/*! 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 23 The human body can survive an ac... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 2: Problem 23

The human body can survive an acceleration trauma incident (sudden stop) if the magnitude of the acceleration is less than 250 m/s\(^{2}\). If you are in an automobile accident with an initial speed of 105 km/h (65 mi/h) and are stopped by an airbag that inflates from the dashboard, over what distance must the airbag stop you for you to survive the crash?

Short Answer

Expert verified
The airbag must stop you over 1.7 meters to survive.

Step by step solution

01

Convert Speed to Meters per Second

The initial speed is given in kilometers per hour (km/h). To use standard units in the calculation, we need to convert this speed to meters per second (m/s). The conversion factor is 1 km/h = 0.27778 m/s.\[ v_i = 105 \text{ km/h} \times \frac{0.27778 \text{ m/s}}{1 \text{ km/h}} = 29.167 \text{ m/s} \]
02

Identify Known Values

We have the following known values:- Initial speed, \( v_i = 29.167 \text{ m/s} \)- Final speed, \( v_f = 0 \text{ m/s} \) (since the body comes to a stop)- Maximum tolerable acceleration, \( a = 250 \text{ m/s}^2 \)
03

Use the Kinematic Equation

We will use the kinematic equation that relates initial velocity, final velocity, acceleration, and distance traveled:\[ v_f^2 = v_i^2 + 2a d \]Plug in the known values where \( v_f = 0 \), solve for \( d \):\[ 0 = (29.167)^2 + 2(-250) d \]
04

Solve for Distance \( d \)

Rearrange the equation to solve for \( d \):\[ 0 = 850.02889 - 500d \]\[ 500d = 850.02889 \]\[ d = \frac{850.02889}{500} \]\[ d = 1.7 \text{ meters} \]Therefore, the airbag must stop the person over a distance of 1.7 meters to ensure survival.

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.

Understanding Acceleration Trauma
When discussing acceleration trauma, it refers to the potential injuries one might sustain during a rapid change in speed. In the context of physics, such trauma can occur when the body undergoes a quick deceleration, like in a car crash. The human body can withstand certain levels of acceleration, but when it exceeds a certain threshold, it can lead to severe injuries.
To put it simply, this concept is about how fast you can slow down without getting hurt. For most people, the body can tolerate decelerations up to 250 m/s虏. Anything above this can lead to what is known as acceleration trauma.
This value is critical in designing safety features, such as airbags. They work by cushioning your impact and extending the time over which your momentum changes, thus reducing the effective acceleration.
Exploring the Kinematic Equation
Kinematic equations are fundamental in physics when analyzing motion. These equations help us understand and calculate how objects move under certain conditions, such as constant acceleration. In our problem, we rely on one particular kinematic equation:
  • \( v_f^2 = v_i^2 + 2a d \)
This equation relates:
  • Initial velocity \( (v_i) \)
  • Final velocity \( (v_f) \)
  • Acceleration \( (a) \)
  • Distance traveled \( (d) \)
By plugging the known values into this equation, we can determine the stopping distance needed to reduce the acceleration to a safe level. In the context of acceleration trauma, it ensures the force experienced by the body is distributed over a longer distance and time, minimizing injury risk.
Understanding how these elements interact provides insight into vehicle safety and the importance of proper crash mechanisms.
Practical Unit Conversion
Unit conversion is a crucial skill in physics to ensure all measurements are compatible. In this exercise, we convert the initial speed from kilometers per hour (km/h) to meters per second (m/s). This step is vital because the kinematic equations require consistent units for speed, distance, and acceleration.
The conversion factor between km/h and m/s is 1 km/h = 0.27778 m/s. For example, our initial speed is 105 km/h. To convert it:
  • \( 105 ext{ km/h} \times 0.27778 = 29.167 ext{ m/s} \)
This conversion ensures that calculations using the kinematic equations operate smoothly, maintaining accuracy and consistency.
Having the correct units allows for more accurate computations and understanding of how different physical quantities relate to one another.

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

A Honda Civic travels in a straight line along a road. The car's distance \(x\) from a stop sign is given as a function of time \(t\) by the equation \(x(t) = \alpha{t^2} - \beta{t^3}\), where \(\alpha =\) 1.50 m/s\(^2\) and \(\beta =\) 0.0500 m/s\(^3\). Calculate the average velocity of the car for each time interval: (a) \(t =\) 0 to \(t =\) 2.00 s; (b) \(t =\) 0 to \(t =\) 4.00 s; (c) \(t =\) 2.00 s to \(t =\) 4.00 s.

A helicopter carrying Dr. Evil takes off with a constant upward acceleration of 5.0 m/s\(^2\). Secret agent Austin Powers jumps on just as the helicopter lifts off the ground. After the two men struggle for 10.0 s, Powers shuts off the engine and steps out of the helicopter. Assume that the helicopter is in free fall after its engine is shut off, and ignore the effects of air resistance. (a) What is the maximum height above ground reached by the helicopter? (b) Powers deploys a jet pack strapped on his back 7.0 s after leaving the helicopter, and then he has a constant downward acceleration with magnitude 2.0 m/s\(^2\). How far is Powers above the ground when the helicopter crashes into the ground?

You throw a small rock straight up from the edge of a highway bridge that crosses a river. The rock passes you on its way down, 6.00 s after it was thrown. What is the speed of the rock just before it reaches the water 28.0 m below the point where the rock left your hand? Ignore air resistance.

Sam heaves a 16-lb shot straight up, giving it a constant upward acceleration from rest of 35.0 m/s\(^2\) for 64.0 cm. He releases it 2.20 m above the ground. Ignore air resistance. (a) What is the speed of the shot when Sam releases it? (b) How high above the ground does it go? (c) How much time does he have to get out of its way before it returns to the height of the top of his head, 1.83 m above the ground?

A rocket carrying a satellite is accelerating straight up from the earth's surface. At 1.15 s after liftoff, the rocket clears the top of its launch platform, 63 m above the ground. After an additional 4.75 s, it is 1.00 km above the ground. Calculate the magnitude of the average velocity of the rocket for (a) the 4.75-s part of its flight and (b) the first 5.90 s of its flight.
See all solutions

Recommended explanations on Physics Textbooks

Geometrical and Physical Optics

Read Explanation

Magnetism

Read Explanation

Torque and Rotational Motion

Read Explanation

Translational Dynamics

Read Explanation

Quantum Physics

Read Explanation

Conservation of Energy and Momentum

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.