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Zorluk: Çok zorStopping Distance Components

A safety compliance officer is reviewing telemetry data from an air-brake equipped commercial vehicle that performed an emergency stop on a wet grade. The vehicle was traveling at a precise velocity of 60 ft/s60\text{ ft/s}. The data log shows a total stopping distance of 375 ft375\text{ ft} from the moment the hazard entered the driver's field of view until the vehicle came to a complete rest. The physical braking distance on the slick surface was measured at 180 ft180\text{ ft}, and the air brake lag distance consumed exactly 30 ft30\text{ ft}. Assuming the driver's perception time perfectly matched the accepted FMCSA average of 1.75 seconds1.75\text{ seconds}, what was the driver's actual physical reaction time, and how does it compare to the standard average commercial reaction time?

  1. A
    1.5 seconds1.5\text{ seconds}; this is 0.75 seconds0.75\text{ seconds} slower than the standard average.
  2. B
    2.75 seconds2.75\text{ seconds}; this is 2.0 seconds2.0\text{ seconds} slower than the standard average.
  3. 1.0 second1.0\text{ second}; this is 0.25 seconds0.25\text{ seconds} slower than the standard average.Cevap
  4. D
    2.0 seconds2.0\text{ seconds}; this is 1.25 seconds1.25\text{ seconds} slower than the standard average.

Cevap

The driver's actual reaction time was 1.0 second, which is 0.25 seconds slower than the standard commercial average.
By subtracting the known components (perception distance of 105 ft, brake lag distance of 30 ft, and braking distance of 180 ft) from the total 375 ft, we find the reaction distance is 60 ft. Traveling at 60 ft/s, a 60 ft distance equals exactly 1.0 second. Since the standard commercial reaction time is 0.75 seconds, the driver was 0.25 seconds slower than average.

Adım Adım Çözüm

1
Identify the four mandatory components of total stopping distance for an air-brake equipped commercial vehicle.
Total Distance = Perception Distance + Reaction Distance + Brake Lag Distance + Braking Distance.
Air brakes have an inherent mechanical lag that hydraulic brakes do not, making it a 4-part equation.
2
Calculate the perception distance using the provided velocity and the standard FMCSA perception time.
60 ft/s×1.75 seconds=105 ft60\text{ ft/s} \times 1.75\text{ seconds} = 105\text{ ft}.
Perception distance is the distance traveled while the brain recognizes the hazard.
3
Set up the algebraic equation with the known values to solve for the missing reaction distance.
375 ft=105 ft+Reaction Distance+30 ft+180 ft375\text{ ft} = 105\text{ ft} + \text{Reaction Distance} + 30\text{ ft} + 180\text{ ft}.
Isolating the unknown variable allows us to find the specific distance traveled during the driver's physical reaction.
4
Solve for the reaction distance.
Reaction Distance = 375−(105+30+180)=375−315=60 ft375 - (105 + 30 + 180) = 375 - 315 = 60\text{ ft}.
Subtracting the sum of the known components from the total yields the distance.
5
Convert the reaction distance into reaction time and compare it to the FMCSA standard.
Reaction Time = 60 ft/60 ft/s=1.0 second60\text{ ft} / 60\text{ ft/s} = 1.0\text{ second}. Standard average is 0.75 seconds0.75\text{ seconds}. Difference = 0.25 seconds0.25\text{ seconds} slower.
Time equals distance divided by velocity.

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