Examkin

Soru

Zorluk: OrtaStopping Distance Components

A commercial motor vehicle is traveling at a sustained speed of 45 mph45\text{ mph} (66 ft/s66\text{ ft/s}) on a dry, level highway. If the driver takes an average perception time of 1.75 seconds1.75\text{ seconds} to see and recognize a hazard, followed by an average reaction time of 0.75 seconds0.75\text{ seconds} to physically apply the foot brake, what is the combined distance the vehicle travels during these two human-factor phases before mechanical braking begins?

  1. A
    116 feet116\text{ feet}
  2. 165 feet165\text{ feet}Cevap
  3. C
    50 feet50\text{ feet}
  4. D
    264 feet264\text{ feet}

Cevap

The combined distance traveled during perception and reaction is 165 feet165\text{ feet}.
Total stopping distance before mechanical application is the sum of perception distance and reaction distance. At 45 mph45\text{ mph} (66 ft/s66\text{ ft/s}), perception time (1.75 seconds1.75\text{ seconds}) accounts for 115.5 feet115.5\text{ feet}, and reaction time (0.75 seconds0.75\text{ seconds}) accounts for 49.5 feet49.5\text{ feet}. Adding both together gives an exact total of 165 feet165\text{ feet} traveled before mechanical braking begins.

Adım Adım Çözüm

1
Calculate the distance traveled during the perception phase
dperception=1.75 s×66 ft/s=115.5 ftd_{\text{perception}} = 1.75\text{ s} \times 66\text{ ft/s} = 115.5\text{ ft}
Perception distance is the distance a vehicle travels from the moment a hazard is visible until the driver mentally perceives and recognizes it.
2
Calculate the distance traveled during the reaction phase
dreaction=0.75 s×66 ft/s=49.5 ftd_{\text{reaction}} = 0.75\text{ s} \times 66\text{ ft/s} = 49.5\text{ ft}
Reaction distance is the distance traveled while the driver moves their foot from the accelerator to press the brake pedal.
3
Sum the two pre-braking distance components
dcombined=115.5 ft+49.5 ft=165.0 ftd_{\text{combined}} = 115.5\text{ ft} + 49.5\text{ ft} = 165.0\text{ ft}
Total pre-braking human-factor distance is the sum of perception distance and reaction distance (or (1.75 s+0.75 s)×66 ft/s=2.50 s×66 ft/s=165 ft(1.75\text{ s} + 0.75\text{ s}) \times 66\text{ ft/s} = 2.50\text{ s} \times 66\text{ ft/s} = 165\text{ ft}).

Anahtar Kavram

Total pre-braking distance consists of perception distance plus reaction distance, which at 45 mph45\text{ mph} (66 ft/s66\text{ ft/s}) over 2.5 seconds2.5\text{ seconds} equals 165 feet165\text{ feet}.
Tahmini Süre:1m 0s
Bu soruyu puanla

Konular

Stopping Distance Components

Bu konudan daha fazla soru