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12 February, 07:16

The velocity of a particle constrained to move along the x-axis as a function of time t is given by:

v (t) = - (15/t_0) sin (t/t_0). a) If the particle is at x=4 m when t = 0, what is its position at t = 4t_0. You will not need the value of t_0 to solve any part of this problem.

b) Denote instantaneous acceleration of this particle by a (t). Evaluate the expression 6 + v (0) t + a (0) t^2/2 at t = 4t_0.

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  1. 12 February, 11:06
    0
    The answer is:

    v = 15/T sin t/T

    then integrate

    x = - 15 cos t/T + c

    if x = 4 at t = 0 then

    4 = - 15 + c so c = 19

    so

    x = 19 - 15 cos t/T

    at t = 4T

    x = 19 - 15 cos 4 = 3.99 or 4

    b)

    a = dv/dt = 15/T^2 cos t/T

    v (0) = 0 since sin 0/T = 0

    a (0) = 15/T^2

    so

    6 + 0 + (7.5/T^2) (16 T^2)

    = 6 + 7.5*16

    = 126 is the answer
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