Question 5 of 7: Moving-vehicle method for volume and travel time

Question 5 — Moving-vehicle method for volume and travel time (a) to (e), 4 marks each — 20 marks

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Approach. Average the four recorded quantities over the eight runs in each direction, then apply Wardrop's two moving-observer relations: the volume in a direction comes from the count of vehicles met on the run against that stream corrected by the net overtaking exchange on the run with it, and the mean travel time of the stream is the test vehicle's own run time corrected by the same exchange.

1. Establish the direction convention. The data table labels the two run sets "Southbound" and "Northbound" while parts (a)–(e) ask for eastbound and westbound results. The two labelled sets are simply the two opposing directions of the study section, and the parts name them in the order the table lists them, so under the paper's Note 1 the convention adopted is
   $$\text{"Southbound" runs} \equiv \text{EASTBOUND direction}, \qquad \text{"Northbound" runs} \equiv \text{WESTBOUND direction}$$
   Nothing in the arithmetic depends on the labels — only on which run set is the "with" run and which the "against" run for each answer.
2. (a) Compute all averages for both sets of trips. Each column is the arithmetic mean of eight values. For the eastbound (first) set the travel times sum to $3.04+2.80+3.15+2.95+3.47+3.51+3.28+3.17 = 25.37$ min, so $\bar t = 25.37/8 = 3.171$ min; the other columns follow the same way.

   Average over 8 runs	Eastbound set ("Southbound" runs)	Westbound set ("Northbound" runs)
   Travel time $t$ (min)	3.171	2.758
   Vehicles met in opposite direction, $M$	111.75	101.75
   Vehicles that overtook the test vehicle, $O$	2.00	1.50
   Vehicles overtaken by the test vehicle, $P$	1.50	1.25
   Net exchange $O - P$	$+0.50$	$+0.25$

   Both net exchanges are positive, meaning the test vehicle travelled slightly slower than the average of the stream it was in — a useful sanity check, since it tells us the travel-time corrections in (d) and (e) must both reduce the run time.
3. State Wardrop's moving-observer relations. For a direction of interest, with $t_a$ the mean run time against that stream, $t_w$ the mean run time with it, $M_a$ the mean number of opposing vehicles met on the against-run, and $y = O_w - P_w$ the mean net exchange on the with-run,
   $$q = \frac{M_a + y}{t_a + t_w}, \qquad \bar t = t_w - \frac{y}{q}$$
   The first relation works because the against-run sweeps past every vehicle in the stream of interest, so $M_a$ counts the whole flow over the combined time $t_a + t_w$; the second removes the difference between the test vehicle's own speed and the stream's.
4. (b) Eastbound traffic volume. For the eastbound stream, the run against it is the westbound run set ($t_a = 2.758$ min, $M_a = 101.75$ vehicles met — those met are eastbound vehicles), and the run with it is the eastbound set ($t_w = 3.171$ min, $y = +0.50$):
   $$q_{EB} = \frac{101.75 + 0.50}{2.758 + 3.171} = \frac{102.25}{5.929} = 17.25\ \text{veh}\,\text{min}^{-1}$$
   $$q_{EB} = 17.25 \times 60 = \boxed{1035\ \text{veh}\,\text{h}^{-1}}$$
5. (c) Westbound traffic volume. Now the roles reverse: the against-run is the eastbound set ($t_a = 3.171$ min, $M_a = 111.75$) and the with-run is the westbound set ($t_w = 2.758$ min, $y = +0.25$):
   $$q_{WB} = \frac{111.75 + 0.25}{3.171 + 2.758} = \frac{112.00}{5.929} = 18.89\ \text{veh}\,\text{min}^{-1} = \boxed{1134\ \text{veh}\,\text{h}^{-1}}$$
   The denominator is the same $t_a + t_w = 5.929$ min in both directions, which is a convenient check: only the numerators differ. The westbound flow is about 10 % heavier, consistent with its faster mean run time and the smaller number of vehicles met by the westbound test car.
6. (d) Average travel time of eastbound traffic. Correcting the test vehicle's own eastbound run time for the net exchange:
   $$\bar t_{EB} = t_w - \frac{y}{q_{EB}} = 3.171 - \frac{0.50}{17.25} = 3.171 - 0.029 = \boxed{3.14\ \text{min}}$$
7. (e) Average travel time of westbound traffic. Likewise,
   $$\bar t_{WB} = 2.758 - \frac{0.25}{18.89} = 2.758 - 0.013 = \boxed{2.74\ \text{min}}$$
   Both corrections are small — about 1 % and 0.5 % — and both act downward, as anticipated in Step 2: the test vehicle was marginally slower than the stream in each direction, so the stream's mean travel time is a little less than the test vehicle's.