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IsTim

  • 4 years ago

A skater travels in a straight line 8.5*10^2m[25 degress N of E] and then 5.6810^2m in a straight line [21 degrees E of N]. The entire motion takes 4.2 min. What is the skater's displacement angle?

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  1. IsTim
    • 4 years ago
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    Figured out the displacement. 1.4*10^3m. The angle's iffy.

  2. JamesJ
    • 4 years ago
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    The amount of time taken is irrelevant. For each leg of the journey figure out the angle of displacement. Call each vector, \( v_1 \) and \( v_2 \). Each vector has an x and a y component. Write \[ v_1 = (x_1,y_1), v_2 = (x_2,y_2) \] Then at the end the skater is at \( v_1 + v_2 = (x_1 + x_2, y_1 + y_2 ) \) and can calculate her angle of displacement by taking the tan of the ratio of the y displacement over the x displacement. It is identical in principle to the last problem you had.

  3. IsTim
    • 4 years ago
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    It is the last problem I had. I just didn't ask for the angle that time. Reposted... Anyways, I usedA=sin^-1(asinB/b). Does that work?

  4. JamesJ
    • 4 years ago
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    You should also draw a diagram so you can see at least directionally where the answer is. No, use tan(y/x)

  5. IsTim
    • 4 years ago
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    |dw:1327905348977:dw|

  6. IsTim
    • 4 years ago
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    Why do I use tan(y/x)? Doesn't that apply only for right-angle triangles?

  7. TuringTest
    • 4 years ago
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    It's nice to use if you get your vector in terms of components, which you haven't done yet -I got the right answer now btw,

  8. IsTim
    • 4 years ago
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    So, Idraw that chart thing?

  9. TuringTest
    • 4 years ago
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    assuming I know what you mean by 'the chart thing' , yeah I do

  10. TuringTest
    • 4 years ago
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    or *I would

  11. IsTim
    • 4 years ago
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    i hafta flip thru my notes now to remember chart thing.

  12. TuringTest
    • 4 years ago
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    V is the displacement vector:\[V_x=850\cos(25^{\circ})+568\cos(90^{\circ}-21^{\circ})\]\[V_y=850\sin(25^{\circ})+568\sin(90^{\circ}-21^{\circ})\]\[|V|=\sqrt{V_x^2+V_y^2}\]\[\theta=\tan^{-1}(\frac{V_x}{V_y})\]

  13. IsTim
    • 4 years ago
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    So I'm allowed doing that? I never learned it like that. I never knew.

  14. TuringTest
    • 4 years ago
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    My way definitely gets the job done. Convert the vector into components, the add them. The new vectors components will be the sum of the old vectors components. You can find the new vectors magnitude from the pythagorean theorem, and its angle from an inverse trig function. I know people have other ways of doing it though.|dw:1327895183170:dw|

  15. IsTim
    • 4 years ago
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    Ok. I'm converting them into components. I'll try doing what you say. But I'll be taking a break. thank you very much though.

  16. TuringTest
    • 4 years ago
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    welcome!

  17. IsTim
    • 4 years ago
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    Turing? You here? I need help again!

  18. IsTim
    • 4 years ago
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    I got 42.2 degrees, but the answer is 24!

  19. anonymous
    • 4 years ago
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    what now?

  20. IsTim
    • 4 years ago
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    I think I'm just gunna quit and move on.

  21. anonymous
    • 4 years ago
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    y

  22. IsTim
    • 4 years ago
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    I worked so long on this simple question. It's wasting my time by now.

  23. anonymous
    • 4 years ago
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    i can try if you have the patience lol

  24. IsTim
    • 4 years ago
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    I can wait. I'll do another question while I wait tho. K?

  25. anonymous
    • 4 years ago
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    k

  26. IsTim
    • 4 years ago
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    WAIT! I FIGURED IT OUT

  27. IsTim
    • 4 years ago
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    Sorry for wasting your time! I'm Sorry!

  28. IsTim
    • 4 years ago
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    Mth?

  29. anonymous
    • 4 years ago
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    o

  30. anonymous
    • 4 years ago
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    lol

  31. anonymous
    • 4 years ago
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    np

  32. IsTim
    • 4 years ago
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    Thank you everyone.

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