Schedule speed for a 1.5 km run (trapezoidal)

Data: $\alpha = 0.8$ km/h/s, $\beta = 3.2$ km/h/s, $V_m/V_a = 1.3$, $D = 1.5$ km, stop $= 26$ s.

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

$$3600D = V_m T - \frac{V_m^2}{2}\left(\frac1\alpha + \frac1\beta\right)$$ $$K = \frac1{0.8} + \frac1{3.2} = 1.5625$$

Since $V_a = 3600D/T$ and $V_m = 1.3V_a$: $V_m T = 1.3\times3600\times1.5 = 7020$.

$$\begin{aligned} 5400 &= 7020 - \frac{V_m^2}{2}\times1.5625 \\ V_m^2 &= \frac{2\times1620}{1.5625} = 2073.6 \\ V_m &= 45.54\ \text{km/h} \end{aligned}$$

Running time and schedule speed

$$\begin{aligned} T &= \frac{7020}{45.54} = 154.16\ \text{s} \\ V_a &= \frac{5400}{154.16} = 35.03\ \text{km/h} \\ T_s &= 154.16 + 26 = 180.16\ \text{s} \\ V_s &= \frac{1.5\times3600}{180.16} = 29.97\ \text{km/h} \end{aligned}$$

Check: $t_1 = 45.54/0.8 = 56.9$ s, $t_3 = 45.54/3.2 = 14.2$ s, free running $= 154.16 - 71.1 = 83.0$ s.

Answer: Schedule speed ≈ 29.97 km/h (≈ 30 km/h); maximum speed 45.5 km/h.