---
title: "Acceleration for 200 t train, stations 777 m apart"
subject: "Utilization of Electrical Energy"
chapter: 4
key: "semester_7/utilization_of_electrical_energy:acceleration-200t-train-777m-stations"
source: https://www.bctengineer.com/q/semester_7/utilization_of_electrical_energy/acceleration-200t-train-777m-stations
site: BCT Engineering
---

# Acceleration for 200 t train, stations 777 m apart

Utilization of Electrical Energy · Chapter 4: Electric Traction

- Asked in: 2079 Baisakh, 8 marks; 2071 Shrawan (asked 2 times)
- 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-200t-train-777m-stations
- Whole chapter: https://www.bctengineer.com/md/semester_7/utilization_of_electrical_energy/chapter-4

## Question

The schedule speed with a 200 tonne train on an electric railway with stations 777 metres apart is 27.2 km per hour and the maximum speed is 20 percent higher than the average running speed. The braking rate is 3.22 km p.h.p.s. and the duration of stop is 20 seconds. Find the acceleration required. Assume a simplified speed-time curve with free running at the maximum speed.

## Short answer

Running time is 82.84 s, average speed 33.77 km/h, max speed 40.52 km/h. Using the trapezoidal distance relation and solving for acceleration yields α ≈ 2.70 km/h/s.

## Answer

Data: $D = 0.777$ km, $V_s = 27.2$ km/h, stop $= 20$ s, $\beta = 3.22$ km/h/s, $V_m = 1.2\,V_a$. Simplified trapezoidal curve (acceleration, free running, braking). The train mass does not affect the kinematics.

#### Running time and speeds
$$
\begin{aligned} T_s &= \frac{0.777}{27.2}\times3600 = 102.84\ \text{s} \\ T &= 102.84 - 20 = 82.84\ \text{s} \\ V_a &= \frac{0.777\times3600}{82.84} = 33.77\ \text{km/h} \\ V_m &= 1.2\times33.77 = 40.52\ \text{km/h} \end{aligned}
$$

#### Trapezoidal relation
$$
D\times3600 = V_m T - \frac{V_m^2}{2}\left(\frac1\alpha + \frac1\beta\right)
$$
$$
\begin{aligned} 2797.2 &= 40.52\times82.84 - \frac{40.52^2}{2}\left(\frac1\alpha + \frac{1}{3.22}\right) \\ \frac1\alpha + \frac{1}{3.22} &= \frac{2(3356.7 - 2797.2)}{1641.9} = 0.6815 \\ \frac1\alpha &= 0.6815 - 0.3106 = 0.3709 \\ \alpha &= 2.70\ \text{km/h/s} \end{aligned}
$$

Check: $t_1 = 40.52/2.696 = 15.03$ s, $t_3 = 40.52/3.22 = 12.58$ s, free run $= 82.84 - 15.03 - 12.58 = 55.23$ s; distance $= 40.52(55.23 + (15.03+12.58)/2)/3600 = 0.777$ km.

**Answer: Required acceleration ≈ 2.70 km/h/s.**

## Check yourself

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

1. How is the actual running time found from the schedule speed?
   A. Total time from D/Vs = 102.84 s, minus the 20 s stop = 82.84 s
   B. Total time is the running time
   C. Multiply the stop time by 3.22
   D. Divide D by Vm
   Answer: A. Schedule speed includes the stop, so T = 102.84 - 20 = 82.84 s. (Key point: Running time)

2. How is Vm obtained?
   A. Vm = 1.2 x Vs = 32.6 km/h
   B. Vm = Va + 20
   C. Vm = 1.2 x Va, with Va = D/T = 33.77 km/h, so Vm = 40.52 km/h
   D. Vm = 3.22 x 20
   Answer: C. The average running speed uses running time only, and the maximum is 20 percent above it. (Key point: Maximum speed)

3. What effect does the 200 tonne mass have on the required acceleration?
   A. Acceleration is proportional to the mass
   B. None; kinematics of the speed-time curve does not depend on mass
   C. It sets the braking rate
   D. It doubles the acceleration
   Answer: B. The acceleration follows from the trapezoid distance relation, D x 3600 = Vm T - (Vm^2/2)(1/α + 1/β); mass only matters for tractive effort. Result α is about 2.70 km/h/s. (Key point: Role of train mass)

## Related in Utilization of Electrical Energy

- [Acceleration for 25 kmph schedule speed, 800 m stations](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aacceleration-25kmph-800m-stations) — Utilization of Electrical Energy, chapter 4
- [Durations of acceleration, coasting and braking (2 km run)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Aquadrilateral-curve-2km-40kmph-periods) — Utilization of Electrical Energy, chapter 4, asked 2×
- [Maximum speed and distance before braking (2.5 km)](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Amax-speed-50kmph-2-5km-trapezoidal) — Utilization of Electrical Energy, chapter 4
- [Acceleration, coasting retardation and schedule speed](https://www.bctengineer.com/question.md?key=semester_7%2Futilization_of_electrical_energy%3Atrain-48kmph-coasting-schedule-speed-stops) — Utilization of Electrical Energy, chapter 4
- [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×

## Related in other subjects

- [Insulator discs per string, 180 MW over 200 km](https://www.bctengineer.com/question.md?key=semester_8%2Ftransmission_and_distribution_system_design%3Ainsulator-discs-180mw-200km) — Transmission and Distribution System Design, chapter 3
- [Floating car method: flow, journey and running speed](https://www.bctengineer.com/question.md?key=semester_7%2Ftransportation_engineering_ii%3Afloating-car-25-5km-flow-speed) — Transportation Engineering II, chapter 1