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anonymous

  • one year ago

Any help? It's velocity and position sort of stuff. http://oi60.tinypic.com/21msxlz.jpg The question is in the link since it's a picture.

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  1. 4n1m0s1ty
    • one year ago
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    ok so lets break this problem down. The first thing you need to find out is where john starts. According to the problem, the mailbox is 1 m in front of him.

  2. anonymous
    • one year ago
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    Right

  3. 4n1m0s1ty
    • one year ago
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    the problem states that the mailbox is the zero location, meaning that it is designated as position 0 m, and that any position that john takes is relative to that position (e.g. at 2 m he is 2 meters in front of the mailbox, - 4 m is 4 meters behind the mailbox.)

  4. 4n1m0s1ty
    • one year ago
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    So if we know that the mailbox is 1 meter in front of john, that means he is 1 m behind the mailbox.

  5. 4n1m0s1ty
    • one year ago
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    So what would be his position at time 0?

  6. anonymous
    • one year ago
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    2m?

  7. 4n1m0s1ty
    • one year ago
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    Not exactly. It would be -1 m. He is 1 m behind the mailbox, so if we use it as a reference point, it would be measured as -1 m.

  8. 4n1m0s1ty
    • one year ago
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    Is that clear or is it still confusing?

  9. anonymous
    • one year ago
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    Oh

  10. anonymous
    • one year ago
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    A little more clear but physics kinda confuses me in general.

  11. 4n1m0s1ty
    • one year ago
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    Well thats ok, its hard at first but once you take the time to actually think through it. It makes a lot of sense.

  12. anonymous
    • one year ago
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    I get why it's -1 for the first one but I don't know how you would figure out the rest though.

  13. 4n1m0s1ty
    • one year ago
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    Ok so next part.

  14. 4n1m0s1ty
    • one year ago
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    We need to examine the velocity graph

  15. 4n1m0s1ty
    • one year ago
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    position tells you where you are, and velocity tells you where you are going

  16. 4n1m0s1ty
    • one year ago
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    So if we look at the velocity of john at t = 0. We find that he is moving at rate of - 2 m/s.

  17. anonymous
    • one year ago
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    The graph goes positive so does that mean he's going away from it?

  18. 4n1m0s1ty
    • one year ago
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    It depends. Just looking at whether its positive or negative isn't enough.

  19. 4n1m0s1ty
    • one year ago
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    His position at t = 0 is -1 m. His velocity is -2 m/s at t =0. The negative velocity and negative position tells you that he will be moving further from the mailbox.

  20. 4n1m0s1ty
    • one year ago
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    |dw:1444358671721:dw|

  21. anonymous
    • one year ago
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    so if I was plotting this the next would be -2?

  22. 4n1m0s1ty
    • one year ago
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    Almost. You have to consider his initial position. If 1 second passed he would move away 2 m from the mailbox, but he started at -1 m. So that would make him be at -3 m at t =1.

  23. anonymous
    • one year ago
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    Or rather -3?

  24. 4n1m0s1ty
    • one year ago
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    yep

  25. 4n1m0s1ty
    • one year ago
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    and now you just repeat what we just did for each time.

  26. anonymous
    • one year ago
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    so the next is -5?

  27. 4n1m0s1ty
    • one year ago
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    At t = 1, he is at -3 m and his velocity is 0 m, so he doesn't move. So at t = 2, he is at the same position.

  28. 4n1m0s1ty
    • one year ago
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    he would be at - 5 m only if he kept the same velocity from before, but he changes velocity at t = 1/

  29. anonymous
    • one year ago
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    So t=3 it would be -4 since he's moving again?

  30. 4n1m0s1ty
    • one year ago
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    Well it would change, but it wouldn't be -4 m

  31. 4n1m0s1ty
    • one year ago
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    Look at t =3, whats the velocity?

  32. anonymous
    • one year ago
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    2

  33. 4n1m0s1ty
    • one year ago
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    I mean, err, t = 2?

  34. anonymous
    • one year ago
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    It's 0

  35. 4n1m0s1ty
    • one year ago
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    Yeah its 2 m/s right? so if he was at -3m, we add positive 2 m after 1 second. So it would be -1 m.

  36. 4n1m0s1ty
    • one year ago
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    Does it make sense?

  37. anonymous
    • one year ago
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    Yeah..sort of

  38. 4n1m0s1ty
    • one year ago
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    Is there something confusing that you don't understand? Don't be afraid to ask :-).

  39. anonymous
    • one year ago
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    So the next is 1?

  40. 4n1m0s1ty
    • one year ago
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    Yep.

  41. anonymous
    • one year ago
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    And the last one is 3

  42. 4n1m0s1ty
    • one year ago
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    The velocity doesn't change for the rest of the problem, so you just need to add 2 to the position after each time.

  43. 4n1m0s1ty
    • one year ago
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    yes

  44. anonymous
    • one year ago
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    Okay. I got that one. Would you be able to help with some more? It's still kinda rocky for me but I'm getting it better.

  45. 4n1m0s1ty
    • one year ago
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    Sure, if it doesn't take too long. But was there anything confusing about how to solve problem?

  46. 4n1m0s1ty
    • one year ago
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    * this problem

  47. anonymous
    • one year ago
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    I just don't really know how to approach these sort of problems.

  48. 4n1m0s1ty
    • one year ago
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    Well do you understand what position is?

  49. 4n1m0s1ty
    • one year ago
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    and velocity?

  50. anonymous
    • one year ago
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    More velocity than position.

  51. anonymous
    • one year ago
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    We just started this stuff so it's all new to me.

  52. 4n1m0s1ty
    • one year ago
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    Ok tell me in you words what you think the 2 are? I need to gauge your understanding.

  53. anonymous
    • one year ago
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    Velocity is the change of the position Positions is where the object is at?

  54. 4n1m0s1ty
    • one year ago
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    Is position relative or absolute? Can I say tell you that I am at 1 meter without any other information and still have that make sense. Also, velocity is the change in position "over time". It is an important distinction because you can have change in position over another variable (outside your course probably). Velocity tells you where you are going (both the speed and direction).

  55. anonymous
    • one year ago
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    It's relative then

  56. 4n1m0s1ty
    • one year ago
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    Yes, so when w approach problems like these (called kinematics problems). We always establish a frame of reference (or like in this problem have it defined for us e.g. the mailbox).

  57. 4n1m0s1ty
    • one year ago
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    When you are ready just start an new thread.

  58. anonymous
    • one year ago
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    Will do.

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