Examkin

Question

Difficulty: MediumStopping Distance Components

A fleet safety audit examines telemetry from a tractor-semitrailer traveling on a dry expressway at a constant speed of 50 mph50\text{ mph} (which equals 2203 ft/s\frac{220}{3}\text{ ft/s}, or approximately 73.33 ft/s73.33\text{ ft/s}). When an obstacle appears ahead, the driver takes 1.75 seconds1.75\text{ seconds} of perception time to recognize the hazard and 0.75 seconds0.75\text{ seconds} of reaction time to press the brake pedal. The vehicle's pneumatic brake system introduces an air brake lag of 0.50 seconds0.50\text{ seconds} before the brake linings contact the drums. How many feet does the commercial vehicle travel from the moment the hazard appears until the brakes actually begin slowing the vehicle down?

Answer: 220 ft

Answer

220 feet
Total stopping distance in an air-brake-equipped vehicle consists of four components: perception distance, reaction distance, brake lag distance, and effective braking distance. Before the vehicle begins decelerating, it travels during perception (1.75 s1.75\text{ s}), reaction (0.75 s0.75\text{ s}), and air lag (0.50 s0.50\text{ s}). Adding these yields an unbraked duration of 3.00 seconds3.00\text{ seconds}. At 50 mph50\text{ mph} (2203 ft/s\frac{220}{3}\text{ ft/s}), the distance covered is 2203 ft/s×3.00 s=220 feet\frac{220}{3}\text{ ft/s} \times 3.00\text{ s} = 220\text{ feet}.

Step-by-Step Solution

1
Determine the total unbraked delay time before friction takes effect.
Total delay time = 1.75 s+0.75 s+0.50 s=3.00 seconds1.75\text{ s} + 0.75\text{ s} + 0.50\text{ s} = 3.00\text{ seconds}
Air-braked commercial vehicles experience three distinct phases before deceleration begins: perception delay, reaction delay, and pneumatic brake lag.
2
Calculate vehicle velocity in feet per second at 50 mph50\text{ mph}.
Velocity = 50 mph×5280 ft3600 s=2203 ft/s≈73.33 ft/s50\text{ mph} \times \frac{5280\text{ ft}}{3600\text{ s}} = \frac{220}{3}\text{ ft/s} \approx 73.33\text{ ft/s}
Distance calculations require consistent units of feet and seconds.
3
Multiply velocity by the total pre-braking elapsed time.
Distance = 2203 ft/s×3.00 s=220 ft\frac{220}{3}\text{ ft/s} \times 3.00\text{ s} = 220\text{ ft}
Because the vehicle maintains cruising speed throughout all three initial phases, multiplying constant speed by cumulative time yields total pre-braking distance.

Key Concept

Air Brake Stopping Distance Components (Pre-Braking Distance)
Estimated Time:1m 0s
Rate this question

Topics

Stopping Distance Components

More questions from this topic