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Homework Help: Position Function of Time

  1. Sep 3, 2009 #1
    New to PF, so here it goes. Oh just a question, did this site use to have a dark skin? I can't remember if that was this or something else.
    The position of a particle as a function of time (in s) is given by C1 + C2t + C3t2. Let C1 = 11.0 m, C2 = 9.5 m/s and C3 = -0.49 m/s2. What is the velocity of the particle at time t = 15.0 s? And what is the particle's acceleration at time t = 15.0 s?

    2. Relevant equations
    Velocity is the derivative of position.(How do you make it look all nice?)
    Acceleration is the 2nd derivative of position.

    3. The attempt at a solution

    Where do I start? I don't even understand the equation that's given. Judging from the question (it's online) it looks like those t are variables and not subscripts.
    Last edited: Sep 3, 2009
  2. jcsd
  3. Sep 3, 2009 #2


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    Staff: Mentor

    Welcome to the PF. The skins have changed several times, so it's possible that a previous skin was dark.

    The "t" in the equation is time, so you would just plug in the time to get the position as a funtion of time. You are correct about the velocity being the derivative of the position, and then the acceleration is the derivative of the velocity. Using LaTex, it looks like this:

    [tex]x(t) = C_1 + C_2 t + C_3 t^2[/tex]

    [tex]v(t) = \frac{dx(t)}{dt} = \frac{d(C_1 + C_2 t + C_3 t^2)}{dt}[/tex]

    [tex]a(t) = \frac{dv(t)}{dt}[/tex]

    So just do the differentiations, and plug in the respective times to get the answers.
  4. Sep 3, 2009 #3
    Just to check my derivatives since I haven't done them in months, I'm assuming that the C's aren't constants, right?

    I don't know the syntax for LaTeX so, the derivative of postion would be:

    C2 + 2C3t

    2nd derivative would be:

  5. Sep 3, 2009 #4


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    Staff: Mentor

    Correct. To see how the LaTex is formed, you can just QUOTE my post, to see the tags that are embedded in it. Also, there is a stickie thread with a LaTex tutorial at the top of the Learning Materials forum:


  6. Sep 3, 2009 #5
    I don't understand why it's saying my answer for the acceleration is wrong. My velocity answer was correct though.

    2C3 = 2(-0.49) = -0.98 m/s2

    Thanks for the sticky, I'll check it out when I finish my homework :P

    Ah never mind, forgot to type in the negative sign -.-