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ap calc

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a particle moves along the x-axis so that its velocity at any time t is equal to or > 0 is given by v(t)= (2pie-5)t-sin(piet)

a.find the acceleration at any time t.

b.find the minimum acceleration of the particle over the interval [0,3]

c.find the maximum velocity of the particle over the interval [0,2]

  • ap calc - ,

    v = (2 pi - 5) t - sin (pi t)
    a)
    a = dv/dt = (2 pi -5) - pi cos (pi t)
    b)
    when t = 0, -pi cos (pi t) is -pi and that is as low as a gets
    or more formally
    da/dt = 0 = pi sin (pi t)
    sin (pi t) = 0 when t = 0 , 1 , 3
    then
    min a = (2 pi -5) - pi = pi - 5

    c)
    v is max or min when a = 0
    (2 pi -5) - pi cos (pi t) = 0
    cos (pi t) = 2 - 5/pi = .4084
    pi t = 1.15 radians
    t = .366
    but pi t can also be -1.15 radians which is + 5.13 radians
    t = 1.63 than will give me the bigger v
    v = (2 pi - 5) t - sin (pi t)
    v = (2 pi -5)(1.63) - sin 1.63 pi
    = 3

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