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Calculus AB

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A particle is moving along the x-axis so that at time t its acceleration is
a(t)=ðcos(ðt)
At time t=1/2, the velocity v of the particle is 1/2.

Find the velocity of the particle at any time t. I think I got that right, as sinðt + C.

Now it wants the minimum velocity of the particle. How do I find that?

  • Calculus AB - ,

    Ugh the pi symbol didn't work. All of the ð should be pi. so it would be a(t) = picos(pit)

    and my answer should be sin(pit) + C

  • Calculus AB - ,

    a(t) = π cos(πt)
    v(t) = sin (πt) + c . where v(t) is velocity
    when t=1/2, v(1/2) = 1/2

    1/2 = sin (π/2) + c
    1/2 = 1 + c
    c = -1/2

    v(t) = sin (πt) - 1/2
    for a min of v(t) , v'(t) = 0 , that is, a(t) = 0

    πcos(πt) = 0
    cos(πt) = 0
    πt = π/2 or πt = 3π/2
    t = 1/2 or t = 3/2

    we already know v(1/2) = 1/2
    we need
    v(3/2) = sin(3π/2) - 1/2
    = -1 - 1/2 = -3/2

    so the minimum velocity is -3/2

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