Soru

Zorluk: Çok zorStopping Distance Components

A commercial driver operating an air-braked combination vehicle at 45 mph45\text{ mph} (66 ft/s66\text{ ft/s}) is maintaining exactly a 4.04.0-second following distance behind a lead vehicle. The lead vehicle strikes an immovable barrier and stops instantaneously. Assuming standard latency times (perception time of 1.75 s1.75\text{ s}, reaction time of 0.75 s0.75\text{ s}, and brake lag of 0.50 s0.50\text{ s}), what is the maximum mechanical braking distance the commercial vehicle can require to completely avoid a collision, and which components consume the rest of the available following distance?

  1. 66 feet66\text{ feet}; the remaining space is consumed by perception distance, reaction distance, and brake lag distance.Cevap
  2. B
    99 feet99\text{ feet}; the remaining space is consumed by perception distance and reaction distance only, as seen in hydraulic brake systems.
  3. C
    181.5 feet181.5\text{ feet}; the remaining space is consumed by reaction distance and brake lag distance, omitting perception distance.
  4. D
    115.5 feet115.5\text{ feet}; the remaining space is consumed by perception distance and brake lag distance, omitting reaction distance.

Cevap

The maximum allowable mechanical braking distance is 66 feet66\text{ feet}; the remaining space is consumed by perception distance, reaction distance, and brake lag distance.
To completely avoid a collision, the total stopping distance must be strictly less than the 264-foot264\text{-foot} following distance (4.0 s×66 ft/s4.0\text{ s} \times 66\text{ ft/s}). For an air-braked vehicle, the pre-braking time is the sum of perception (1.75 s1.75\text{ s}), reaction (0.75 s0.75\text{ s}), and brake lag (0.50 s0.50\text{ s}), totaling 3.003.00 seconds. During this latency, the vehicle travels 198 feet198\text{ feet} (3.00 s×66 ft/s3.00\text{ s} \times 66\text{ ft/s}). Subtracting this pre-braking distance from the total available space leaves exactly 66 feet66\text{ feet} for mechanical braking.

Adım Adım Çözüm

1
Calculate the total physical space available to stop.
4.0 s×66 ft/s=264 feet4.0\text{ s} \times 66\text{ ft/s} = 264\text{ feet}.
The 4.04.0-second following distance represents the total gap between the commercial vehicle and the lead vehicle before the hazard occurs.
2
Sum the time for all pre-braking components specific to air-braked vehicles.
1.75 s (perception)+0.75 s (reaction)+0.50 s (brake lag)=3.00 s1.75\text{ s} \text{ (perception)} + 0.75\text{ s} \text{ (reaction)} + 0.50\text{ s} \text{ (brake lag)} = 3.00\text{ s}.
Unlike hydraulic brakes, air brakes require accounting for a fourth component (brake lag) before the mechanical brakes engage and begin slowing the vehicle.
3
Calculate the pre-braking distance traveled during this latency period.
3.00 s×66 ft/s=198 feet3.00\text{ s} \times 66\text{ ft/s} = 198\text{ feet}.
The vehicle continues to travel at its original speed of 66 ft/s66\text{ ft/s} while the driver perceives, reacts, and the air system pressurizes.
4
Determine the maximum allowable mechanical braking distance.
264 feet−198 feet=66 feet264\text{ feet} - 198\text{ feet} = 66\text{ feet}.
Subtracting the pre-braking distance from the total available following distance reveals exactly how much space remains for the physical deceleration of the vehicle.

Anahtar Kavram

Air Brake Total Stopping Distance Components
Bu soruyu puanla