--- title: "Crest, average and schedule speed; speed-time curve" subject: "Utilization of Electrical Energy" chapter: 4 key: "semester_7/utilization_of_electrical_energy:crest-average-schedule-speed-42kmph" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/crest-average-schedule-speed-42kmph site: BCT Engineering --- # Crest, average and schedule speed; speed-time curve Utilization of Electrical Energy · Chapter 4: Electric Traction - Asked in: 2072 Kartik - Video: [Crest Speed, Average Speed, Schedule Speed. Chapter:-2 Lecture:-1 , Traction system.](https://www.youtube.com/watch?v=Cx-aq-mj_2o) — Aarchita Technical Education - 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/crest-average-schedule-speed-42kmph - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-4 ## Question What do you mean by crest speed, average speed and schedule speed? An electric train has an average speed of 42 Kmph on a level track between stops 1400 m apart. It is accelerated at 1.7 Kmphps and braked at 3.3 Kmphps. Draw speed time curve for the run. ## Short answer Crest speed is the maximum speed; average speed = distance/running time; schedule speed = distance/(running+stop time). For the train: running time 120 s, crest speed 52.07 km/h (acceleration 30.63 s, free run 73.59 s, braking 15.78 s). ## Answer #### Crest, average and schedule speed - **Crest speed ($V_m$):** the maximum speed reached by the train during a run. - **Average speed ($V_a$):** distance between two stops divided by the actual running time (stop time not included). $V_a = \dfrac{\text{distance}}{\text{running time}}$ - **Schedule speed ($V_s$):** distance between two stops divided by running time plus stop time. $V_s = \dfrac{\text{distance}}{\text{running time} + \text{stop time}}$, so $V_s < V_a < V_m$. #### Numerical: speed-time curve for the run **Given:** $V_a = 42$ km/h, $D = 1400$ m $= 1.4$ km, $\alpha = 1.7$ km/h/s, $\beta = 3.3$ km/h/s. A simplified trapezoidal curve is assumed. Running time: $$ T = \frac{3600 \times 1.4}{42} = 120\ \text{s} $$ With $K = \frac{1}{2}\left(\frac{1}{1.7}+\frac{1}{3.3}\right) = 0.4456$: $$ \begin{aligned} V_m &= \frac{T}{2K} - \sqrt{\frac{T^2}{4K^2} - \frac{3600D}{K}} \\ &= \frac{120}{0.8913} - \sqrt{\left(\frac{120}{0.8913}\right)^2 - \frac{3600\times1.4}{0.4456}} \\ &= 134.64 - \sqrt{18128.5 - 11309.7} = 134.64 - 82.58 \\ &= 52.07\ \text{km/h} \end{aligned} $$ Times and distances: | Period | Time | Distance | |---|---|---| | Acceleration | $t_1 = 52.07/1.7 = 30.63$ s | 0.221 km | | Free run | $t_2 = 120 - 30.63 - 15.78 = 73.59$ s | 1.064 km | | Braking | $t_3 = 52.07/3.3 = 15.78$ s | 0.114 km | | Total | 120 s | 1.4 km | ```text V (km/h) 52.07 | _________________ | / \ | / \ | / \ | / \ 0 +-/-------------------------\--- t (s) 0 30.63 104.22 120 ``` **Answer:** Crest speed $= 52.07$ km/h; acceleration 30.63 s, free run 73.59 s, braking 15.78 s. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Which gives the correct ordering and definitions? A. Va < Vs < Vm B. Vs < Va < Vm; schedule speed includes stop time, average does not C. Vs = Va always D. Vm < Va Answer: B. Average speed uses running time only; schedule speed adds stop time, so it is lower. (Key point: Definitions) 2. What is T for Va = 42 km/h, D = 1.4 km? A. 120 s B. 42 s C. 60 s D. 84 s Answer: A. T = 3600 x 1.4 / 42 = 120 s. (Key point: Running time) 3. What is the crest speed result? A. 42 km/h B. 82.58 km/h C. 30.63 km/h D. 52.07 km/h Answer: D. Solving the quadratic gives Vm = 134.64 - 82.58 = 52.07 km/h. 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