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Math Help - Motion problems

  1. #1
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    Motion problems

    Can someone please show me the steps needed to achieve the answers to this question?

    A helicopter holding a 70kg package suspended from a 5m long rope accelerates upward at 5.2m/s^2. When the velocity of the helicopter is 30m/s up, the rope is cut. The helicopter continues to accelerate at 5.2m/s^2 up. find the distance between the helicopter and the package 2 seconds after the rope is cut.

    I know that's going to be a lot of work, but any help is much appreciated. =)
    Last edited by mr fantastic; April 9th 2010 at 06:05 PM. Reason: Moved a question to another subforum.
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  2. #2
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    Quote Originally Posted by Hellooo View Post
    Can someone please show me the steps needed to achieve the answers to this question?

    A helicopter holding a 70kg package suspended from a 5m long rope accelerates upward at 5.2m/s^2. When the velocity of the helicopter is 30m/s up, the rope is cut. The helicopter continues to accelerate at 5.2m/s^2 up. find the distance between the helicopter and the package 2 seconds after the rope is cut.

    I know that's going to be a lot of work, but any help is much appreciated. =)
    What have you tried? Where are you stuck?
    Last edited by mr fantastic; April 9th 2010 at 06:06 PM.
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  3. #3
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    I'm completely stuck on the question.
    Last edited by mr fantastic; April 9th 2010 at 06:07 PM.
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  4. #4
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    Quote Originally Posted by Hellooo View Post
    Can someone please show me the steps needed to achieve the answers to this question?

    A helicopter holding a 70kg package suspended from a 5m long rope accelerates upward at 5.2m/s^2. When the velocity of the helicopter is 30m/s up, the rope is cut. The helicopter continues to accelerate at 5.2m/s^2 up. find the distance between the helicopter and the package 2 seconds after the rope is cut.

    I know that's going to be a lot of work, but any help is much appreciated. =)
    Package: Initial velocity = 30 m/s, acceleration = -9.8 m/s^2, t = 2 s, Displacement = x1 = ?

    Helicopter: Iinitial velcocity = 30 m/s, acceleration = 5.2 m/s^2, t = 2 s, Displacement = x2 = ?

    Distance between parcel and helicopter = x2 + 5 - x1 (the 5 comes from the rope)
    Last edited by mr fantastic; April 9th 2010 at 06:27 PM. Reason: Corrected a small error
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  5. #5
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    So dx=30m, correct? Thank you very much for the quick replies =) and sorry for double posting.
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  6. #6
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    Quote Originally Posted by Hellooo View Post
    So dx=30m, correct? [snip]
    I don't think so. Please show your calculations and answers for x1 and x2.
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  7. #7
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    dy1=0.5(-9.8)(2)^2+30(2)
    =40.4m

    dy2=0.5(5.2)(2)^2+40(2)
    =91.2m

    dy2=91.2+5=96.2m

    dy=96.2m-40.4m=50.8m?
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  8. #8
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    Quote Originally Posted by Hellooo View Post
    dy1=0.5(-9.8)(2)^2+30(2)
    =40.4m Mr F says: Correct.

    dy2=0.5(5.2)(2)^2+40(2)
    =91.2m Mr F says: No. What is the 40 doing there? Fix this, put it into the working below and you will get the correct answer.

    dy2=91.2+5=96.2m

    dy=96.2m-40.4m=50.8m?
    ..
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  9. #9
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    i thought we use v2 to find the displacement of the helicopter, v2=5.2 x2 +30=40.4m/s up. Is that incorrect? Thanks for being patient with me =)
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  10. #10
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    Quote Originally Posted by Hellooo View Post
    i thought we use v2 to find the displacement of the helicopter, v2=5.2 x2 +30=40.4m/s up. Is that incorrect? Thanks for being patient with me =)
    You're using a fomula that looks something like x = v_0 t + \frac{1}{2} a t^2. Do you understand the meaning of each symbol in this formula? In particular, do you know what v_0 means?
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  11. #11
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    oh sorry for my misunderstanding, so dy2 should be 0.5(5.2)(2)^2+30(2)=70.4? Thanks for correcting me once again
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  12. #12
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    Quote Originally Posted by Hellooo View Post
    oh sorry for my misunderstanding, so dy2 should be 0.5(5.2)(2)^2+30(2)=70.4? Thanks for correcting me once again
    Yes.
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