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anonymous

  • one year ago

A particle confined to motion along the x-axis moves with constant acceleration from x = 2.0 m to x = 8.0 m during a 2.5-s time interval. The velocity of the particle at x = 8.0 m is 2.8 m/s. What is the acceleration during this time interval?

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  1. anonymous
    • one year ago
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    \[x _{f}=x _{0}+vt+\frac{ 1 }{ 2 }at^2\]

  2. anonymous
    • one year ago
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    What do I use as acceleration? I know that the answer is -1.0 m, I just do not see it yet

  3. anonymous
    • one year ago
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    Acceleration unit is m/s^2. What makes you think you know the answer? I solved it and it was not 1

  4. anonymous
    • one year ago
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    nevermind, I got it. .32 m/s^2

  5. anonymous
    • one year ago
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    Um, the answer is -1.0m is the answer? Are you sure? The units should be m/s^2 Either that is the wrong solution or the solution you saw is another one.

  6. anonymous
    • one year ago
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    I was looking at the other problem for some reason, this one i did get with the formula :) Thanks!

  7. anonymous
    • one year ago
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    Ah, I see. Nice job No problem!

  8. anonymous
    • one year ago
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    I thought I had copied the other problem here that I cant figure out haha

  9. anonymous
    • one year ago
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    Oh no wonder. Actually the answer is -0.33 not positive 0.32. Forgot to tell you

  10. anonymous
    • one year ago
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    Why do you get a negative answer? I got positive .032... and all the answers that I get to choose from are positive as well

  11. anonymous
    • one year ago
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    Oh sorry. Because the correct formula was \[x_{f}=x _{0}+vt-\frac{ 1 }{ 2 }at^2\] I wrote the wrong formula and you solved it?

  12. anonymous
    • one year ago
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    I had done that one in a different way and got the same result :) I can show you hold on. Also, when I copied the formula I looked at my notes from class too

  13. anonymous
    • one year ago
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    Nope, the first formula is -0.33 m/s^2 The second formula is 0.33m/s^2

  14. anonymous
    • one year ago
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    I write even the simplest thing on the notes, so that when going back I know where things came from. Even if I can do it in my head..... I know its long sorry

  15. anonymous
    • one year ago
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    Could have told me you used a different formula. I wrote the first formula wrong, but it is better to use mines, since it just 1 formula. It is faster.

  16. anonymous
    • one year ago
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    Yes it is better and way faster. I did it both ways and I always got .32 m/s^2

  17. anonymous
    • one year ago
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    Which formula? Mines or yours?

  18. anonymous
    • one year ago
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    yours... here is what I did with yours,

  19. anonymous
    • one year ago
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    Yea, it is a shortcut. That is why they made it separate formula.

  20. sohailiftikhar
    • one year ago
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    as he said the acc is constant you it means increase in velocity in equal interval of time is same ..

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