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Finding average acceleration over an time interal

  1. Feb 9, 2009 #1
    1. The problem statement, all variables and given/known data

    An object moving with a constant acceleration travels 13 m in 7.8 s. The acceleration is then changed and the object covers an additional 34 m in 1.7 s. What is the average acceleration of the object over the entire interval described?

    2. Relevant equations

    Xf1 = 1/2(vo +vf)t
    a=(vo +vf)/t

    3. The attempt at a solution

    vf1 =2(13)/7.8
    vo =2(34)(1.7) - 44.2
    a = (115.6-44.2)9.5

    im not sure how to solve this problem
     
  2. jcsd
  3. Feb 9, 2009 #2
    take it back to orginial equation... you can not break down like that...
    here is my answer for you: 0.52m/s/s (starting at 0)
    however please consider the starting speed...
     
  4. Feb 9, 2009 #3
    can you show me what equation u used?
     
  5. Feb 9, 2009 #4
    lol, this has nothing to do with acceleration equation...
    you just simply add 2 total distance together and divid them by total time to get average v, then v/t=a
     
  6. Feb 9, 2009 #5
    if u add the total distance and time, you only get an average velocity of one interval. The motion changes velocity.
     
  7. Feb 9, 2009 #6
    do you mean average accerleration over entire time or something else? i am not quit getting it...
     
  8. Feb 9, 2009 #7
    average acceleration over the entire interval
     
  9. Feb 9, 2009 #8
    then my answer should be right...
     
  10. Feb 9, 2009 #9
    you said average...
     
  11. Feb 9, 2009 #10
    i solved it. by finding the first velocity d/t. then the second velocity is 2(d2/t2) - v1. then acceleration is vf - vo over time
     
  12. Feb 10, 2009 #11
    dude i told you....
     
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