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Chapter 2: Problem 38

The "reaction time" of the average automobile driver is about 0.7 s. (The reaction time is the interval between the perception of a signal to stop and the application of the brakes.) If an automobile can slow down with an acceleration of 12.0 \(\mathrm{ft} / \mathrm{s}^{2}\) , compute the total distance covered in coming to a stop after a signal is observed (a) from an initial velocity of 15.0 \(\mathrm{mi} / \mathrm{h}(\mathrm{in}\) a school zone) and (b) from an initial velocity of 55.0 \(\mathrm{mi} / \mathrm{h}\)

Short Answer

Expert verified
(a) 35.57 ft, (b) 327.85 ft

Step by step solution

01

Convert Initial Velocity to ft/s

First, we need to convert the initial velocities from miles per hour (mi/h) to feet per second (ft/s). To do this, we use the conversion factor: 1 mi/h = 1.46667 ft/s.For (a) 15.0 mi/h:\[ 15.0 \text{ mi/h} \times 1.46667 \frac{\text{ft}}{\text{s}} \approx 22.0 \text{ ft/s} \]For (b) 55.0 mi/h:\[ 55.0 \text{ mi/h} \times 1.46667 \frac{\text{ft}}{\text{s}} \approx 80.7 \text{ ft/s} \]
02

Calculate Distance During Reaction Time

The distance covered during the reaction time is calculated using the formula: \[ d_{reaction} = v_i \cdot t_{reaction} \]Where \( v_i \) is the initial velocity and \( t_{reaction} = 0.7 \text{ s} \).For (a) 22.0 ft/s:\[ d_{reaction} = 22.0 \text{ ft/s} \times 0.7 \text{ s} = 15.4 \text{ ft} \]For (b) 80.7 ft/s:\[ d_{reaction} = 80.7 \text{ ft/s} \times 0.7 \text{ s} = 56.5 \text{ ft} \]
03

Calculate Stopping Distance After Braking

The stopping distance after the brakes are applied is calculated using the formula:\[ d_{braking} = \frac{{v_i^2}}{{2a}} \]where \( v_i \) is the initial velocity and \( a \) is the deceleration (12.0 ft/s\(^2\)).For (a) 22.0 ft/s:\[ d_{braking} = \frac{{(22.0 \text{ ft/s})^2}}{{2 \times 12.0 \text{ ft/s}^2}} = \frac{484}{24} = 20.17 \text{ ft} \]For (b) 80.7 ft/s:\[ d_{braking} = \frac{{(80.7 \text{ ft/s})^2}}{{2 \times 12.0 \text{ ft/s}^2}} = \frac{6512.49}{24} = 271.35 \text{ ft} \]
04

Calculate Total Stopping Distance

Finally, we add the distance covered during the reaction time and the distance covered after braking to find the total stopping distance.For (a):\[ d_{total} = d_{reaction} + d_{braking} = 15.4 \text{ ft} + 20.17 \text{ ft} = 35.57 \text{ ft} \]For (b):\[ d_{total} = d_{reaction} + d_{braking} = 56.5 \text{ ft} + 271.35 \text{ ft} = 327.85 \text{ ft} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Reaction Time
Reaction time is a crucial factor in driving as it determines how quickly a driver can respond to a stimulus such as a stop signal. It is the interval between perceiving a signal and the physical act of pressing the brake pedal. For the average driver, this time is approximately 0.7 seconds. During this period, the vehicle continues to travel at its initial speed, covering a certain distance known as the reaction distance.
  • Understanding reaction time helps in calculating the total stopping distance.
  • It is influenced by several factors, including driver alertness, road conditions, and distractions.
  • The longer the reaction time, the greater the distance traveled before the vehicle begins to decelerate.
The formula to calculate the distance covered during reaction time is: \[ d_{reaction} = v_i \cdot t_{reaction} \]where \( v_i \) is the initial velocity and \( t_{reaction} \) is the reaction time.
Deceleration
Deceleration refers to the rate at which a vehicle slows down. When a driver steps on the brakes, the vehicle's velocity decreases over time. This slowing down is expressed as negative acceleration or deceleration. It is an important concept in physics that impacts how quickly a vehicle can come to a stop.
  • In our exercise, the deceleration is given as 12.0 ft/s虏.
  • A higher deceleration value means the car will stop more quickly.
  • It is impacted by factors such as brake efficiency, tire conditions, and road surface.
By using the formula \[ a = \frac{(v_f - v_i)}{t} \], where \( v_f \) is the final velocity (0 in the case of stopping), \( v_i \) is the initial velocity, and \( t \) is the time taken to stop, one can understand how deceleration affects stopping distance.
Stopping Distance
Stopping distance is the total distance a vehicle travels before it comes to a complete halt. It comprises two main components: the distance traveled during the driver's reaction time, and the distance traveled during the braking process. Calculating stopping distance is vital for road safety, ensuring that drivers maintain a safe distance from other vehicles.
  • It reveals the required distance to avoid collisions after a signal is spotted.
  • The total stopping distance is the sum of reaction distance and braking distance.
For the braking distance, the formula \[ d_{braking} = \frac{v_i^2}{2a} \] is used, where \( v_i \) is the initial velocity in ft/s, and \( a \) is the deceleration in ft/s虏. This shows the importance of initial speed and deceleration in stopping a vehicle.
Velocity Conversion
Velocity conversion is essential in solving physics problems involving speeds and distances, especially when different units are involved. In many cases, speed might be given in miles per hour (mi/h), but calculations require the use of feet per second (ft/s).
  • Converting velocity aids in utilizing consistent units across different parts of a problem.
  • In our problem, 1 mi/h is converted to 1.46667 ft/s.
The conversion process involves using a conversion factor, making it crucial to ensure that calculations are consistent and accurate. Knowing how to convert velocities correctly ensures that subsequent calculations for reaction distance and stopping distance are reliable.
  • For example, converting 15.0 mi/h for (a) to 22.0 ft/s.
  • And 55.0 mi/h for (b) approximately 80.7 ft/s.
Correct unit conversion is the first step toward applying formulas and solving physics problems accurately.

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Most popular questions from this chapter

That's a lot of hot air! A hot-air balloonist, rising vertically with a constant speed of 5.00 \(\mathrm{m} / \mathrm{s}\) releases a sandbag at the instant the balloon is 40.0 \(\mathrm{m}\) above the ground. (See Figure \(2.52 . )\) After it is released, the sandbag encounters no appreciable air drag. (a) Compute the position and velocity of the sandbag at 0.250 s and 1.00 s after its release. (b) How many seconds after its release will the bag strike the ground? (c) How fast is it moving as it strikes the ground? (d) What is the greatest height above the ground that the sandbag reaches? (e) Sketch graphs of this bag's acceleration, velocity, and vertical position as functions of time.

\(\cdot\) If a pilot accelerates at more than 4\(g\) , he begins to "gray out," but not completely lose consciousness. (a) What is the shortest time that a jet pilot starting from rest can take to reach Mach 4 (four times the speed of sound) without graying out? (b) How far would the plane travel during this period of acceleration? (Use 331 \(\mathrm{m} / \mathrm{s}\) for the speed of sound in cold air.)

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In \(1954,\) Roger Bannister became the first human to run a mile in less than 4 minutes. Suppose that a runner on a straight track covers a distance of 1.00 mi in exactly 4.00 min. What is his average speed in (a) \(\mathrm{mi} / \mathrm{h},\) (b) \(\mathrm{ft} / \mathrm{s},\) and \((\mathrm{c}) \mathrm{m} / \mathrm{s}\) ?

A runner covers one lap of a circular track 40.0 \(\mathrm{m}\) in diameter in 62.5 s. For that lap, what were her average speed and average velocity? (b) If she covered the first half-lap in 28.7 s, what were her average speed and average velocity for that half-lap?
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