point: Saturation flow) 4. Solving 0.43K^2 - 60K + 1000 = 0 gives two densities (19.35 and 120.18 veh/km). Why two? A. One is a calculation error B. Flow of 1000 is above capacity C. A flow below capacity occurs in both uncongested and congested states D. One root is for cars and one for trucks Answer: C. The q-k parabola is symmetric about Km, so a flow below Qmax occurs once on the stable branch (51.7 km/h) and once on the congested branch (8.3 km/h). 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