--- title: "Durations of acceleration, coasting and braking (2 km run)" subject: "Utilization of Electrical Energy" chapter: 4 key: "semester_7/utilization_of_electrical_energy:quadrilateral-curve-2km-40kmph-periods" source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/quadrilateral-curve-2km-40kmph-periods site: BCT Engineering --- # Durations of acceleration, coasting and braking (2 km run) Utilization of Electrical Energy · Chapter 4: Electric Traction - Asked in: 2080 Baisakh, 8 marks; 2071 Chaitra (asked 2 times) - Video: [ELECTRIC TRACTION - PART - 06 - IMPORTANT TERMS IN TRACTION AND TWO PROBLEMS](https://www.youtube.com/watch?v=Ae-00H5FX0M) — zenmurali Sachithanandam - 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/quadrilateral-curve-2km-40kmph-periods - Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-4 ## Question A train is required to run between two stations 2 Km apart at an average speed of 40 km/hr. The run is to be made to a simplified quadrilateral speed time curve. If the maximum speed is to be limited to 60 km/hr, acceleration to 2 km/hr/sec, coasting retardation to 0.15 km/hr/sec and braking retardation to 3 km/hr/sec. Determine the duration of acceleration, coasting and braking periods. ## Short answer Running time is 180 s. Solving time and distance equations for a quadrilateral speed-time curve gives peak speed 54.9 km/h and coast end speed 33.7 km/h. Duration: acceleration 27.4 s, coasting 141.3 s, braking 11.2 s. ## Answer Data: $D = 2$ km, $V_{avg} = 40$ km/h, $V_{max} \le 60$ km/h, $\alpha = 2$, $\beta_c = 0.15$, $\beta = 3$ km/h/s. Running time: $$ T = \frac{D}{V_{avg}} = \frac{2}{40}\times3600 = 180\ \text{s} $$ #### Check with 60 km/h reached If the train accelerates to 60 km/h ($t_1 = 30$ s), coasts to $V_2$ and brakes, the time condition $30 + \frac{60-V_2}{0.15} + \frac{V_2}{3} = 180$ gives $V_2 = 39.47$ km/h, but the distance covered would be 2.21 km, more than 2 km. So the train does **not** need to reach 60 km/h; the peak speed $V_1$ is found from both conditions (and checked $\le 60$). ```text V V1 /\ / \___ coasting (-0.15) / ----___ V2 / accel 2 |\ brake 3 /________________|_\____> t 0 t1 t2 t3 ``` #### Equations (speed in km/h, time in s; distance in km = area/3600) $$ \begin{aligned} \text{Time:}\quad & \frac{V_1}{2} + \frac{V_1 - V_2}{0.15} + \frac{V_2}{3} = 180 \\ \text{Distance:}\quad & \frac{V_1^2}{2\times2} + \frac{V_1^2 - V_2^2}{2\times0.15} + \frac{V_2^2}{2\times3} = 2\times3600 = 7200 \end{aligned} $$ From the time equation: $V_2 = \dfrac{7.1667\,V_1 - 180}{6.3333}$. Substituting in the distance equation and solving (numerically): $$ V_1 = 54.88\ \text{km/h}, \quad V_2 = 33.68\ \text{km/h} $$ $V_1 < 60$ km/h, so the speed limit is respected. #### Periods $$ \begin{aligned} t_1 &= \frac{V_1}{\alpha} = \frac{54.88}{2} = 27.44\ \text{s} \\ t_2 &= \frac{V_1 - V_2}{\beta_c} = \frac{54.88 - 33.68}{0.15} = 141.33\ \text{s} \\ t_3 &= \frac{V_2}{\beta} = \frac{33.68}{3} = 11.23\ \text{s} \end{aligned} $$ Check: $27.44 + 141.33 + 11.23 = 180$ s. Distances: acceleration $= \frac{54.88\times27.44}{2\times3600} = 0.209$ km; coasting $= \frac{(54.88+33.68)\times141.33}{2\times3600} = 1.738$ km; braking $= \frac{33.68\times11.23}{2\times3600} = 0.053$ km; total = 2.0 km. **Answer: acceleration 27.4 s, coasting 141.3 s, braking 11.2 s (peak speed 54.9 km/h, braking starts at 33.7 km/h).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. A 2 km run at an average speed of 40 km/h takes how long? A. 150 s B. 120 s C. 200 s D. 180 s Answer: D. T = D/Vavg = 2/40 h = 0.05 h = 180 s. (Key point: Running time) 2. Why does the train not reach the 60 km/h limit? A. Acceleration is too low B. Reaching 60 km/h and meeting the time would cover 2.21 km, more than 2 km, so the peak speed is lower C. Braking is too slow D. Coasting is not allowed above 50 km/h Answer: B. 60 km/h is only a limit; the time and distance equations together give V1 = 54.88 km/h, which respects the limit. (Key point: Peak speed not 60 km/h) 3. How are V1 and V2 found? A. From the time equation alone B. From the distance equation alone C. By assuming V2 = V1/2 D. By solving the time equation V1/2 + (V1-V2)/0.15 + V2/3 = 180 together with the distance (area) equation = 7200 Answer: D. Two unknowns need two equations; the area of the quadrilateral is the distance (2 km = 7200 in km/h-s units). (Key point: Equations used) 4. With V1 = 54.88 and V2 = 33.68 km/h, what is the coasting duration? A. 141.3 s B. 27.4 s C. 11.2 s D. 180 s Answer: A. t2 = (V1 - V2)/0.15 = 21.2/0.15 = 141.3 s; the periods 27.4 + 141.3 + 11.2 total 180 s. 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