--- title: "Maximum speed and distance before braking (2.5 km)" subject: "Utilization of Electrical Energy" chapter: 4 key: "semester_7/utilization_of_electrical_energy:max-speed-50kmph-2-5km-trapezoidal" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/max-speed-50kmph-2-5km-trapezoidal site: BCT Engineering --- # Maximum speed and distance before braking (2.5 km) Utilization of Electrical Energy · Chapter 4: Electric Traction - Asked in: 2073 Chaitra, 6 marks - Video: [Speed Time Curve & Analysis for Main Line Service |Electric Traction & Control|](https://www.youtube.com/watch?v=Nhu7eWWMQ1Q) — Anish Vora - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/max-speed-50kmph-2-5km-trapezoidal - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-4 ## Question A train runs an average speed of 50 km/hr between stations situated 2.5 km apart. Train accelerates at 2 km/hr/s and retards at 3 km/hr/s. Find its maximum speed assuming simplified trapezoidal speed time curve. Draw the speed time curve for the run and calculate also the distance travelled by it before the brakes applied. ## Short answer Running time is 180 s. Using the trapezoidal distance equation, the maximum speed is 57.71 km/h. The distance travelled before braking is 2.346 km, comprising 28.85 s acceleration and 131.91 s free run. ## Answer For a simplified trapezoidal curve, the distance between stops equals the area under the speed-time curve. **Given:** $V_a = 50$ km/h, $D = 2.5$ km, $\alpha = 2$ km/h/s, $\beta = 3$ km/h/s. #### Running time $$ \begin{aligned} T &= \frac{3600\,D}{V_a} = \frac{3600 \times 2.5}{50} = 180\ \text{s} \end{aligned} $$ #### Maximum (crest) speed For a trapezoidal curve (speeds in km/h, time in s, $D$ in km): $$ D = \frac{V_m T}{3600} - \frac{V_m^2}{7200}\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) $$ Let $K = \frac{1}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) = \frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}\right) = 0.4167$. Then $$ \begin{aligned} V_m &= \frac{T}{2K} - \sqrt{\frac{T^2}{4K^2} - \frac{3600\,D}{K}} \\ &= \frac{180}{0.8333} - \sqrt{\frac{180^2}{0.6944} - \frac{3600 \times 2.5}{0.4167}} \\ &= 216 - \sqrt{46656 - 21600} = 216 - 158.29 \\ &= 57.71\ \text{km/h} \end{aligned} $$ #### Times of each period $$ \begin{aligned} t_1 &= \frac{V_m}{\alpha} = \frac{57.71}{2} = 28.85\ \text{s} \\ t_3 &= \frac{V_m}{\beta} = \frac{57.71}{3} = 19.24\ \text{s} \\ t_2 &= T - t_1 - t_3 = 180 - 28.85 - 19.24 = 131.91\ \text{s} \end{aligned} $$ #### Distance before brakes are applied Distance covered during braking: $$ D_3 = \frac{V_m t_3}{2 \times 3600} = \frac{57.71 \times 19.24}{7200} = 0.154\ \text{km} $$ $$ D_1 + D_2 = D - D_3 = 2.5 - 0.154 = 2.346\ \text{km} $$ #### Speed-time curve ```text V (km/h) 57.7 | ______________________ | / \ | / \ | / \ | / \ 0 +/------------------------------\---- t (s) 0 28.85 160.76 180 ``` **Answer:** Maximum speed $V_m = 57.71$ km/h; distance travelled before brakes are applied $= 2.346$ km (accelerating 28.85 s, free run 131.91 s, braking 19.24 s). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is running time found? A. T = D x Va B. T = Va / D C. T = 2.5 x 50 D. T = 3600 D / Va = 180 s Answer: D. T = 3600 x 2.5 / 50 = 180 s. (Key point: Running time) 2. Which is Vm? A. 50 km/h B. 28.85 km/h C. 57.71 km/h D. 131.9 km/h Answer: C. Solving the trapezoidal area equation gives Vm = 216 - 158.29 = 57.71 km/h. (Key point: Maximum speed) 3. How is distance before brakes are applied found? A. Braking distance 0.154 km B. Total distance minus braking distance: 2.5 - 0.154 = 2.346 km C. Vm x t2 only D. Va x T Answer: B. Braking distance = Vm t3/7200 = 0.154 km, subtracted from D. 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