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Homework Help: Initial Value Problem

  1. Nov 3, 2005 #1
    The problem:

    The acceleration of a particle moving back and forth on a line is a = (d^2)s/d(t^2) = (pi)^2 cos(pi)(t) m/sec^2 for all t. If s = 0 and v = 8 m/sec when t = 0, find s when t = 1 sec.

    My work:

    (d^2)s/d(t^2) = (pi)^2 cos(pi)(t)

    ds/dt = (pi)^2 sin(pi)(1) + C

    ds/dt = (pi)^2(0) + C

    0 = C

    Then..

    ds/dt = (pi)^2 sin(pi)(t)

    s = -(pi)^2 cos (pi) (8)

    0 = -(pi)^2(1) + C

    (pi)^2 = C

    So my answer turned out to be (pi)^2 meters. I'm not so sure that that's the correct answer though.

    Any help is appreciated.
     
  2. jcsd
  3. Nov 3, 2005 #2
    well, oook.

    First of all, you start with acceleration as a function of time yes? [tex] a(t) = \pi^2 cos(\pi t)[/tex] when you integrate [tex]\frac{d^2 s}{dt^2}=\pi^2 cos(\pi t)[/tex] this is an equation for velocity as a function of time [tex] v(t)= \frac {ds}{dt}[/tex] right?

    so, first when you integrate, [tex]cos(\pi t) dt[/tex] its the opposite of the chain rule, so you have to divide by the coefficient of t, in this case, pi. so you get,
    [tex]v(t) = \frac{ds}{dt}=\pi sin(\pi t) +C[/tex]

    then to solve for C you plug in the given values for v and t and solve.
    Then because you want s(t), you integrate once more. and you solve for C again with the given initial values of s and t. then you plug in t=1 into your final equation and solve for s.
     
  4. Nov 3, 2005 #3

    Hurkyl

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    A quick notational fix for the original poster:

    The notation:

    [tex]\frac{d^2 s}{dt^2}[/tex]

    means

    [tex]\frac{d^2 s}{(dt)^2}[/tex]

    or equivalently,

    [tex]\left( \frac{d}{dt} \right)^2 s[/tex]
     
  5. Nov 3, 2005 #4
    oh man, you scared me when i saw your post after mine... i thought i'd done something wrong! whew.
     
  6. Nov 3, 2005 #5

    Hurkyl

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    Nah, your work looks fine! :smile: The only thing I might consider complaining about is that you did virtually all the work for the OP.
     
  7. Nov 3, 2005 #6
    Ah, I think I've got it now. I just hope I won't confuse this with anything else on my test tomorrow. :P

    Thanks!
     
  8. Nov 4, 2005 #7

    HallsofIvy

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    Sedm: another notational point. Many of us might be inclined to read
    cos(pi)(t) as {cos(pi)} t, in which case, your original calculation would be correct: the derivative of that would be {cos(pi)} but NOT the anti-derivative! Clearly you meant cos(pi t). The anti-derivative of that is
    (1/pi) sin(pi t). I don't know why had "1" in place of t.

    Hurkyl: ?? What??
    [tex]\frac{d^2 s}{dt^2}[/tex] is the second derivative. It definitely is not
    [tex]\frac{d^2 s}{(dt)^2}[/tex]
    (I'm not even sure what that could mean!)
    Yes, you could write that as
    [tex]\left( \frac{d}{dt} \right)^2 s[/tex]
    but you had better make it clear that that is NOT
    [tex]\left(\frac{ds}{dt}\right)^2[/tex]!
     
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