Question

Difficulty: Very hardSpace Ahead and Following Distance

Two commercial drivers are traveling on a clear, dry highway at a steady speed of 55 mph55\text{ mph}. Driver A is operating a 4040-foot straight truck, while Driver B is operating an 8080-foot combination vehicle. According to standard commercial vehicle space management formulas, Driver B is required to maintain exactly double the forward following distance (measured in seconds) compared to Driver A.

Answer: Answer

Answer

False
The correct answer is False. While an 8080-foot vehicle is twice as long as a 4040-foot vehicle, the required following distance at speeds over 40 mph40\text{ mph} includes a flat 1-second1\text{-second} addition for safety. Driver A requires 5 seconds5\text{ seconds} (4 seconds4\text{ seconds} for length + 1 second1\text{ second} for speed), while Driver B requires 9 seconds9\text{ seconds} (8 seconds8\text{ seconds} for length + 1 second1\text{ second} for speed). Because the 1-second1\text{-second} speed penalty is a constant addition rather than a multiplier, 9 seconds9\text{ seconds} is not exactly double 5 seconds5\text{ seconds}.

Step-by-Step Solution

1
Calculate the base following distance for Driver A (4040-foot vehicle).
4010=4 seconds\frac{40}{10} = 4\text{ seconds}.
The base standard is 1 second1\text{ second} of following distance for every 10 feet10\text{ feet} of vehicle length.
2
Apply the high-speed safety modifier for Driver A.
4 seconds+1 second=5 seconds4\text{ seconds} + 1\text{ second} = 5\text{ seconds}.
Because the vehicle is traveling at 55 mph55\text{ mph} (which is ≥40 mph\ge 40\text{ mph}), exactly 1 additional second1\text{ additional second} must be added.
3
Calculate the base following distance for Driver B (8080-foot vehicle).
8010=8 seconds\frac{80}{10} = 8\text{ seconds}.
Applying the same base standard of 1 second1\text{ second} per 10 feet10\text{ feet} of vehicle length.
4
Apply the high-speed safety modifier for Driver B.
8 seconds+1 second=9 seconds8\text{ seconds} + 1\text{ second} = 9\text{ seconds}.
The vehicle is also traveling at 55 mph55\text{ mph}, requiring the identical flat 1-second1\text{-second} addition.
5
Compare Driver B's total required time to Driver A's total required time.
9 seconds≠2×5 seconds9\text{ seconds} \neq 2 \times 5\text{ seconds}.
The statement claims the required time is exactly double, which is mathematically false due to the non-proportional constant added to both.

Key Concept

Commercial Vehicle Following Distance Formula and High-Speed Modifier Proportionality
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