---
title: "Schedule speed for a 1.5 km run (trapezoidal)"
subject: "Utilization of Electrical Energy"
chapter: 4
key: "semester_7/utilization_of_electrical_energy:schedule-speed-trapezoidal-1-5km-run"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/schedule-speed-trapezoidal-1-5km-run
site: BCT Engineering
---

# Schedule speed for a 1.5 km run (trapezoidal)

Utilization of Electrical Energy · Chapter 4: Electric Traction

- Asked in: 2079 Bhadra, 8 marks
- 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-1-5km-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 0.8 km/h/s and 3.2 km/h/s respectively. If the ratio of maximum to average speed is 1.3 and time for stops 26 seconds, find the schedule speed for a run of 1.5 km. Assume simplified trapezoidal speed-time curve.

## Short answer

Using the trapezoidal relation 3600D=VmT-(Vm^2/2)K with K=1/0.8+1/3.2=1.5625 and Vm/Va=1.3, solve Vm=45.54 km/h, running time 154.16 s. Total time with 26 s stop is 180.16 s, schedule speed=(1.5×3600)/180.16=29.97 km/h (≈30 km/h).

## Answer

Data: $\alpha = 0.8$ km/h/s, $\beta = 3.2$ km/h/s, $V_m/V_a = 1.3$, $D = 1.5$ km, stop $= 26$ s.

#### Trapezoidal relation (speeds in km/h, time in s)
$$
3600D = V_m T - \frac{V_m^2}{2}\left(\frac1\alpha + \frac1\beta\right)
$$
$$
K = \frac1{0.8} + \frac1{3.2} = 1.5625
$$
Since $V_a = 3600D/T$ and $V_m = 1.3V_a$: $V_m T = 1.3\times3600\times1.5 = 7020$.
$$
\begin{aligned} 5400 &= 7020 - \frac{V_m^2}{2}\times1.5625 \\ V_m^2 &= \frac{2\times1620}{1.5625} = 2073.6 \\ V_m &= 45.54\ \text{km/h} \end{aligned}
$$

#### Running time and schedule speed
$$
\begin{aligned} T &= \frac{7020}{45.54} = 154.16\ \text{s} \\ V_a &= \frac{5400}{154.16} = 35.03\ \text{km/h} \\ T_s &= 154.16 + 26 = 180.16\ \text{s} \\ V_s &= \frac{1.5\times3600}{180.16} = 29.97\ \text{km/h} \end{aligned}
$$
Check: $t_1 = 45.54/0.8 = 56.9$ s, $t_3 = 45.54/3.2 = 14.2$ s, free running $= 154.16 - 71.1 = 83.0$ s.

**Answer: Schedule speed ≈ 29.97 km/h (≈ 30 km/h); maximum speed 45.5 km/h.**

## Check yourself

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

1. Given Vm/Va = 1.3, what is Vm x T?
   A. 3600 x D = 5400
   B. 1.3 x D = 1.95
   C. 7020 x 26
   D. 1.3 x 3600 x D = 7020
   Answer: D. Va = 3600D/T so Vm T = 1.3 x 3600 x 1.5 = 7020. (Key point: Using the speed ratio)

2. What is the maximum speed?
   A. 29.97 km/h
   B. 35.03 km/h
   C. 45.54 km/h
   D. 56.9 km/h
   Answer: C. 5400 = 7020 - (Vm^2/2)(1.5625) gives Vm^2 = 2073.6, Vm = 45.54 km/h. (Key point: Crest speed)

3. How is schedule speed found after T = 154.16 s?
   A. Divide distance by T only
   B. Add 26 s stop time: 1.5 x 3600 / 180.16 = 29.97 km/h
   C. Divide by 26 s
   D. Multiply Va by 1.3
   Answer: B. Schedule speed includes stop time; Vs = 5400/180.16 = 29.97 km/h. (Key point: Schedule speed)

## Close variants set in other exams

- [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×
- [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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