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Crest speed for 60 kmph schedule over 6 km

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.

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.

A simplified trapezoidal speed-time curve is used.

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

Schedule time and running time

Ts=3600×660=360 sT=360−60=300 s\begin{aligned} T_s &= \frac{3600 \times 6}{60} = 360\ \text{s} \\ T &= 360 - 60 = 300\ \text{s} \end{aligned}

(Average speed Va=3600×6/300=72V_a = 3600 \times 6 / 300 = 72 km/h.)

Crest speed

D=VmT3600−Vm27200(1α+1β)D = \frac{V_m T}{3600} - \frac{V_m^2}{7200}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)

With K=12(12+13)=0.4167K = \frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}\right) = 0.4167:

Vm=T2K−T24K2−3600DK=3000.8333−3602−3600×60.4167=360−129600−51840=360−278.85=81.15 km/h\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: t1=81.15/2=40.57t_1 = 81.15/2 = 40.57 s, t3=81.15/3=27.05t_3 = 81.15/3 = 27.05 s, free run t2=300−40.57−27.05=232.38t_2 = 300 - 40.57 - 27.05 = 232.38 s.

Answer: Crest speed Vm≈81.15V_m \approx 81.15 km/h.

✓ Answer checked by Claude Opus 5.5 (high effort)

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