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Math Help - moving particle

  1. #1
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    moving particle

    A particle is moving so that the acceleration at time t is:
    a(t) = 24t^2 i - 12t j

    The initial position and velocity are r(0) = i and v(0) = i - j respectively. Find r(t), the position at time t.

    please help me with this problem.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by rcmango View Post
    A particle is moving so that the acceleration at time t is:
    a(t) = 24t^2 i - 12t j

    The initial position and velocity are r(0) = i and v(0) = i - j respectively. Find r(t), the position at time t.

    please help me with this problem.
    Personally i like the vector notation better, so i will work in that and then convert it to i,j notation. You should be able to reproduce each step in i,j notation if you so desire

    a(t) = <24t^2 , 12t>
    => v(t) = int(a(t)) = <8t^3 + C , 6t^2 + D>
    But v(0) = <1,1>
    so v(0) = <1,1> = < 0 + C , 0 + D>
    => C=D=1
    so v(t) = <8t^3 + 1 , 6t^2 + 1>

    r(t) = int(v(t)) = < 2t^4 + t + C , 2t^3 + t + D>
    but r(0) = <1,0>
    so r(0) = <1,0> = <0 + 0 + C , 0 + 0 + D>
    => C = 1, D = 0

    so r(t) = <2t^4 + t + 1 , 2t^3 + t>
    = (2t^4 + t + 1)i + (2t^3 + t)j
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  3. #3
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    Hello, rcmango!

    A particle is moving so that the acceleration at time t is:
    . . a(t) .= .24ti - 12tj

    The initial position and velocity are: .r(0) = i and v(0) = i - j, respectively.
    Find r(t), the position at time t.
    We will integrate twice, and use the initial conditions to evaluate the constants.


    Velocity is the integral of the acceleration: .v(t) .= .
    a(t) dt
    . . v(t) .= .
    (24ti - 12tj) dt .= .8ti - 6tj + C1

    Since v(0) = i - j: .80i - 60j + C1 .= .i - j . . C1 .= .i - j

    . . Hence: .v(t) .= .8ti - 6tj + i - j . . v(t) .= .(8t - 1)i - (6t + 1)j


    Position is the integral of velocity: .r(t) .= .
    v(t) dt
    . . r(t) .= .
    [(8t - 1)i - (6t + 1)j] dt .= .(2t^4 - t)i - (2t + t)j + C2

    Since r(0) = i: .(20^4 - 0)i - (20 + 0)j + C2 .= .i . . C2 = i

    . . Hence: .r(t) .= .(2t^4 - t)i - (2t + t)j + i


    Therefore: .r(t) .= .(2t^4 - t + 1)i - (2t + t)j


    Hmmm, too slow once again . . .
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  4. #4
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Soroban View Post
    Hello, rcmango!

    We will integrate twice, and use the initial conditions to evaluate the constants.


    Velocity is the integral of the acceleration: .v(t) .= .
    a(t) dt
    . . v(t) .= .
    (24ti - 12tj) dt .= .8ti - 6tj + C1

    Since v(0) = i - j: .80i - 60j + C1 .= .i - j . . C1 .= .i - j

    . . Hence: .v(t) .= .8ti - 6tj + i - j . . v(t) .= .(8t - 1)i - (6t + 1)j


    Position is the integral of velocity: .r(t) .= .
    v(t) dt
    . . r(t) .= .
    [(8t - 1)i - (6t + 1)j] dt .= .(2t^4 - t)i - (2t + t)j + C2

    Since r(0) = i: .(20^4 - 0)i - (20 + 0)j + C2 .= .i . . C2 = i

    . . Hence: .r(t) .= .(2t^4 - t)i - (2t + t)j + i


    Therefore: .r(t) .= .(2t^4 - t + 1)i - (2t + t)j


    Hmmm, too slow once again . . .
    sorry Soroban, i didnt realize you were replying to it until after i had finished doing so. you're aesthetic math symbols are always welcome however
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  5. #5
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    awesome explanations, thanks for all the work!
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