/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 During an auto accident, the veh... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 32

During an auto accident, the vehicle's air bags deploy and slow down the passengers more gently than if they had hit the windshield or steering wheel. According to safety standards, the bags produce a maximum acceleration of \(60 g,\) but lasting only \(36 \mathrm{~ms}\) (or less). How far (in meters) does a person travel in coming to a complete stop in \(36 \mathrm{~ms}\) at a constant acceleration of \(60 \mathrm{~g} ?\)

Short Answer

Expert verified
The person travels approximately 0.381 meters.

Step by step solution

01

Identify the acceleration in \\( m/s^2 \\)

First, convert the given maximum acceleration from units of gravity \( g \) to meters per second squared (\( m/s^2 \)). The acceleration due to gravity \( g \) is approximately \( 9.81 \, m/s^2 \). Thus, the maximum acceleration is calculated as follows: \[ a = 60 \times 9.81 = 588.6 \, m/s^2 \]
02

Identify the initial velocity and final velocity

Since the problem states that the person comes to a complete stop, the final velocity \( v_f \) is \( 0 \, m/s \). Let's assume that the initial velocity \( v_i \) is not known, but we can use kinematic equations to solve for the distance.
03

Determine the time interval in seconds

The time given is in milliseconds \( 36 \, ms \). Convert this to seconds: \[ t = 36 \, ms = 36 \times 10^{-3} \, s = 0.036 \, s \]
04

Use the kinematic equation to find distance

We can use the kinematic equation that relates initial velocity \( v_i \), final velocity \( v_f \), acceleration \( a \), time \( t \), and distance \( d \): \[ v_f = v_i + a \times t \] However, since we need to find the distance traveled, we will use the equation: \[ d = v_i \times t + \frac{1}{2} a \times t^2 \] \( v_i \) is not needed since it becomes zero when the full acceleration is applied, thus \[ d = \frac{1}{2} \times 588.6 \times (0.036)^2 \] Compute this to find the distance.
05

Calculate the distance

Simplify and calculate the previously derived formula: \[ d = \frac{1}{2} \times 588.6 \times (0.036)^2 \approx 0.381 \text{ meters} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

kinematic equations
Kinematic equations are a set of mathematical formulas used to describe the motion of objects. They are particularly useful when dealing with problems involving constant acceleration, where factors like velocity, time, and displacement need to be connected. In the context of the exercise involving airbag deployment, kinematic equations help to determine the distance a person travels while coming to a stop.
  • The core equations include relationships between initial and final velocities, acceleration, time, and displacement.
  • For the given problem, the relevant equation is: \[ d = v_i \times t + \frac{1}{2} a \times t^2 \]This equation allows you to solve for the distance \( d \), based on initial velocity \( v_i \), time \( t \), and acceleration \( a \).
  • Kinematic equations provide a structured approach to analyze how different factors in motion influence each other under constant acceleration.
By applying the appropriate kinematic equation, you can find out how far a person travels while decelerating with the aid of the airbag.
constant acceleration
Constant acceleration refers to a steady and unchanging rate of change in velocity over time. This is a key condition that simplifies many motion problems because it allows us to use kinematic equations confidently. In the exercise, the airbag decelerates the passenger with a constant acceleration of 60 times the gravitational acceleration, which equals 588.6 m/s虏.
  • Because the acceleration is constant, the equations become simpler, reducing the need for calculus-based approaches.
  • A constant acceleration implies that each second, the velocity changes by the same amount, providing a predictable pattern of motion.
  • When dealing with safety devices like airbags, understanding and calculating constant acceleration helps ensure that the forces exerted on passengers are within safe limits.
Thus, constant acceleration serves as a cornerstone concept for analyzing motion in a controlled and predictable manner, especially in vehicle safety designs.
vehicle safety
Vehicle safety is a comprehensive field that focuses on minimizing harm to passengers during accidents. A crucial aspect of vehicular safety is the management of forces exerted on passengers during rapid deceleration, such as in crashes.
  • Features like seat belts and airbags are designed to work in harmony to protect occupants by managing the motion and forces exerted during a collision.
  • Devices like airbags help in redistributing the passenger's deceleration over time to reduce the risk of injury, making rapid stops less harmful.
  • Understanding motion physics and vehicle dynamics allows engineers to design systems that both prevent accidents and minimize injury severity when they occur.
By integrating concepts such as constant acceleration and using tools like kinematic equations, vehicle safety can be optimized to enhance overall protection for passengers.
airbags
Airbags are a critical component of modern vehicle safety systems, designed to protect passengers during collisions by providing a cushioning effect. They deploy in milliseconds, slowing down the passenger's movement more gently than a direct impact with the car's interior.
  • Airbags inflate during a crash, usually within 30 to 40 milliseconds, utilizing chemical reactions that rapidly generate gas.
  • They spread the stopping force over a larger area of the body, decreasing the risk of localized injury or direct impacts with harder surfaces.
  • The deceleration provided by an airbag reduces the intense forces experienced by passengers, by extending the deceleration time, reducing the potential for severe injury.
Through innovative engineering and application of physics principles such as constant acceleration, airbags serve as highly effective safety features that significantly enhance crash survival rates.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two coconuts fall freely from rest at the same time, one from a tree twice as high as the other. (a) If the coconut from the taller tree reaches the ground with a speed \(V,\) what will be the speed (in terms of \(V\) ) of the coconut from the other tree when it reaches the ground? (b) If the coconut from the shorter tree takes time \(T\) to reach the ground, how long (in terms of \(T\) ) will it take the other coconut to reach the ground?

The "reaction time" of the average automobile driver is about \(0.7 \mathrm{~s}\). (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 \(15.0 \mathrm{mi} / \mathrm{h}\) (in a school zone) and (b) from an initial velocity of \(55.0 \mathrm{mi} / \mathrm{h}\).

Astronauts on our moon must function with an acceleration due to gravity of \(1.62 g .\) (a) If an astronaut can throw a certain wrench \(12.0 \mathrm{~m}\) vertically upward on earth, how high could he throw it on our moon if he gives it the same starting speed in both places? (b) How much longer would it be in motion (going up and coming down) on the moon than on earth?

Suppose that you drop a marble from the top of the Burj Khalifa building in Dubai, which is about \(830 \mathrm{~m}\) tall. If you ignore air resistance, (a) how long will it take for the marble to hit the ground? (b) How fast will it be moving just before it hits?

On March \(27,2004,\) the United States successfully tested the hypersonic X-43A scramjet, which flew at Mach 7 (seven times the speed of sound) for 11 s. (a) At this rate, how many minutes would it take such a scramjet to carry passengers the approximately \(5000 \mathrm{~km}\) from San Francisco to New York? (Use the speed of sound at \(\left.0^{\circ} \mathrm{C}, 331 \mathrm{~m} / \mathrm{s} .\right)\) (b) How many kilometers did the scramjet travel during its 11 s test?
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Radiation

Read Explanation

Kinematics Physics

Read Explanation

Linear Momentum

Read Explanation

Waves Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.