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Summersnow8

  • one year ago

I need help on all of these problems, sometimes using this website: http://phet.colorado.edu/sims/vector-addition/vector-addition_en.html. The questions are attached below. @zepdrix

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  1. Kash_TheSmartGuy
    • one year ago
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    For the first few questions.

  2. Kash_TheSmartGuy
    • one year ago
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    https://www.mathsisfun.com/algebra/vectors.html

  3. Kash_TheSmartGuy
    • one year ago
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    https://www.khanacademy.org/math/precalculus/vectors-precalc

  4. Kash_TheSmartGuy
    • one year ago
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    http://www.physicsclassroom.com/class/vectors/Lesson-1/Vectors-and-Direction

  5. Kash_TheSmartGuy
    • one year ago
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    Maybe, those will help^^

  6. Summersnow8
    • one year ago
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    it didn't

  7. Summersnow8
    • one year ago
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  8. Summersnow8
    • one year ago
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    attempt at #7

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  9. Kash_TheSmartGuy
    • one year ago
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    sry, :(

  10. zepdrix
    • one year ago
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    link is broken :(

  11. Summersnow8
    • one year ago
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    @zepdrix http://phet.colorado.edu/sims/vector-addition/vector-addition_en.html

  12. zepdrix
    • one year ago
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    so uhhhh which one you up to right now? 0_o by the way, i would recommend using `style 2`, that one makes the most sense.

  13. Summersnow8
    • one year ago
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    @zepdrix I did #1, that's as far as I got. Everything else I attempted but didn't work

  14. zepdrix
    • one year ago
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    Ok let's use trig to explain problem 2, as they have requested. If the `x component` is `twice the size` of the `y component` we can write that relationship like this: \(\large\rm x=2y\)

  15. Summersnow8
    • one year ago
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    does that solve it?

  16. zepdrix
    • one year ago
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    Mmm that's the right idea, yes! But they want this generalized for ANY vector. You showed that it works for a specific set of lengths, 11 and 22.

  17. zepdrix
    • one year ago
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    So what we would rather do is call our \(\large\rm \color{orangered}{x=2y}\) and our \(\large\rm y=y\) x is twice the length of y, that's how we got that relationship. So again plug this into your inverse tangent.\[\large\rm \tan^{-1}\left(\frac{y}{\color{orangered}{x}}\right)=\tan^{-1}\left(\frac{y}{\color{orangered}{2y}}\right)\]

  18. zepdrix
    • one year ago
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    Do you see where this is going? :)

  19. Summersnow8
    • one year ago
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    yes, I understand what you are implying

  20. zepdrix
    • one year ago
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    11 divided by 22 can be simplified to 1/2. The same thing will happen with our setup here! We can cancel the y's and see that the argument simplifies to 1/2. Again, you'll get 26.6 degrees, but now we've generalized it. yay team

  21. Summersnow8
    • one year ago
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    okay, haha, yes I understand, now on to the next one :)

  22. zepdrix
    • one year ago
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    mmmm number 3 should be pretty straight forward.

  23. Summersnow8
    • one year ago
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    but what would it look like?

  24. Summersnow8
    • one year ago
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    like this?

  25. zepdrix
    • one year ago
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    if i have some vector V and then I add to that some vector -V, they'll give me 0, ya? So draw a vector going off in some direction. And draw another vector going in the opposite direction. Illustrated here as an example:|dw:1436388162556:dw|

  26. zepdrix
    • one year ago
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    Yah I think yours will work also. But it'll be easier to setup if both tails start from the origin.

  27. zepdrix
    • one year ago
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    i mean, if both tails are touching.

  28. zepdrix
    • one year ago
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    i dont really understand how to use the `show sum` tool. trying to figure that out...

  29. zepdrix
    • one year ago
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    oh oh ok i get it now

  30. zepdrix
    • one year ago
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    actually, your example was fine, ya just go with that :) neeeeext

  31. zepdrix
    • one year ago
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    On your picture that you posted, if you click the "show sum" button, you should notice that nothing happens. That's because the sum is zero, so there is nothing to display

  32. Summersnow8
    • one year ago
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    so my example works or no?

  33. zepdrix
    • one year ago
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    ya

  34. Summersnow8
    • one year ago
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    okay cool. next

  35. zepdrix
    • one year ago
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    4 is the same idea... this should be pretty straight forward. We just showed that we can add two vectors to get zero. So now we want to add three vectors and end up with only the first one...... meaning that `the other two vectors added to zero`. ya? ;o

  36. Summersnow8
    • one year ago
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    when I try to press the show sum button it doesnt come up

  37. Summersnow8
    • one year ago
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    no, I dont really get 4 either

  38. zepdrix
    • one year ago
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    |dw:1436388637712:dw|So for problem 3, we did something like this. Two vectors added together, gave us nothing, zero.

  39. zepdrix
    • one year ago
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    |dw:1436388727454:dw|

  40. zepdrix
    • one year ago
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    Now for this next part, we want to add 3 vectors together. A+B+C and end up with only A.

  41. zepdrix
    • one year ago
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    So we could draw A ..... anywhere else.|dw:1436388787394:dw|

  42. zepdrix
    • one year ago
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    If they want us to "describe it" though, instead of graphing it... let's umm.... Let's just say that vector C is the negative of vector B.

  43. Summersnow8
    • one year ago
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    okay.....

  44. zepdrix
    • one year ago
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    \(\large\rm \color{orangered}{\vec C=-\vec B}\) Therefore adding some vectors A, B and C: \(\large\rm \vec A+\vec B+\color{orangered}{\vec C}=?\)

  45. zepdrix
    • one year ago
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    Can you fill in the blank? :)

  46. Summersnow8
    • one year ago
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    A?

  47. Summersnow8
    • one year ago
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    That is how I would describe it?

  48. zepdrix
    • one year ago
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    Well, don't forget the middle step though.\[\large\rm \vec A+\vec B+\color{orangered}{\vec C}=\vec A+\vec B+\color{orangered}{-\vec B}\]Which yes, you can go a little further and say,\[\large\rm =\vec A+\vec 0=\vec A\]Because B and -B cancel out.

  49. Summersnow8
    • one year ago
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    oh okay, that makes sense

  50. zepdrix
    • one year ago
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    how bout 5, figure that one out? :d comeonnnn that one is eassyyyy :D

  51. Summersnow8
    • one year ago
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    no......?

  52. zepdrix
    • one year ago
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    you `could` get all fancy and try to make the y's add up to zero. That seems tedious though. let's just make it really easy on ourselves and draw horizontal lines. Example: Let's say I want three vectors to add up to `20 in the x direction` `0 in the y direction`. Well I could do something like this:|dw:1436389431399:dw|\[\large\rm \vec A+\color{royalblue}{\vec B}+\vec C=5x+0y+\color{royalblue}{5x+0y}+10x+0y\]\[\large\rm =20x+0y\]

  53. zepdrix
    • one year ago
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    i gotta go :3 i be back later maybe

  54. Summersnow8
    • one year ago
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    okay, thanks for the help so far

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