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Zorluk: ZorStopping Distance Components

A commercial motor vehicle equipped with an air brake system is traveling at a steady speed of 50 mph50\text{ mph} (73.33 ft/s73.33\text{ ft/s}) on a dry, level highway. The driver requires an average perception time of 1.75 seconds1.75\text{ seconds} to detect a roadway obstruction and a reaction time of 0.75 seconds0.75\text{ seconds} to move their foot to the brake pedal. The air brake system exhibits an air lag time of 0.50 seconds0.50\text{ seconds} before the mechanical brakes actuate. After the brakes engage, the vehicle travels an actual braking distance of 170 feet170\text{ feet} before coming to a complete stop. What is the total stopping distance of the vehicle, in feet, from the initial hazard perception to the complete stop?

Cevap: 390 feet

Cevap

The total stopping distance is 390 feet390\text{ feet}.
Total stopping distance in an air-brake commercial motor vehicle is the sum of four distinct phases: Perception Distance (128.33 ft128.33\text{ ft}), Reaction Distance (55.00 ft55.00\text{ ft}), Air Brake Lag Distance (36.67 ft36.67\text{ ft}), and Vehicle Braking Distance (170.00 ft170.00\text{ ft}). Adding these four segments together gives exactly 390 feet390\text{ feet}.

Adım Adım Çözüm

1
Calculate vehicle travel rate in feet per second
50 mph×5280 ft3600 s=2203 ft/s≈73.333 ft/s50\text{ mph} \times \frac{5280\text{ ft}}{3600\text{ s}} = \frac{220}{3}\text{ ft/s} \approx 73.333\text{ ft/s}
Distance calculations during constant-speed pre-braking phases require speed in feet per second.
2
Compute perception distance over 1.75 seconds1.75\text{ seconds}
dperception=73.333 ft/s×1.75 s=128.333 ftd_{\text{perception}} = 73.333\text{ ft/s} \times 1.75\text{ s} = 128.333\text{ ft}
Perception distance is the distance traveled from seeing the hazard until recognizing it.
3
Compute driver reaction distance over 0.75 seconds0.75\text{ seconds}
dreaction=73.333 ft/s×0.75 s=55.000 ftd_{\text{reaction}} = 73.333\text{ ft/s} \times 0.75\text{ s} = 55.000\text{ ft}
Reaction distance is the distance traveled while physically moving the foot from throttle to brake pedal.
4
Compute air brake lag distance over 0.50 seconds0.50\text{ seconds}
dlag=73.333 ft/s×0.50 s=36.667 ftd_{\text{lag}} = 73.333\text{ ft/s} \times 0.50\text{ s} = 36.667\text{ ft}
Unlike hydraulic systems, air brake systems introduce an additional delay for air to flow through the lines and push the brake shoes against the drums.
5
Sum all four stopping distance components
Dtotal=128.333 ft+55.000 ft+36.667 ft+170.000 ft=390.00 ftD_{\text{total}} = 128.333\text{ ft} + 55.000\text{ ft} + 36.667\text{ ft} + 170.000\text{ ft} = 390.00\text{ ft}
Total stopping distance for air brake vehicles equals Perception Distance + Reaction Distance + Brake Lag Distance + Braking Distance.

Anahtar Kavram

Four Components of Air Brake Total Stopping Distance
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Stopping Distance Components

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