Acceleration, coasting retardation and schedule speed

Assumption: braking at 3.3 km/h/s from the speed at end of coasting (the braking time then comes out as 11.6 s; the stated 10 s does not match this rate, see the note at the end).

(i) Acceleration

$$\alpha = \frac{48}{24} = 2\ \text{km/h/s}$$

(ii) Coasting retardation

Resistance 50 N/t, effective mass $1.1W$:

$$\begin{aligned} 277.8\times1.1\,\beta_c &= 50 \\ \beta_c &= \frac{50}{305.58} = 0.1636\ \text{km/h/s} \end{aligned}$$

(iii) Schedule speed (20 s stops)

Speed at end of coasting:

$$V_2 = 48 - 0.1636\times60 = 38.18\ \text{km/h}$$

Braking time: $t_3 = 38.18/3.3 = 11.57$ s.

$$\begin{aligned} T_s &= 95.57 + 20 = 115.57\ \text{s} \\ V_s &= \frac{939.55}{115.57}\times3.6 = 29.27\ \text{km/h} \end{aligned}$$

Effect of 15 s stops

$$\begin{aligned} T_s &= 95.57 + 15 = 110.57\ \text{s} \\ V_s &= \frac{939.55}{110.57}\times3.6 = 30.59\ \text{km/h} \end{aligned}$$

Increase $= 30.59 - 29.27 = 1.32$ km/h (about 4.5 %).

Answer: (i) 2 km/h/s, (ii) 0.164 km/h/s, (iii) 29.27 km/h; with 15 s stops the schedule speed rises to 30.59 km/h.

Note: if the braking period is taken as exactly 10 s from 38.18 km/h, the distance is 931.2 m and the schedule speeds are 29.41 km/h (20 s stops) and 30.76 km/h (15 s stops); the conclusion is the same.