--- title: "Linear speed-density: free speed 90, jam 150" subject: "Transportation Engineering II" chapter: 1 key: "semester_7/transportation_engineering_ii:linear-speed-density-free-90-jam" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/linear-speed-density-free-90-jam site: BCT Engineering --- # Linear speed-density: free speed 90, jam 150 Transportation Engineering II · Chapter 1: Traffic Engineering - Asked in: 2079 Bhadra, 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-free-90-jam - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1 ## Question Assuming the linear speed-density relationship, the mean free speed is observed to be 90 kmph near zero density and the corresponding jam density is 150 veh/hr. Average length of vehicle is 6.1 m. a) Write down the speed-density and flow density equation. b) Draw v-k, v-q and q-v diagram indicating the critical values. c) Compute speed and density corresponding to flow of 800 veh/hr. d) Compute the avg headways, spacing, clearance and gap when the flow is maximum. ## Short answer Speed-density is v = 90 - 0.6k and flow-density is q = 90k - 0.6k^2; critical values are km = 75 veh/km, vm = 45 km/h, qmax = 3375 veh/h. For q = 800 veh/h, stable flow gives k = 9.5 veh/km and v = 84.3 km/h; congested gives k = 140.5 veh/km and v = 5.7 km/h. At maximum flow, headway = 1.07 s, spacing = 13.33 m, clearance = 7.23 m, gap = 0.58 s. ## Answer (Jam density is a number of vehicles per kilometre, so "veh/hr" in the question is read as veh/km.) #### (a) Speed-density and flow-density equations For a linear (Greenshields) relation, speed falls from the free-flow speed $v_f$ at zero density to zero at jam density $k_j$: $$ v = v_f\left(1-\frac{k}{k_j}\right) = 90\left(1-\frac{k}{150}\right) = 90 - 0.6\,k $$ Since $q = v\,k$: $$ q = v_f\,k\left(1-\frac{k}{k_j}\right) = 90\,k - 0.6\,k^2 $$ #### (b) Diagrams and critical values Setting $dq/dk = 0$ gives $k_m = k_j/2$, $v_m = v_f/2$ and $q_{max} = v_f k_j/4$: $$ \begin{aligned} k_m &= 150/2 = 75\ \text{veh/km}\\ v_m &= 90/2 = 45\ \text{km/h}\\ q_{max} &= \frac{90 \times 150}{4} = 3375\ \text{veh/h} \end{aligned} $$ ```text v-k diagram (straight line) v (km/h) 90 |* | * 45 |- - - * <- at q max | * +-----------*--- k (veh/km) 0 75 150 q-k diagram (parabola) q (veh/h) 3375 | _ * _ <- q max | * * | * * +--*-------|-------*-- k 0 75 150 v-q diagram (half parabola) v (km/h) 90 |* | * 45 |- - - * <- q max (nose) | * | * (congested branch) +--------|--------- q (veh/h) 0 3375 ``` #### (c) Speed and density for a flow of 800 veh/h Substitute in $q = v_f k - (v_f/k_j)k^2$: $$ 0.6k^2 - 90k + 800 = 0 \quad\Rightarrow\quad k = \frac{90 \pm \sqrt{90^2 - 4(0.6)(800)}}{2(0.6)} $$ $$ k_1 = 9.49\ \text{veh/km},\quad k_2 = 140.51\ \text{veh/km} $$ - Uncongested (stable) state: $k = 9.5$ veh/km and $v = 90(1-9.49/150) = 84.3$ km/h. - Congested (forced) state: $k = 140.5$ veh/km and $v = 5.7$ km/h. **Answer: for 800 veh/h, speed = 84.3 km/h at density = 9.5 veh/km (stable flow); the congested alternative is 5.7 km/h at 140.5 veh/km.** #### (d) Headway, spacing, clearance and gap at maximum flow At $q_{max} = 3375$ veh/h, $k_m = 75$ veh/km, $v_m = 45$ km/h: $$ \begin{aligned} \text{Average headway}\ h &= \frac{3600}{q_{max}} = \frac{3600}{3375} = 1.07\ \text{s}\\ \text{Average spacing}\ s &= \frac{1000}{k_m} = \frac{1000}{75} = 13.33\ \text{m}\\ \text{Clearance}\ &= s - L = 13.33 - 6.1 = 7.23\ \text{m}\\ \text{Gap (time)}\ &= h - \frac{L}{v_m} = 1.07 - \frac{6.1}{12.500} = 0.58\ \text{s} \end{aligned} $$ **Answer: headway = 1.07 s, spacing = 13.33 m, clearance = 7.23 m, gap = 0.58 s.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For linear Greenshields with vf = 90 km/h and kj = 150 veh/km, the speed-density equation is: A. v = 90 - 0.6k B. v = 90 - 1.67k C. v = 150 - 0.6k D. v = 90k - 0.6k^2 Answer: A. v = vf(1 - k/kj) = 90 - 0.6k. (Key point: Speed-density equation) 2. At maximum flow, the density, speed and flow are: A. k = 150, v = 90, q = 0 B. k = 75 veh/km, v = 45 km/h, q = 3375 veh/h C. k = 75 veh/km, v = 90 km/h, q = 6750 veh/h D. k = 45 veh/km, v = 75 km/h, q = 3375 veh/h Answer: B. km = kj/2, vm = vf/2 and qmax = vf kj/4 = 3375. (Key point: Critical values) 3. Why does q = 800 veh/h give two answers for density? A. The q-k curve is a parabola, so one flow occurs in both stable and congested states B. The speed is not unique C. Calculation error D. Jam density is doubled Answer: A. The quadratic gives k = 9.49 (stable, 84.3 km/h) and k = 140.5 veh/km (congested, 5.7 km/h). (Key point: Two states for a flow) 4. Average headway at q_max = 3375 veh/h is: A. 0.58 s B. 1.07 s C. 13.3 s D. 3.375 s Answer: B. h = 3600/q = 3600/3375 = 1.07 s. (Key point: Headway at maximum flow) 5. With spacing 13.33 m, L = 6.1 m and vm = 12.5 m/s, clearance and gap are: A. Clearance 7.23 m, gap 0.58 s B. Clearance 13.33 m, gap 1.07 s C. Clearance 19.4 m, gap 1.55 s D. Clearance 7.23 s, gap 0.58 m Answer: A. Clearance = s - L = 7.23 m; gap = h - L/v = 1.07 - 0.49 = 0.58 s. (Key point: Clearance and gap) ## Close variants set in other exams - [Linear speed-density: free speed 80, jam 120](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Alinear-speed-density-free-80-jam) — Transportation Engineering II, chapter 1, asked 2× ## Related in Transportation Engineering II - [Linear speed-density: free speed 80, jam 130](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Alinear-speed-density-free-80-130) — Transportation Engineering II, chapter 1 - [Linear speed-density V = 60 - 0.43K](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Alinear-speed-density-v-60-minus) — Transportation Engineering II, chapter 1