Durations of acceleration, coasting and braking (2 km run)

Data: $D = 2$ km, $V_{avg} = 40$ km/h, $V_{max} \le 60$ km/h, $\alpha = 2$, $\beta_c = 0.15$, $\beta = 3$ km/h/s.

Running time:

$$T = \frac{D}{V_{avg}} = \frac{2}{40}\times3600 = 180\ \text{s}$$

Check with 60 km/h reached

If the train accelerates to 60 km/h ($t_1 = 30$ s), coasts to $V_2$ and brakes, the time condition $30 + \frac{60-V_2}{0.15} + \frac{V_2}{3} = 180$ gives $V_2 = 39.47$ km/h, but the distance covered would be 2.21 km, more than 2 km. So the train does not need to reach 60 km/h; the peak speed $V_1$ is found from both conditions (and checked $\le 60$).

 V  V1 /\
      /  \___  coasting (-0.15)
     /       ----___ V2
    / accel 2       |\ brake 3
   /________________|_\____> t
   0   t1      t2      t3

Equations (speed in km/h, time in s; distance in km = area/3600)

$$\begin{aligned} \text{Time:}\quad & \frac{V_1}{2} + \frac{V_1 - V_2}{0.15} + \frac{V_2}{3} = 180 \\ \text{Distance:}\quad & \frac{V_1^2}{2\times2} + \frac{V_1^2 - V_2^2}{2\times0.15} + \frac{V_2^2}{2\times3} = 2\times3600 = 7200 \end{aligned}$$

From the time equation: $V_2 = \dfrac{7.1667\,V_1 - 180}{6.3333}$. Substituting in the distance equation and solving (numerically):

$$V_1 = 54.88\ \text{km/h}, \quad V_2 = 33.68\ \text{km/h}$$

$V_1 < 60$ km/h, so the speed limit is respected.

Periods

$$\begin{aligned} t_1 &= \frac{V_1}{\alpha} = \frac{54.88}{2} = 27.44\ \text{s} \\ t_2 &= \frac{V_1 - V_2}{\beta_c} = \frac{54.88 - 33.68}{0.15} = 141.33\ \text{s} \\ t_3 &= \frac{V_2}{\beta} = \frac{33.68}{3} = 11.23\ \text{s} \end{aligned}$$

Check: $27.44 + 141.33 + 11.23 = 180$ s.

Distances: acceleration $= \frac{54.88\times27.44}{2\times3600} = 0.209$ km; coasting $= \frac{(54.88+33.68)\times141.33}{2\times3600} = 1.738$ km; braking $= \frac{33.68\times11.23}{2\times3600} = 0.053$ km; total = 2.0 km.

Answer: acceleration 27.4 s, coasting 141.3 s, braking 11.2 s (peak speed 54.9 km/h, braking starts at 33.7 km/h).