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The acceleration of a particle at a time t moving along the x-axis is give by: a(t) = 4e^(2t). At the instant when t=0, the particle is at the point x=2, moving with velocity v(t)=-2. Find the position of the particle at t=1/2

if you could show me how to get that please

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  1. postition= integral v dt
    v= int a dt
    find v first, you know the constant of integration from at t=2

    Then integrate again to get position.

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    bobpursley
  2. a(t) = 4e^(2t)
    so v(t) = 2e^(2t) + C (I integrated, since dv/dt = a)
    when t=0, v=-2
    -2 = 2e^0 + c
    c = -4 and then
    v)t) = 2e^(2t) - 4
    since ds/dt = v
    s = e^(2t) - 4t + k
    when t=0, s=2 (I assume that is what you meant by x=2)
    2 = e^0 - 4(0) + k
    k = 1
    then s(t) = e^(2t) - 4t + 1
    s(1/2) = e^1 - 2 + 1
    = e - 1

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