Schedule speed for a 2 km run (trapezoidal)

Data: $\alpha = 0.8$ km/h/s, $\beta = 3.2$ km/h/s, $V_m/V_a = 1.3$, $D = 2$ km, stop $= 26$ s.

Trapezoidal relation

$$D = \frac{1}{3600}\left[V_m T - \frac{V_m^2}{2}\left(\frac1\alpha + \frac1\beta\right)\right]$$ $$K = \frac1\alpha + \frac1\beta = \frac1{0.8} + \frac1{3.2} = 1.25 + 0.3125 = 1.5625$$

Average speed $V_a = \dfrac{3600D}{T}$, so $V_m T = 1.3\times3600D = 1.3\times7200 = 9360$.

$$\begin{aligned} 7200 &= 9360 - \frac{V_m^2}{2}\times1.5625 \\ V_m^2 &= \frac{2\times2160}{1.5625} = 2764.8 \\ V_m &= 52.58\ \text{km/h} \end{aligned}$$

Times and speeds

$$\begin{aligned} T &= \frac{9360}{52.58} = 178.01\ \text{s} \\ V_a &= \frac{7200}{178.01} = 40.45\ \text{km/h} \\ T_s &= 178.01 + 26 = 204.01\ \text{s} \\ V_s &= \frac{2\times3600}{204.01} = 35.29\ \text{km/h} \end{aligned}$$

Check: $t_1 = 52.58/0.8 = 65.7$ s, $t_3 = 52.58/3.2 = 16.4$ s, free run $= 95.9$ s.

Answer: Schedule speed = 35.3 km/h (maximum speed 52.6 km/h).