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Help Solve This Linear Equation

  1. Oct 9, 2009 #1
    1. I thought I was doing it right, but Quest said I have to wrong answer. Can some one solve it in steps and produce a final answer so I can compare?

    Tan 58=(vi-ayt)/axt

    where vi=5.6
    ay=-9.8
    ax=2.1

    solve for t.

    I got 0.87 for t.




    Thank you.
     
  2. jcsd
  3. Oct 9, 2009 #2

    rl.bhat

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    Can you post the actual problem? You have given formula without any information.
     
  4. Oct 9, 2009 #3
    Initially (at time t = 0) a particle is moving vertically at 5.6 m/s and and horizontally at 0 m/s. The particle accelerates horizontally at 2.1 m/s2 . The acceleration of gravity is 9.8 m/s2 . At what time will the particle be traveling at 58◦ with respect to the horizontal?

    Sorry about that.
     
  5. Oct 9, 2009 #4

    rl.bhat

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    The horizontal distance traveled = 1/2*ax*t^2
    Similarly what is the vertical distance traveled?
     
  6. Oct 9, 2009 #5
    d=1/2*ay*t^2

    ?
     
  7. Oct 9, 2009 #6

    rl.bhat

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    No.
    y = Vo*t - 0.5*g*t^2
     
  8. Oct 9, 2009 #7
    Okay, so what do I do with those equations? My prof said it should be as easy as Tan 58=(vi-ayt)/axt
    and whether you need to break that down more or not, I don't know.
     
  9. Oct 9, 2009 #8
    What did you do to get t = 0.87?

    If you substitute it into the equation you get tan 58 = -1.60, while in reality tan 58 = 1.60,
    so I think there's a sign error somewhere
     
  10. Oct 9, 2009 #9
    vi-at=tan 58 (at)
    -(-9.8)t=tan58(2.1)t-5.6
    9.8t=3.36t-5.6
    6.44t=-5.6
    t=0.87

    but time cant be negative..?
     
  11. Oct 9, 2009 #10

    rl.bhat

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    This formula is wrong. It should be
    Tanθ = (2vi -ayt)/axt
     
  12. Oct 9, 2009 #11
    so i get -1.74=t

    do we just take the absolute value for the time? or have I made another mistake
     
  13. Oct 9, 2009 #12
    you replace a_x and a_y both by a here wich is confusing.
    You made a sign error when moving 5.6 to the right side.
    Another problem that i can see now is that when replacing [itex] v_y [/itex] with[itex] (v_i-a_yt) [/itex] to get your original equation this minus sign already accounted for the fact that gravity would be in the negative y direction. So you should substitute g = 9.8 here, and not g = -9.8
    If your prof really gave you that exact equation with [itex] a_y [/itex] instead of g that would
    be highly misleading, if not wrong.
     
  14. Oct 9, 2009 #13
    Yes, and as a result I have spent a week being confused by this problem...
     
  15. Oct 9, 2009 #14

    rl.bhat

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    Instead of giving the answer, it will be better to show the actual calculations by substituting the values. You can rewrite the expression as
    tan58 = Vo/ax*t - ay/ax. Now solve for t.
     
  16. Oct 9, 2009 #15
    Where do you get the 2 from? The equation comes from [tex] tan(58) = \frac { v_y} {v_x} [/tex]

    you just substitute [itex] v_y = v_i - gt [/itex] and [itex] v_x = a_x t [/itex] into it
     
  17. Oct 9, 2009 #16
    okay, so using that I get t=0.426s
     
  18. Oct 9, 2009 #17

    rl.bhat

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    In the problem, the time at which position of the particle makes an angle 58 degrees to the horizontal, is required.
     
  19. Oct 10, 2009 #18
    That is not how I interpret this sentence

     
  20. Oct 10, 2009 #19

    rl.bhat

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    In that case it should be " the net velocity in the direction 58 degrees to the horizontal." And I feel, in the problem position is more appropriate than the velocity.
     
  21. Oct 10, 2009 #20
    Instead of giving the answer, it will be better to show the actual calculations by substituting the values. You can rewrite the expression as
    tan58 = Vo/ax*t - ay/ax. Now solve for t.

    This formula gave me the right answer.
     
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