Examkin

Question

Difficulty: HardSpeed Control and Following Distance

When a driver doubles their vehicle speed from 30 mph30\text{ mph} to 60 mph60\text{ mph}, doubling their time-based following distance from 3 seconds3\text{ seconds} to 6 seconds6\text{ seconds} keeps the physical space (distance in feet) between the two vehicles exactly the same.

Answer: Answer

Answer

The statement is False. Doubling both the speed and the time gap quadruples the physical following distance in feet, increasing it from 132 feet132\text{ feet} at 30 mph30\text{ mph} (3 seconds3\text{ seconds}) to 528 feet528\text{ feet} at 60 mph60\text{ mph} (6 seconds6\text{ seconds}).
The statement is false because distance equals velocity multiplied by time (D=v×tD = v \times t). When speed doubles (2v2v) and time interval doubles (2t2t), the resulting distance quadruples (4D4D). At 30 mph30\text{ mph}, 3 seconds3\text{ seconds} represents 132 feet132\text{ feet}, while at 60 mph60\text{ mph}, 6 seconds6\text{ seconds} represents 528 feet528\text{ feet}.

Step-by-Step Solution

1
Calculate the physical distance covered in feet during a 3-second3\text{-second} following distance at 30 mph30\text{ mph}.
At 30 mph30\text{ mph}, a vehicle travels 44 ft/s44\text{ ft/s}. Over 3 seconds3\text{ seconds}, the distance is 44 ft/s×3 s=132 feet44\text{ ft/s} \times 3\text{ s} = 132\text{ feet}.
Converting miles per hour to feet per second (1 mph≈1.467 ft/s1\text{ mph} \approx 1.467\text{ ft/s}) establishes the baseline spatial gap.
2
Calculate the physical distance covered in feet during a 6-second6\text{-second} following distance at 60 mph60\text{ mph}.
At 60 mph60\text{ mph}, a vehicle travels 88 ft/s88\text{ ft/s}. Over 6 seconds6\text{ seconds}, the distance is 88 ft/s×6 s=528 feet88\text{ ft/s} \times 6\text{ s} = 528\text{ feet}.
Doubling speed doubles the distance traveled per second.
3
Compare the initial and final physical distances in feet.
528 feet=4×132 feet528\text{ feet} = 4 \times 132\text{ feet}. The physical spatial gap quadruples rather than remaining constant.
Spatial distance scales with both speed and time interval (D=v×tD = v \times t), so doubling both variables multiplies the total spatial distance by four.

Key Concept

Speed, Time, and Distance Relationship in Following Distance
Rate this question

Topics

Speed Control and Following Distance

More questions from this topic