---
title: "Acceleration for 25 kmph schedule speed, 800 m stations"
subject: "Utilization of Electrical Energy"
chapter: 4
key: "semester_7/utilization_of_electrical_energy:acceleration-25kmph-800m-stations"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/acceleration-25kmph-800m-stations
site: BCT Engineering
---

# Acceleration for 25 kmph schedule speed, 800 m stations

Utilization of Electrical Energy · Chapter 4: Electric Traction

- Asked in: 2072 Chaitra, 8 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/acceleration-25kmph-800m-stations
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-4

## Question

An electric train has a schedule speed of 25 kmph between stations 800 m apart. The duration of station stop is 20 seconds, the maximum speed is 20% higher than average running speed and the braking retardation is 3 kmphps. Calculate the rate of acceleration required to operate this service.

## Short answer

Schedule time 115.2 s, running time 95.2 s, average speed 30.25 km/h, crest speed 36.30 km/h. From D = (VmT/3600) - (Vm^2/7200)(1/α+1/β), solving gives required acceleration α ≈ 1.85 km/h/s.

## Answer

A simplified trapezoidal speed-time curve is assumed.

**Given:** $V_s = 25$ km/h, $D = 800$ m $= 0.8$ km, stop time $= 20$ s, $V_m = 1.2\,V_a$, $\beta = 3$ km/h/s.

#### Schedule time and running time
$$
\begin{aligned}
T_s &= \frac{3600 \times 0.8}{25} = 115.2\ \text{s} \\
T &= T_s - \text{stop time} = 115.2 - 20 = 95.2\ \text{s}
\end{aligned}
$$

#### Average and crest speed
$$
\begin{aligned}
V_a &= \frac{3600 \times 0.8}{95.2} = 30.25\ \text{km/h} \\
V_m &= 1.2 \times 30.25 = 36.30\ \text{km/h}
\end{aligned}
$$

#### Acceleration
For the trapezoidal curve:
$$
D = \frac{V_m T}{3600} - \frac{V_m^2}{7200}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)
$$
$$
\begin{aligned}
0.8 &= \frac{36.30 \times 95.2}{3600} - \frac{36.30^2}{7200}\left(\frac{1}{\alpha}+\frac{1}{3}\right) \\
0.8 &= 0.96 - 0.18303\left(\frac{1}{\alpha}+0.3333\right) \\
\frac{1}{\alpha}+0.3333 &= \frac{0.16}{0.18303} = 0.8741 \\
\frac{1}{\alpha} &= 0.5408 \quad\Rightarrow\quad \alpha = 1.849\ \text{km/h/s}
\end{aligned}
$$

Check: $t_1 = 36.30/1.849 = 19.63$ s, $t_3 = 36.30/3 = 12.10$ s, free run $= 95.2 - 19.63 - 12.10 = 63.47$ s (positive, so the trapezoidal curve is valid).

**Answer:** Required acceleration $\alpha \approx 1.85$ km/h/s.

## Check yourself

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

1. How is running time found from schedule speed?
   A. T = 115.2 + 20
   B. Ts = 3600D/Vs = 115.2 s, minus 20 s stop = 95.2 s
   C. T = D/Vs only
   D. T = 20 s
   Answer: B. Schedule time includes the stop, so subtract it. (Key point: Running time)

2. How is Vm found?
   A. Va = 3600 x 0.8/95.2 = 30.25, then Vm = 1.2 Va = 36.3 km/h
   B. Vm = 25 x 1.2
   C. Vm = 1.2 x Vs only
   D. Vm = 0.8 x 95.2
   Answer: A. Average speed uses running time, not schedule time. (Key point: Crest speed)

3. What is the required acceleration?
   A. About 1.85 km/h/s
   B. About 3 km/h/s
   C. About 0.54 km/h/s
   D. About 36.3 km/h/s
   Answer: A. 1/alpha = 0.5408 gives alpha = 1.849 km/h/s. (Key point: Acceleration)

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