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Air-Bag Injuries. 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\(g\) that lasts for 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\(g?\)

Short Answer

Expert verified
The distance is approximately 0.381 meters.

Step by step solution

01

Understand the Problem

We need to find the distance a person travels while experiencing a deceleration of 60\(g\) over a time period of 36 milliseconds before coming to a stop. Deceleration means that the velocity decreases until it reaches zero.
02

Convert Units

We often use acceleration due to gravity \(g = 9.81 \, \text{m/s}^2\). Given that the acceleration is 60\(g\), convert it into \(\text{m/s}^2\): \[ a = 60 \times 9.81 = 588.6 \, \text{m/s}^2 \]The time given is 36 milliseconds, which is 0.036 seconds.
03

Use the Equation of Motion

To find the distance traveled under constant acceleration, use the equation of motion: \[ s = ut + \frac{1}{2} a t^2 \]Where:- \(s\) is the distance traveled.- \(u\) is the initial velocity (unknown, but non-zero).- \(a\) is the acceleration (588.6 m/s²).- \(t\) is the time (0.036 s).The final velocity \(v\) is 0, so we need to find the initial velocity \(u\) using another equation: \[ v = u + at \] Since \(v = 0\), we can rearrange to find \(u\): \[ u = -at = -588.6 \times 0.036 \] \[ u = -21.1776 \, \text{m/s}\]The negative sign here indicates that this is a deceleration (since velocity decreases).
04

Calculate Distance

Now substitute \(u = 21.1776 \, \text{m/s}\), \(a = -588.6 \, \text{m/s}^2\), and \(t = 0.036 \, \text{s}\) into the motion equation:\[ s = 21.1776 \, \times \, 0.036 + \frac{1}{2} \times (-588.6) \, \times \, (0.036)^2 \]Calculate the two parts separately: 1. \(21.1776 \, \times \, 0.036 = 0.7623936\)2. \(\frac{1}{2} \times -588.6 \, \times \, (0.036)^2 = -0.3811968\)Add them together: \[ s = 0.7623936 - 0.3811968 = 0.3811968 \, \text{m}\]
05

Final Distance

Therefore, the distance the person travels before coming to a stop is approximately 0.381 meters.

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Key Concepts

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

Understanding Physics Problems
When faced with a physics problem, like the air-bag injuries scenario, it's important to first grasp the situation and what exactly is being asked. In this case, we're examining how a passenger in a car accident slows down due to the deployment of an airbag. The key is to understand the relationship between acceleration and deceleration and how these affect motion over time.

Physics problems often involve deciphering which concepts and formulas apply. You begin by converting the information given into usable data. For instance, the problem provides acceleration in terms of 'g', the acceleration due to gravity. This must be converted to meters per second squared (m/s²) for calculations.

In this scenario, understanding that 1 'g' equals 9.81 m/s² is crucial. By multiplying this by 60, we convert the problem's conditions to 588.6 m/s². The time of 36 milliseconds also needs converting to seconds (0.036 seconds), as equations of motion require units to be consistent.
Equations of Motion
Equations of motion are the backbone of solving kinematic problems involving constant acceleration. They allow us to find unknown quantities such as distance, time, velocity, and acceleration.

In the air-bag scenario, we use the equation \[ s = ut + \frac{1}{2} a t^2 \] where:
  • \(s\) is the distance travelled.
  • \(u\) is the initial velocity which is to be determined from the context.
  • \(a\) is the constant acceleration (588.6 m/s² in this case).
  • \(t\) is the time period (0.036 s).
To find the initial velocity \(u\), we use another equation: \[ v = u + at \] Setting final velocity \(v\) to 0 allows us to solve for \(u\), yielding \(-21.1776 \text{ m/s}.\) The negative sign is typical, signaling deceleration. These equations are standard tools in physics for linking various motion attributes during constant acceleration.
Impact of Deceleration
Deceleration is simply negative acceleration, occurring when an object slows down. In this problem, it's shown as 60 times the gravitational acceleration, or \(-588.6 \text{ m/s}^2.\)

Deceleration plays a key role in safety features like airbags because they help reduce injuries by increasing the time it takes to stop, thus decreasing the force experienced by passengers.

When dealing with deceleration in motion equations, maintaining awareness of the direction of acceleration is crucial. Substituting into the standard equation \[ s = ut + \frac{1}{2} a t^2 \],not only gives distance but also reveals impact management's importance through controlled deceleration. The system safely brings a car passenger to a stop over a small distance (0.381 meters in this case), a critical safety development in modern automobiles.

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Most popular questions from this chapter

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