--- title: "Acceleration, coasting retardation and schedule speed" subject: "Utilization of Electrical Energy" chapter: 4 key: "semester_7/utilization_of_electrical_energy:train-48kmph-coasting-schedule-speed-stops" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/train-48kmph-coasting-schedule-speed-stops site: BCT Engineering --- # Acceleration, coasting retardation and schedule speed Utilization of Electrical Energy · Chapter 4: Electric Traction - Asked in: 2081 Bhadra, 8 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/train-48kmph-coasting-schedule-speed-stops - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-4 ## Question An electric train accelerates uniformly from rest to a speed of 48 km/hour in 24 seconds. The coasting period is 60 seconds against a constant resistance of 50 N/tonne and is braked to rest at 3.3 km/hour/second in 10 seconds. Calculate (i) The acceleration (ii) Coasting retardation (iii) The schedule speed, if the station stops are of 20 second duration. What would be the effect on schedule speed of reducing the station stops to 15 seconds duration, other conditions remaining same? Consider 10% for rotational inertia. ## Short answer Acceleration is 2 km/h/s, coasting retardation is 0.164 km/h/s. With 20 s stops the schedule speed is 29.27 km/h; reducing stops to 15 s raises it to 30.59 km/h, an increase of 1.32 km/h. ## Answer Assumption: braking at 3.3 km/h/s from the speed at end of coasting (the braking time then comes out as 11.6 s; the stated 10 s does not match this rate, see the note at the end). #### (i) Acceleration $$ \alpha = \frac{48}{24} = 2\ \text{km/h/s} $$ #### (ii) Coasting retardation Resistance 50 N/t, effective mass $1.1W$: $$ \begin{aligned} 277.8\times1.1\,\beta_c &= 50 \\ \beta_c &= \frac{50}{305.58} = 0.1636\ \text{km/h/s} \end{aligned} $$ #### (iii) Schedule speed (20 s stops) Speed at end of coasting: $$ V_2 = 48 - 0.1636\times60 = 38.18\ \text{km/h} $$ Braking time: $t_3 = 38.18/3.3 = 11.57$ s. | Period | Time (s) | Distance (m) | |---|---|---| | Acceleration | 24 | $\frac{48}{3.6}\times\frac{24}{2} = 160.0$ | | Coasting | 60 | $\frac{48+38.18}{2\times3.6}\times60 = 718.19$ | | Braking | 11.57 | $\frac{38.18}{3.6}\times\frac{11.57}{2} = 61.36$ | | Total | 95.57 | 939.55 | $$ \begin{aligned} T_s &= 95.57 + 20 = 115.57\ \text{s} \\ V_s &= \frac{939.55}{115.57}\times3.6 = 29.27\ \text{km/h} \end{aligned} $$ #### Effect of 15 s stops $$ \begin{aligned} T_s &= 95.57 + 15 = 110.57\ \text{s} \\ V_s &= \frac{939.55}{110.57}\times3.6 = 30.59\ \text{km/h} \end{aligned} $$ Increase $= 30.59 - 29.27 = 1.32$ km/h (about 4.5 %). **Answer: (i) 2 km/h/s, (ii) 0.164 km/h/s, (iii) 29.27 km/h; with 15 s stops the schedule speed rises to 30.59 km/h.** Note: if the braking period is taken as exactly 10 s from 38.18 km/h, the distance is 931.2 m and the schedule speeds are 29.41 km/h (20 s stops) and 30.76 km/h (15 s stops); the conclusion is the same. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. How is coasting retardation found from a resistance of 50 N/t with 10 percent rotational inertia? A. 277.8 x 1.1 x βc = 50, so βc = 0.164 km/h/s B. βc = 50/9.81 C. βc = 50 x 1.1 D. βc = 48/60 Answer: A. During coasting only resistance retards the effective mass 1.1W, giving 0.1636 km/h/s. (Key point: Coasting retardation) 2. What is the speed at the end of the coasting period? A. 48 km/h B. 38.18 km/h C. 28.8 km/h D. 45 km/h Answer: B. V2 = 48 - 0.1636 x 60 = 38.18 km/h. (Key point: Speed at end of coasting) 3. How is schedule speed obtained? A. Total distance divided by running time only B. Average of 48 and 38.18 C. Total distance divided by running time plus stop time (95.57 + 20 s), about 29.27 km/h D. Maximum speed divided by two Answer: C. Schedule speed includes the halt: 939.55 m / 115.57 s = 29.27 km/h. (Key point: Schedule speed) 4. What is the effect of reducing the stop from 20 s to 15 s? A. No change B. Schedule speed falls C. Schedule speed rises to about 30.59 km/h because the cycle time shortens D. Running time shortens Answer: C. Distance is unchanged but total time falls to 110.57 s, increasing the schedule speed by about 1.3 km/h. 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