---
title: "Crest speed for 60 kmph schedule over 6 km"
subject: "Utilization of Electrical Energy"
chapter: 4
key: "semester_7/utilization_of_electrical_energy:crest-speed-60kmph-6km-trapezoidal"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/crest-speed-60kmph-6km-trapezoidal
site: BCT Engineering
---

# Crest speed for 60 kmph schedule over 6 km

Utilization of Electrical Energy · Chapter 4: Electric Traction

- Asked in: 2070 Asar
- 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/crest-speed-60kmph-6km-trapezoidal
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-4

## Question

A train has schedule speed of 60 km per hour between the stops which are 6 km apart. Determine the crest speed over the run assuming trapezoidal speed curve. The train accelerates at 2 km per hour per sec and retards at 3 km per hour per sec. Duration of stops is 60 second.

## Short answer

Using trapezoidal speed-time curve: schedule time = 3600×6/60 = 360 s, running time = 360 - 60 = 300 s. Distance formula D = (Vm×T)/3600 - (Vm²/7200)(1/α + 1/β) with α=2, β=3 km/h/s. Solving quadratic gives crest speed Vm ≈ 81.15 km/h.

## Answer

A simplified trapezoidal speed-time curve is used.

**Given:** $V_s = 60$ km/h, $D = 6$ km, stop time $= 60$ s, $\alpha = 2$ km/h/s, $\beta = 3$ km/h/s.

#### Schedule time and running time
$$
\begin{aligned}
T_s &= \frac{3600 \times 6}{60} = 360\ \text{s} \\
T &= 360 - 60 = 300\ \text{s}
\end{aligned}
$$
(Average speed $V_a = 3600 \times 6 / 300 = 72$ km/h.)

#### Crest speed
$$
D = \frac{V_m T}{3600} - \frac{V_m^2}{7200}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)
$$
With $K = \frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}\right) = 0.4167$:
$$
\begin{aligned}
V_m &= \frac{T}{2K} - \sqrt{\frac{T^2}{4K^2} - \frac{3600D}{K}} \\
&= \frac{300}{0.8333} - \sqrt{360^2 - \frac{3600 \times 6}{0.4167}} \\
&= 360 - \sqrt{129600 - 51840} = 360 - 278.85 \\
&= 81.15\ \text{km/h}
\end{aligned}
$$

Check: $t_1 = 81.15/2 = 40.57$ s, $t_3 = 81.15/3 = 27.05$ s, free run $t_2 = 300 - 40.57 - 27.05 = 232.38$ s.

**Answer:** Crest speed $V_m \approx 81.15$ km/h.

## Check yourself

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

1. How is T found from Vs = 60 km/h and 60 s stop?
   A. Ts = 360 s, T = 300 s
   B. T = 360 s
   C. T = 60 s
   D. T = 420 s
   Answer: A. Subtract stop time from schedule time. (Key point: Running time)

2. What is Va?
   A. 60 km/h
   B. 81.15 km/h
   C. 50 km/h
   D. 72 km/h
   Answer: D. Va = 3600 x 6/300 = 72 km/h. (Key point: Average speed)

3. What is Vm?
   A. 72 km/h
   B. 60 km/h
   C. 278.9 km/h
   D. About 81.15 km/h
   Answer: D. Vm = 360 - sqrt(129600 - 51840) = 81.15 km/h. (Key point: Crest speed)

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