---
title: "Schedule speed for a 2 km run (trapezoidal)"
subject: "Utilization of Electrical Energy"
chapter: 4
key: "semester_7/utilization_of_electrical_energy:schedule-speed-trapezoidal-2km-run"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/schedule-speed-trapezoidal-2km-run
site: BCT Engineering
---

# Schedule speed for a 2 km run (trapezoidal)

Utilization of Electrical Energy · Chapter 4: Electric Traction

- Asked in: 2082 Baisakh, 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/schedule-speed-trapezoidal-2km-run
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-4

## Question

An electric train is to have acceleration and retardation of 0.8 km/h/s and 3.2 km/h/s respectively. If the ratio of maximum to average speed is 1.3, find schedule speed for a run of 2 km. Time taken at stops = 26 seconds. Assume simplified trapezoidal speed-time curve.

## Short answer

Using trapezoidal curve with distance 2 km, K=1.5625, solving gives maximum speed 52.6 km/h and running time 178 s. Adding 26 s stop gives schedule speed ≈ 35.3 km/h.

## Answer

Data: $\alpha = 0.8$ km/h/s, $\beta = 3.2$ km/h/s, $V_m/V_a = 1.3$, $D = 2$ km, stop $= 26$ s.

#### Trapezoidal relation
$$
D = \frac{1}{3600}\left[V_m T - \frac{V_m^2}{2}\left(\frac1\alpha + \frac1\beta\right)\right]
$$
$$
K = \frac1\alpha + \frac1\beta = \frac1{0.8} + \frac1{3.2} = 1.25 + 0.3125 = 1.5625
$$
Average speed $V_a = \dfrac{3600D}{T}$, so $V_m T = 1.3\times3600D = 1.3\times7200 = 9360$.
$$
\begin{aligned} 7200 &= 9360 - \frac{V_m^2}{2}\times1.5625 \\ V_m^2 &= \frac{2\times2160}{1.5625} = 2764.8 \\ V_m &= 52.58\ \text{km/h} \end{aligned}
$$

#### Times and speeds
$$
\begin{aligned} T &= \frac{9360}{52.58} = 178.01\ \text{s} \\ V_a &= \frac{7200}{178.01} = 40.45\ \text{km/h} \\ T_s &= 178.01 + 26 = 204.01\ \text{s} \\ V_s &= \frac{2\times3600}{204.01} = 35.29\ \text{km/h} \end{aligned}
$$
Check: $t_1 = 52.58/0.8 = 65.7$ s, $t_3 = 52.58/3.2 = 16.4$ s, free run $= 95.9$ s.

**Answer: Schedule speed = 35.3 km/h (maximum speed 52.6 km/h).**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. With Vm/Va = 1.3 and D = 2 km, what is the value of Vm x T?
   A. 7200
   B. 5538
   C. 9360
   D. 4000
   Answer: C. Va = 3600D/T gives Va T = 7200, so Vm T = 1.3 x 7200 = 9360. (Key point: Using Vm/Va ratio)

2. With 1/α + 1/β = 1.5625, which equation gives Vm?
   A. 7200 = 9360 - (Vm^2/2)(1.5625), so Vm = 52.58 km/h
   B. Vm = 7200/1.5625
   C. Vm = 1.3 x 40
   D. Vm^2 = 7200 x 1.5625
   Answer: A. Substituting into the trapezoid relation gives Vm^2 = 2 x 2160/1.5625 = 2764.8. (Key point: Solving for Vm)

3. After T = 178 s running time, how is the schedule speed obtained?
   A. Va = 40.45 km/h is the schedule speed
   B. Divide D by running time plus 26 s stop (204 s), giving about 35.3 km/h
   C. Multiply Va by 1.3
   D. Use 178 + 26 as seconds per km
   Answer: B. Schedule speed includes the halt time: 7200/204.01 = 35.3 km/h, lower than the average running speed of 40.45 km/h. (Key point: Schedule speed)

## Close variants set in other exams

- [Schedule speed for a 1.5 km run (trapezoidal)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aschedule-speed-trapezoidal-1-5km-run) — Utilization of Electrical Energy, chapter 4, asked 3×
- [Schedule speed for a 2.5 km run (trapezoidal)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aschedule-speed-trapezoidal-2-5km-run) — Utilization of Electrical Energy, chapter 4, asked 3×

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