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Air bag injuries

  1. Feb 1, 2012 #1
    1. The problem statement, all variables and given/known data

    During an auto accident, the vehicle's air bags deploy and slowdown the passengers more gently than if they had hit the windshieldor steering wheel. According to safety standards, the bags producea maximum acceleration of 60, but lasting for only 36 (or less).

    How far (in meters) does a person travel in coming to a completestop in 36 at a constant accelerationof 60?


    2. Relevant equations

    s=1/2a(t)^2

    3. The attempt at a solution

    What I first did was analyze what I have. I know I have an acceleration of -588m/s^2 ( due to the ax=-g) also I have the time which is 0.036s.

    I initially wanted to use the forumula vx2=v0x2 +2ax(x-x0)

    Here is where I constantly get confused in these types of problems I know one of the velocities is going to be zero but how do I know which one to make zero.

    So what I wanted to do was aquire one of the velocities using vx=v0x+at. Then plug it into the above equation however the solution manual chose to use the following equation s=1/2a(t)^2.

    Can someone explain to me why my thought process is wrong and why, and how they aquired the above equation.

    Thank you
     
  2. jcsd
  3. Feb 1, 2012 #2

    Andrew Mason

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    Science Advisor
    Homework Helper

    36 what? 60 what? What are the units? Where did you get your formula? The units are not correct. vx^2 has units of distance^3/time and ax has dimensions of distance^2/time^2

    Assume the maximum acceleration (deceleration) is constant over the maximum stopping time. How would you determine initial speed from that? What is the final speed?

    You may find it is simpler to use s = .5at^2

    AM
     
    Last edited: Feb 1, 2012
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