Acceleration for 25 kmph schedule speed, 800 m stations

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.