Moving observer: bicycle racer on 800 m bikeway

The moving observer method gives the flow and mean travel time of lane A from counts made by the rider moving with and against the stream.

Reading of the table: the cars met while riding southward are the northbound stream of lane A ($N_s$). While riding northward, the cars overtaking her ($N_o$) and the cars she overtakes ($N_p$) are counted. Following the order of the question text, the column "overtaken" is read as cars that overtook her ($N_o$: 10, 25, 15, 18) and "overtaking" as cars she overtook ($N_p$: 74, 41, 5, 9).

Average values (4 runs)

Travel time on each run $t = 0.8/v$: 90.0, 84.7, 90.0, 87.3 s, giving a mean of $t_w = t_a = 88.0$ s $= 1.467$ min $= 0.02444$ h.

$$\begin{aligned} \bar N_s &= \frac{117+93+30+70}{4} = 77.50\\ \bar N_o &= 17.00,\quad \bar N_p = 32.25,\quad y = \bar N_o - \bar N_p = -15.25 \end{aligned}$$

Flow

$$q = \frac{\bar N_s + y}{t_w + t_a} = \frac{77.50 + (-15.25)}{2(0.02444)} = 1273\ \text{veh/h}$$

Mean travel time of lane A

$$\bar t = t_w - \frac{y}{q} = 0.02444 - \frac{(-15.25)}{1273} = 0.03642\ \text{h} = 131.1\ \text{s}$$

Mean speed $= 0.8/0.03642 = 22.0$ km/h.

Answer: average flow of lane A = 1273 veh/h; average travel time = 131.1 s (2.19 min) over the 800 m, i.e. a stream speed of 22.0 km/h.

Note: if the column headings are read literally the other way round ("overtaking" = cars that overtook her, $\bar N_o = 32.25$; "overtaken" = cars she overtook, $\bar N_p = 17.00$), then $y = +15.25$, $q = (77.50 + 15.25)/0.04889 = 1897$ veh/h and $\bar t = 0.02444 - 15.25/1897 = 0.01640$ h $= 59.1$ s (stream speed 48.8 km/h). State the reading you use in the exam.