Acceleration for 200 t train, stations 777 m apart

Data: $D = 0.777$ km, $V_s = 27.2$ km/h, stop $= 20$ s, $\beta = 3.22$ km/h/s, $V_m = 1.2\,V_a$. Simplified trapezoidal curve (acceleration, free running, braking). The train mass does not affect the kinematics.

Running time and speeds

$$\begin{aligned} T_s &= \frac{0.777}{27.2}\times3600 = 102.84\ \text{s} \\ T &= 102.84 - 20 = 82.84\ \text{s} \\ V_a &= \frac{0.777\times3600}{82.84} = 33.77\ \text{km/h} \\ V_m &= 1.2\times33.77 = 40.52\ \text{km/h} \end{aligned}$$

Trapezoidal relation

$$D\times3600 = V_m T - \frac{V_m^2}{2}\left(\frac1\alpha + \frac1\beta\right)$$ $$\begin{aligned} 2797.2 &= 40.52\times82.84 - \frac{40.52^2}{2}\left(\frac1\alpha + \frac{1}{3.22}\right) \\ \frac1\alpha + \frac{1}{3.22} &= \frac{2(3356.7 - 2797.2)}{1641.9} = 0.6815 \\ \frac1\alpha &= 0.6815 - 0.3106 = 0.3709 \\ \alpha &= 2.70\ \text{km/h/s} \end{aligned}$$

Check: $t_1 = 40.52/2.696 = 15.03$ s, $t_3 = 40.52/3.22 = 12.58$ s, free run $= 82.84 - 15.03 - 12.58 = 55.23$ s; distance $= 40.52(55.23 + (15.03+12.58)/2)/3600 = 0.777$ km.

Answer: Required acceleration ≈ 2.70 km/h/s.