Maximum speed and distance before braking (2.5 km)

For a simplified trapezoidal curve, the distance between stops equals the area under the speed-time curve.

Given: $V_a = 50$ km/h, $D = 2.5$ km, $\alpha = 2$ km/h/s, $\beta = 3$ km/h/s.

Running time

$$\begin{aligned} T &= \frac{3600\,D}{V_a} = \frac{3600 \times 2.5}{50} = 180\ \text{s} \end{aligned}$$

Maximum (crest) speed

For a trapezoidal curve (speeds in km/h, time in s, $D$ in km):

$$D = \frac{V_m T}{3600} - \frac{V_m^2}{7200}\left(\frac{1}{\alpha} + \frac{1}{\beta}\right)$$

Let $K = \frac{1}{2}\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) = \frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}\right) = 0.4167$. Then

$$\begin{aligned} V_m &= \frac{T}{2K} - \sqrt{\frac{T^2}{4K^2} - \frac{3600\,D}{K}} \\ &= \frac{180}{0.8333} - \sqrt{\frac{180^2}{0.6944} - \frac{3600 \times 2.5}{0.4167}} \\ &= 216 - \sqrt{46656 - 21600} = 216 - 158.29 \\ &= 57.71\ \text{km/h} \end{aligned}$$

Times of each period

$$\begin{aligned} t_1 &= \frac{V_m}{\alpha} = \frac{57.71}{2} = 28.85\ \text{s} \\ t_3 &= \frac{V_m}{\beta} = \frac{57.71}{3} = 19.24\ \text{s} \\ t_2 &= T - t_1 - t_3 = 180 - 28.85 - 19.24 = 131.91\ \text{s} \end{aligned}$$

Distance before brakes are applied

Distance covered during braking:

$$D_3 = \frac{V_m t_3}{2 \times 3600} = \frac{57.71 \times 19.24}{7200} = 0.154\ \text{km}$$ $$D_1 + D_2 = D - D_3 = 2.5 - 0.154 = 2.346\ \text{km}$$

Speed-time curve

 V (km/h)
 57.7 |     ______________________
      |    /                      \
      |   /                        \
      |  /                          \
      | /                            \
    0 +/------------------------------\---- t (s)
      0   28.85                  160.76  180

Answer: Maximum speed $V_m = 57.71$ km/h; distance travelled before brakes are applied $= 2.346$ km (accelerating 28.85 s, free run 131.91 s, braking 19.24 s).