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Acceleration car crash g's experienced?

  1. Sep 25, 2011 #1
    acceleration car crash g's experienced??

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

    Acceleration is sometimes expressed in multiples of g where g=9.8m/s^2 is the acceleration due to the earths gravity. In a car crash the cars velocity may go from 26m/s to 0m/s in 15s. how many g"s are experienced on average by the driver??
    a, 23- i know this is wrong
    b, 18g
    c,22g
    d,23g

    2. Relevant equations

    we have a, vi, vf and t
    so something like v=v0+at

    3. The attempt at a solution
    v=26m/s+ 9.8m/s^2*15s=173??
     
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  3. Sep 25, 2011 #2
    Re: acceleration car crash g's experienced??

    The general equation you have is correct, yes. However, look what you know you have. Do you know the final velocity? How about the initial velocity? Does the problem give you the time it takes for the change in velocity?

    Also, look at the given problem in your book or on the assignment. The number seems a little off for time. Are you sure it is not 0.15s? 15s would be more like someone stepping on the brakes of a car, rather than an impact.
     
  4. Sep 25, 2011 #3
    Re: acceleration car crash g's experienced??

    o haha. i guess i typed the time wrong. so that helps a bit but i redid it and got 27.
     
  5. Sep 25, 2011 #4
    Re: acceleration car crash g's experienced??

    Okay, look at the first half of the hint I gave. Did you rearrange the equation based on what you know? What does it look like when you do that?
     
  6. Sep 25, 2011 #5
    Re: acceleration car crash g's experienced??

    okay well. im not really sure what the question is asking me to find.It says g's. i know final and initial velocity and the time it takes to change. but how do i set that up?
     
  7. Sep 25, 2011 #6

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    Re: acceleration car crash g's experienced??

    Yep. That's the equation you need.
    However, you don't have "a" yet.
    "a" is what you need to calculate with this formula.

    v0 is the same as vi.
    And v is the same as vf.

    So you have:
    vf = vi + a t


    In this formula you substituted g=9.8 for the acceleration.
    But that's not right.
    "g" is not the acceleration.

    And v is not 173 after 15 seconds.
    v is vf.

    Can you substitute the final velocity vf, and then calculate "a" using your formula?
     
  8. Sep 25, 2011 #7

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    Re: acceleration car crash g's experienced??

    Yes, it says g's.
    This means that after you have calculated "a", you need to divide it by "g" to get an answer in g's.
     
  9. Sep 25, 2011 #8
    Re: acceleration car crash g's experienced??

    okay so then my equation is vf-vi/t=a?? because then you get -17.68 which is pretty close to 18 but its negative.
     
  10. Sep 25, 2011 #9

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    Re: acceleration car crash g's experienced??

    It is negative because the speed is decreasing to zero.
    However, the acceleration acting on the driver is the magnitude of this acceleration and as such positive.

    And you should check your numbers.
    If I use vi = 26 m/s
    and t = 15 s
    then I'm getting a = -26 / 15 / 9.8 = -0.1768 g

    So you seem to have a couple of decimal points in the wrong place...
     
  11. Sep 25, 2011 #10
    Re: acceleration car crash g's experienced??

    okay but theres no decimal points in the answer choices??
     
  12. Sep 25, 2011 #11

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    Re: acceleration car crash g's experienced??

    First, are you sure your numbers 26 m/s and 15 s are correct?
    You remarked earlier that you made a typo, but you did not clarify.
    Is it perhaps .15 s? Which is 0.15 s.

    Then, when you write: "b, 18g"
    Could that perhaps be: "b ,18g" which is actually "b 0.18g"?
     
  13. Sep 25, 2011 #12
    Re: acceleration car crash g's experienced??

    yes it was .15s but theres still no decimals in the answers
     
  14. Sep 25, 2011 #13

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    Re: acceleration car crash g's experienced??

    So redo the calculation with t = 0.15 s?
     
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