Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Rectilinear Kinematics Help

  1. Jan 19, 2010 #1
    1. The problem statement, all variables and given/known data

    The acceleration of a particle is described by the following:
    a=(2t -1) m/s^2
    So= 1m
    Vo= 2m/s

    a) velocity of the particle at 6 seconds

    b) its position (Sf) at six seconds

    2. Relevant equations


    3. The attempt at a solution
    Using integration I found velocity to be 32 m/s, (dv=apartdt)

    Now i'm stuck on part b and since i have yet to learn how to take a double integral to directly relate the acceleration of the particle to its position, I assume it must have something to do with avg velocity or avg acceleration... nonetheless im stuck on part b and have been for quite some time.
  2. jcsd
  3. Jan 19, 2010 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Your answer is not correct, and the part you have in parentheses seems to be meaningless. Can you show your integration steps?

    Edit: Actually your answer is fine. Nevermind.

    Not really. This is only a "double integral" in the sense that you have to integrate twice in succession. (In mathematics, a double integral actually means something else entirely that is not relevant here). To see why you have to integrate twice, consider the following argument. If:

    [tex] v(t) = \frac{ds(t)}{dt} [/tex]


    [tex] a(t) = \frac{dv(t)}{dt} = \frac{d}{dt}\left(\frac{ds(t)}{dt} \right) = \frac{d^2s(t)}{dt^2} [/tex]

    So acceleration is the second derivative of position (w.r.t. time). In order to calculate the acceleration as function of time, you must differentiate the position function twice. It follows that, since integration is the inverse operation of differentiation, then to go back to position from acceleration, you would have to integrate twice. A more explicit way of showing this would be:

    [tex] \frac{dv(t)}{dt} = a(t) \Rightarrow v(t) = \int a(t) \, dt [/tex]

    [tex] \frac{ds(t)}{dt} = v(t) \Rightarrow s(t) = \int v(t) \, dt = \int \left(\int a(t) \, dt \right) \, dt [/tex]

    Why would you think that?
    Last edited: Jan 19, 2010
  4. Jan 19, 2010 #3
    i understand that the dv= a(t)dt can be integrated to acquire
    v(t)= [tex]\int[/tex]a(t) dt

    However this produces 32 m/s if taken definitely (with 6 and 0 upper and lower limits respectively)... to integrate this again seems unusual since it is a constant (it almost seems to disregard the fact that v is variable with respect to time). This i guess is where i became confused. Is it legal to take the integral of 32m/s as the velocity with respect to time ?
  5. Jan 19, 2010 #4


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    No no...you integrate the function v(t) to get s(t). That's the definition. You don't integrate v(6 s)! You integrate v(t) to get s(t) and then find the position you want by evaluating the resulting position function there (i.e. you calculate s(6 s)). This is no different from what you did to find the velocity at that time in part a!
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook