Math

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the acceleration of a particle is defined by the equation a= 4-t^2 with an initial velocity of two millimeters per second. Find the exact equation for the position of the particle if its initial position was zero.

  • Math -

    if a = 4 - t^2 , then
    v = 4t - (1/3)t^3 + c
    given: when t=0, v = 2 mm/s
    2 = 0 - 0 + c
    c = 2

    v = 4t - (1/3)t^2 + 2

    s = 2t^2 - (1/12)t^4 + 2t + k
    when t=0 , s = 0 , so k = 0

    s = 2t^2 - (1/12)t^4 + 2t

    check by taking the derivative twice

  • Math -

    x"(t)=d²x/da²=a=4-t^2
    Integrate two times and substitute the initial conditions to find the required equation:

    x'(t)
    =∫a dt
    =∫(4-t^2)dt
    =4t-(t^3/3)+C1
    From x'(t)=2 mm-t^-1, we get
    2 mm-t^-1 = 4(0)-(0^3/3)+C1
    =>
    C1=2mm-t^-1

    x(t)=∫x'(t)dt
    =∫(4t-(t^3/3)+2) dt
    =2t²-(1/12)t^4+2t+C2

    x(0)=0 => C2=0
    =>

    x(t)=2t² -(1/12)t^4 +2t
    (in millimetres)

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