A commercial truck equipped with an air brake system is traveling at a constant speed of 4… — Commercial Driver's License (CDL) General Knowledge Exam

A commercial truck equipped with an air brake system is traveling at a constant speed of $45\text{ mph}$ ( $66\text{ ft/s}$ ). The driver spots a stationary hazard ahead and initiates an emergency stop. The pre-braking sequence before the mechanical foundation brakes physically begin slowing the wheels comprises three distinct phases: a perception time of $1.75\text{ seconds}$ , a driver reaction time of $0.75\text{ seconds}$ , and an air brake system lag time of $0.50\text{ seconds}$ . What is the total distance traveled by the truck, in feet, during these three combined phases before physical braking begins?

Answer: 198 feet

Answer

198 feet

Before the foundation brakes physically begin slowing the vehicle, the truck travels through perception distance (

66\text{ ft/s} \times 1.75\text{ s} = 115.5\text{ ft}

), reaction distance (

66\text{ ft/s} \times 0.75\text{ s} = 49.5\text{ ft}

), and air brake lag distance (

66\text{ ft/s} \times 0.50\text{ s} = 33.0\text{ ft}

). Summing these distances gives

115.5 + 49.5 + 33.0 = 198\text{ feet}

(

66\text{ ft/s} \times 3.00\text{ s} = 198\text{ ft}

Step-by-Step Solution

Sum the durations of all three phases preceding mechanical braking.

Total pre-braking duration = 3.00 seconds

Perception time (1.75 seconds), reaction time (0.75 seconds), and air brake lag time (0.50 seconds) all occur while the vehicle continues moving at its initial cruising speed.

Multiply the vehicle velocity (66 ft/s) by the total pre-braking duration (3.00 s).

Total pre-braking distance = 198 feet

Distance is the product of constant velocity and elapsed time (

d = v \times t

) before deceleration forces are applied to the wheels.

Key Concept

Air brake stopping distance before actual braking begins includes perception distance, reaction distance, and brake lag distance.

Estimated Time:1m 30s

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