A particle moves with acceleration function a(t) = 5 + 4t − 2t^2 4.?

Its initial velocity is v(0) = 2m/s and initial position is s(0) = 10 m. Find its position after t seconds.

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  • Juan
    Lv 6
    7 years ago
    Favorite Answer

    a(t) = 5 + 4t -2t^2

    a(t) = dv/dt

    dv/dt = 5 + 4t -2t^2

    dv = (5 + 4t - 2t^2) dt

    v = 5t + 2t^2 - (2/3)t^3 + c_1

    v(t = 0) = 2 m/s

    v = 5*0 + 2*0^2 - (2/3)*0^3 + c_1 = 2 m/s

    c_1 = 2 m/s

    v = 5t + 2t^2 - (2/3)(t^3) + 2

    v = ds/dt

    ds/dt = 2 + 5t + 2t^2 - (2/3)(t^3)

    ds = [2 + 5t + 2t^2 - (2/3)(t^3)] dt

    s = 2t + (5/2)t^2 + (2/3)t^3 - (1/6)(t^4) + c_2

    s(t = 0) = 2t + (5/2)t^2 + (2/3)t^3 - (1/6)(t^4) + c_2 = 10

    s(t = 0) = 2*0 + (5/2)*0^2 + (2/3)*0^3 - (1/6)(0^4) + c_2 = 10 m

    c_2 = 10 m

    s = 2*t + (5/2)*t^2 + (2/3)*t^3 - (1/6)(t^4) + 10

    s = 10 + 2*t + (5/2)*t^2 + (2/3)*t^3 - (1/6)(t^4)

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