Schedule speed for a 2.5 km run (trapezoidal)

Data: $\alpha = 1.2$ km/h/s, $\beta = 3.8$ km/h/s, $V_m/V_a = 1.6$, $D = 2.5$ km, stop $= 45$ s.

Trapezoidal relation (speeds in km/h, time in s)

$$3600D = V_m T - \frac{V_m^2}{2}\left(\frac1\alpha + \frac1\beta\right)$$ $$K = \frac1{1.2} + \frac1{3.8} = 0.8333 + 0.2632 = 1.0965$$

With $V_a = 3600D/T$ and $V_m = 1.6V_a$: $V_m T = 1.6\times3600\times2.5 = 14400$.

$$\begin{aligned} 9000 &= 14400 - \frac{V_m^2}{2}\times1.0965 \\ V_m^2 &= \frac{2\times5400}{1.0965} = 9849.6 \\ V_m &= 99.25\ \text{km/h} \end{aligned}$$

Running time and schedule speed

$$\begin{aligned} T &= \frac{14400}{99.25} = 145.10\ \text{s} \\ V_a &= \frac{9000}{145.10} = 62.03\ \text{km/h} \\ T_s &= 145.10 + 45 = 190.10\ \text{s} \\ V_s &= \frac{2.5\times3600}{190.10} = 47.34\ \text{km/h} \end{aligned}$$

Check: $t_1 = 99.25/1.2 = 82.7$ s, $t_3 = 99.25/3.8 = 26.1$ s, free running $= 145.10 - 108.8 = 36.3$ s (positive, so the curve is valid).

Answer: Schedule speed ≈ 47.3 km/h (maximum speed 99.2 km/h).