---
title: "Schedule speed for a 2.5 km run (trapezoidal)"
subject: "Utilization of Electrical Energy"
chapter: 4
key: "semester_7/utilization_of_electrical_energy:schedule-speed-trapezoidal-2-5km-run"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/schedule-speed-trapezoidal-2-5km-run
site: BCT Engineering
---

# Schedule speed for a 2.5 km run (trapezoidal)

Utilization of Electrical Energy · Chapter 4: Electric Traction

- Asked in: 2073 Shrawan
- Video: [Analysis of Trapezoidal Speed Time Curve |Relationship between Principal Quantities|](https://www.youtube.com/watch?v=C7XvkElrij4) — 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-2-5km-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 braking retardation of 1.2 km/h/s and 3.8 km/h/s respectively. If the ratio of maximum to average speed is 1.6 and time for stop 45 seconds, find the schedule speed from a run of 2.5 km. Assume simplified trapezoidal speed time curve.

## Short answer

Using trapezoidal relation with K=1/1.2+1/3.8=1.0965 and Vm/Va=1.6, solve Vm=99.25 km/h, running time 145.10 s. Total time with 45 s stop is 190.10 s, schedule speed=(2.5×3600)/190.10=47.34 km/h (≈47.3 km/h).

## Answer

Data: $\alpha = 1.2$ km/h/s, $\beta = 3.8$ km/h/s, $V_m/V_a = 1.6$, $D = 2.5$ km, stop $= 45$ s.

#### Trapezoidal relation (speeds in km/h, time in s)
$$
3600D = V_m T - \frac{V_m^2}{2}\left(\frac1\alpha + \frac1\beta\right)
$$
$$
K = \frac1{1.2} + \frac1{3.8} = 0.8333 + 0.2632 = 1.0965
$$
With $V_a = 3600D/T$ and $V_m = 1.6V_a$: $V_m T = 1.6\times3600\times2.5 = 14400$.
$$
\begin{aligned} 9000 &= 14400 - \frac{V_m^2}{2}\times1.0965 \\ V_m^2 &= \frac{2\times5400}{1.0965} = 9849.6 \\ V_m &= 99.25\ \text{km/h} \end{aligned}
$$

#### Running time and schedule speed
$$
\begin{aligned} T &= \frac{14400}{99.25} = 145.10\ \text{s} \\ V_a &= \frac{9000}{145.10} = 62.03\ \text{km/h} \\ T_s &= 145.10 + 45 = 190.10\ \text{s} \\ V_s &= \frac{2.5\times3600}{190.10} = 47.34\ \text{km/h} \end{aligned}
$$
Check: $t_1 = 99.25/1.2 = 82.7$ s, $t_3 = 99.25/3.8 = 26.1$ s, free running $= 145.10 - 108.8 = 36.3$ s (positive, so the curve is valid).

**Answer: Schedule speed ≈ 47.3 km/h (maximum speed 99.2 km/h).**

## Check yourself

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

1. Which relation is used to find Vm?
   A. D = Vm x T
   B. 3600D = Vm T + Vm^2
   C. 3600D = Vm T - (Vm^2/2)(1/alpha + 1/beta)
   D. D = Va x Ts
   Answer: C. The area of the trapezium gives this relation with speeds in km/h and times in s. (Key point: Trapezoidal relation)

2. What is Vm x T for Vm/Va = 1.6, D = 2.5 km?
   A. 9000
   B. 14400
   C. 5400
   D. 2250
   Answer: B. Vm T = 1.6 x 3600 x 2.5 = 14400. (Key point: Vm T value)

3. After T = 145.1 s and stop 45 s, what is Vs?
   A. About 47.3 km/h
   B. About 62.0 km/h
   C. About 99.2 km/h
   D. About 71 km/h
   Answer: A. Vs = 2.5 x 3600 / 190.1 = 47.34 km/h; it is lower than Va = 62.03. (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 km run (trapezoidal)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aschedule-speed-trapezoidal-2km-run) — Utilization of Electrical Energy, chapter 4, asked 3×

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