Question 4 of 7: Moving-vehicle method (5 × 4 marks)

Question 4: Moving-vehicle method (5 × 4 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.

Given. Eight test-car runs in each direction, each recording the travel time $t$, the number met head-on $M$, the number that overtook the test car $O$, and the number the test car overtook $P$.

Find. The run averages, the volume in each direction, and the average travel time of the traffic (not of the test car) in each direction.

Approach. Average each column, then apply Wardrop’s two relations — taking care that the MET count used for a direction comes from the run made against it.

1. Part (a) — average every column. Each average is the column sum over the eight runs. Westbound: $\bar t_w = 37.60/8 = 4.700$ min, $\bar M_w = 1336/8 = 167.000$, $\bar O_w = 24/8 = 3.00$, $\bar P_w = 20/8 = 2.50$. Eastbound: $\bar t_e = 32.80/8 = 4.100$ min, $\bar M_e = 1216/8 = 152.000$, $\bar O_e = 20/8 = 2.50$, $\bar P_e = 16/8 = 2.00$.

Part (a): run averages

Run direction	Mean travel time (min)	Mean met M	Mean overtaking O	Mean overtaken P	Net overtaking O − P
Westbound	4.700	167.000	3.00	2.50	+0.50
Eastbound	4.100	152.000	2.50	2.00	+0.50

2. Set up Wardrop’s volume relation, and cross the directions. The volume in a direction is $$q = \frac{M_a + O_w - P_w}{t_a + t_w}$$ where the subscript $w$ means the run made with the direction of interest and $a$ the run made against it. The vehicles “met” while driving against a direction are the vehicles travelling in that direction, which is why $M$ must be taken from the opposing run. Using the same-direction $M$ is the single easiest way to lose this question, and it produces a plausible answer that nothing else contradicts.
3. Part (b) — westbound volume. The against-run for westbound is the eastbound run, so $M_a = \bar M_e = 152.000$, while $O$ and $P$ come from the westbound run itself. The denominator is the sum of the two mean run times, $4.100 + 4.700 = 8.800$ min: $$q_{WB} = \frac{152.000 + 3.00 - 2.50}{8.800} = \frac{152.50}{8.800} = 17.330\ \text{veh/min} = \boxed{1039.8\ \text{veh/h}}$$
4. Part (c) — eastbound volume. Now the roles reverse: $M_a = \bar M_w = 167.000$ from the westbound run, with the eastbound $O$ and $P$: $$q_{EB} = \frac{167.000 + 2.50 - 2.00}{8.800} = \frac{167.50}{8.800} = 19.034\ \text{veh/min} = \boxed{1142.0\ \text{veh/h}}$$
5. Part (d) — average westbound travel time. The test car’s own run time is not the travel time of the traffic: if more vehicles overtook the test car than it overtook, the test car was slower than the stream and its run time overstates the true mean. Wardrop’s correction subtracts the net overtaking divided by the volume, $$\bar t = t_w - \frac{O_w - P_w}{q}$$ so, working in veh/min so the units match, $$\bar t_{WB} = 4.700 - \frac{3.00 - 2.50}{17.330} = 4.700 - 0.0289 = \boxed{4.671\ \text{min}}\ (280.3\ \text{s})$$
6. Part (e) — average eastbound travel time. $$\bar t_{EB} = 4.100 - \frac{2.50 - 2.00}{19.034} = 4.100 - 0.0263 = \boxed{4.074\ \text{min}}\ (244.4\ \text{s})$$ In both directions the net overtaking is a positive 0.5 vehicles, so the stream was marginally faster than the test car and both corrections are small and downward — a sign of a well-conducted survey, in which the driver held the “average” speed closely.
7. Sanity checks. The two-way volume is $1039.8 + 1142.0 = 2181.8$ veh/h, split 47.7 per cent westbound and 52.3 per cent eastbound — a mild directional imbalance, consistent with an off-peak or shoulder-of-peak survey. The westbound stream takes 0.597 min (35.9 s) longer over the same section despite carrying less traffic, which is the kind of asymmetry that signal progression in one direction produces. As a check on the method, using the westbound MET count for the westbound volume would give 1142.0 veh/h — numerically the eastbound answer, and 102 veh/h from the correct value.