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Math Help - time, velocity, position of a particle

  1. #1
    Junior Member LexiRae's Avatar
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    time, velocity, position of a particle

    at t time a particle is moving along the x axis at position x. the relationship between x and t is given by tx=x^2 +8. at x = 2 the velocity of the particle is:
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  2. #2
    Member Mentia's Avatar
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    Did you mean to write t(x) = x^2 + 8 instead of tx = x^2 + 8? I hope so because that makes it easier.

    Then:

    t=x^{2}+8

    then

    x=\pm \sqrt[ ]{ t-8}

    then

     velocity=\frac{dx }{dt }=\pm \frac{1}{2}\frac{1}{\sqrt[]{t-8}} = \frac{1}{2x}

    Then if x = 2, velocity = 1/4

    Seem reasonable?
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  3. #3
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    Quote Originally Posted by Mentia View Post
    Did you mean to write t(x) = x^2 + 8 instead of tx = x^2 + 8? I hope so because that makes it easier.

    Then:

    t=x^{2}+8

    then

    x=\pm \sqrt[ ]{ t-8}

    then

     velocity=\frac{dx }{dt }=\pm \frac{1}{2}\frac{1}{\sqrt[]{t-8}} = \frac{1}{2x}

    Then if x = 2, velocity = 1/4

    Seem reasonable?
    I think simpler: from t= x^2+ 8, 1= 2x dx/dt. At x= 2, 1= 4 dx/dt so v= dx/dt= 1/4.
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