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Homework Help: Position functions

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data

    The motion of a particle is described by the position function

    s=((1/3)t^3)-(29t) - (1296/t), t>=1,
    where t is the time in seconds and s is the displacement in metres

    a) What is the velocity when acceleration is zero?
    b) Find the displacement when v = 68m/s
    c) When is the acceleration -90m/s^2?

    2. Relevant equations

    3. The attempt at a solution

    a) Well, if the velocity is ((t^2)-(29)+(1296t^-2), and the acceleration is (2t^4 - 2592)/t^3, then wouldn't the velocity be 0 when acceleration is 0?

    b) v(68)= 4595

    c)-90 = (2t^4 - 2592)/t^3 ? (Doesnt seem to work)

    I know Im doing something terribly wrong
    Last edited: Feb 22, 2009
  2. jcsd
  3. Feb 22, 2009 #2


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    Homework Helper
    Gold Member

    Hi DontEvenTry, welcome to PF!:smile:

    Position, velocity, and acceleration are usually vectors; before we proceed, are you sure the position is given as the scalar function you've written above and not as a vector?
  4. Feb 22, 2009 #3


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    Science Advisor

    gabbagabbahey, I would interpret this as "one-dimensional" motion. The only distinction between "velocity" and "speed" in one dimension is that velocity can be negative (and this clearly can, for example, v(1) is negative) while speed is always non-negative.

    Why would you think that? Surely you don't think "acceleration 0" means "sitting still". A car going at a constant speed of 90 mph has "acceleration 0". You are correct that the acceleration is (2t^4- 2592)/t^3. Now solve (2t^4- 2592)/t^3= 0 and put that value of t into the formula for v.

    ??? The problem did not ask for v, it asked for "displacement"- the distance moved. Since v= dx/dt, [itex]x= \int v dt[/itex].

    Why not? This is the same as -90t^3= 2t^4- 2592 or 2t^4+ 90t^3- 2592. Solve that. (Hint: it has a simple integer solution.)

    Yes, you are giving up too easily- that's terribly wrong!
    How do I know there is a "simple integer root"? I tried. Since cubics and quartics can be horrible to solve in general, I guessed (hoped) that there was a simple integer root and just tried putting integer values into the formula.
    Last edited by a moderator: Feb 22, 2009
  5. Feb 22, 2009 #4
    Alright, for a) i got 71m/s
    b) 272m
    c) 3sec
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