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An Interesting Acceleration

  1. Feb 22, 2015 #1
    1. The problem statement, all variables and given/known data
    A particle moving from a point [itex] A [/itex] to a point [itex] B [/itex], [itex] 1 [/itex] meter away, travels in a straight line in such a way so that its acceleration is proportional to the distance left to point [itex] B [/itex]. If the particle arrives at point [itex] B [/itex] in [itex] 1 [/itex] second, how long did it take for the particle to reach the point halfway to point [itex] B [/itex]?

    2. Relevant equations
    I suppose we need that the acceleration is the second-derivative of position.

    3. The attempt at a solution
    So we know that [itex] a(t) = \frac{d^2p}{dt^2} = k(1 - p(t)) [/itex] (and [itex] a(1) = 0 [/itex] and [itex] p(0) = 0 [/itex] and [itex] p(1) = 1 [/itex]), but I don't know how to solve this differential equation. Once [itex] p(t) [/itex] is found, [itex] t [/itex] can be found by equating [itex] p(t) = \frac{1}{2} [/itex].
     
    Last edited: Feb 22, 2015
  2. jcsd
  3. Feb 22, 2015 #2

    TSny

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    You can try writing the differential equation in terms of a new dependent variable y that is defined in terms of p. Can you see how to define y(t) in terms of p(t) so that you get a simpler differential equation?
     
  4. Feb 22, 2015 #3

    ehild

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    substitute 1-p(t) by u. What equation do you get for u? Are you familiar with it?
     
  5. Feb 23, 2015 #4
    Define [itex] y(t) = p(t) - 1 [/itex]. Then [itex] \frac{d^2y}{dt^2} = \frac{d^2p}{dt^2} [/itex] and the differential equation becomes:
    [itex] \frac{d^2y}{dt^2} = -ky [/itex]
    Auxiliary equation is [itex] r^2 + k = 0 [/itex] so [itex] r = \pm i \sqrt{k} [/itex]. Then [itex] y = e^a(c_1 cos(bt) + c_2 sin(bt)) = e^{(0)}(c_1 cos(t\sqrt{k}) + c_2 sin(t\sqrt{k})) = c_1 cos(t\sqrt{k}) + c_2 sin(t\sqrt{k}) [/itex]. Then [itex] p(t) = y(t) + 1 = c_1 cos(t\sqrt{k}) + c_2 sin(t\sqrt{k}) + 1 [/itex]. I think I can figure the rest out myself. Thanks all!
     
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