1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    A quick notational fix for the original poster:

    The notation:

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


    [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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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

  8. Nov 4, 2005 #7


    User Avatar
    Staff Emeritus
    Science Advisor

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Initial Value Problem
  1. Initial value problem (Replies: 7)

  2. Initial-value problem (Replies: 2)