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Acceleration homework help

  1. Oct 7, 2007 #1
    1. The problem statement, all variables and given/known data

    An object undergoes constant acceleration, initially at rest, then it travels 5 meters in the first second. What additional distance will be covered in the next second?

    3. The attempt at a solution

    Is it right that it covers 15 meters the next second? I think it should cover 10 meters the next second. Well, if I use kinematic equation to find out the acceleration, a=10m/s^2 when time lasts two seconds, but i think the question asks you the next second, so it lasts only 1 second,too despite it's the 2nd second, so I use my common sense, the acceleration for next second should be 5m/s^2, too. I got the additional distance is 10 meters. Which one is right?
     
  2. jcsd
  3. Oct 7, 2007 #2

    Doc Al

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    I'm not telling! :wink: But the acceleration doesn't change. Apply that same kinematic equation twice: for t = 1 and for t = 2.
     
  4. Oct 7, 2007 #3
    I mean the question asks me for the additional distance in the next second. It's the 2nd second, but the time still lasts for one second. so I would choose to use t=1. If I use t=2, then a=10, but it is from time 0-2. It's different.
     
  5. Oct 7, 2007 #4

    Doc Al

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    If you are starting from the beginning, you'd use the distance from t = 1 to t = 2. But if you want to start counting 1 sec after the first 5 m, then don't forget that it is no longer starting from rest.

    Maybe it would to clearer to answer these questions:
    (1) Where is it after 1 second? (Answer is given: x = 5 m.)
    (2) Where is it after 2 seconds?

    Once you answer these you can see the additional distance it travel during that second second.
     
  6. Oct 7, 2007 #5
    wait, how come two different formulas got two different results. a from the time 0-1, should be 5m/s^2, vf=vi+at, a=5 (I put vf=5 m/s), but for this formula d=vit+at^2/2, a=10? which one is right?
     
  7. Oct 7, 2007 #6

    Doc Al

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    vf at t = 1 is not 5 m/s. 5 m/s is the average speed for the first second. vf at t = 1 is 10 m/s.
     
  8. Oct 7, 2007 #7
    OHH. I was confused with the basic conception. Okay, I got it. Then, I think the additional distance is 10m not 15m. 15m is the total distance of 2 seconds. But question asks you ADDITIONAL DISTANCE, so I think is 10m (t=1, 1-2)
     
  9. Oct 7, 2007 #8

    Doc Al

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    How did you determine this?
     
  10. Oct 7, 2007 #9
    wait, I got 15 before was because I calculated the final speed vf=15, and then 1*15=15, it's wrong I think because average speed may not be 15. Then, a=10, d=vi+at^2/2 d=5+10*1/2=10m right?
     
  11. Oct 7, 2007 #10

    Doc Al

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    If you are starting at the beginning of the second second (at t = 1), what's vi?

    Why not start at the beginning (vi = 0) and just find out where this thing is at t=1 and t=2?
     
  12. Oct 7, 2007 #11
    ohh. vi=10 at the beginning of second second!! 10+5 so it should be 15.
     
  13. Oct 7, 2007 #12

    Doc Al

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    Good!

    You could also have used: D = 1/2 a t^2 starting from t = 0

    At t = 1; D = 1/2 a (1) = 5 m
    At t = 2; D = 1/2 a (4) which means it must be 4 times further, or D = 20 m
     
  14. Oct 7, 2007 #13
    Yes, and then subtract them. Thank you.
     
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