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Homework Help: Final velocity after distance given acceleration with respect to time

  1. Jul 23, 2014 #1
    1. The problem statement, all variables and given/known data
    I am given an equation for the acceleration of an object with respect to time a(t). I know that it will be accelerating over a certain distance d, after starting at rest. How do I find the final velocity that this object has reached after travelling the distance d?

    2. Relevant equations
    The given equation for a(t) is:
    a(t) = A(eBt-eCt)2
    Where A, B, and C are known constants, and t is time.

    3. The attempt at a solution
    I know that to find this final velocity I would use this equation:
    vf = ∫a(t)dt from 0 to te, the time at which the object reaches the distance d. I am stuck trying to find te, or at least some conversion to acceleration with respect to position a(x), so that I could integrate from 0 to d.
  2. jcsd
  3. Jul 23, 2014 #2


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    Homework Helper

    I haven't learned much calculus, but I am curious, why can't you solve for [itex]t_e[/itex] like this?


    (I don't know if that's the correct way to write the second integral)

    Thank you if someone replies, and sorry if this wasn't helpful.
  4. Jul 23, 2014 #3


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    Gold Member

    Integrated, you're not quite right on that function. It works in this case, but only because it starts from rest.

    In general:

    ##\vec{v} = \frac{d\vec{x}}{dt}##
    ##\vec{a} = \frac{d\vec{v}}{dt} = \frac{d^2\vec{x}}{dt^2}##

    And backwards it's

    ##\vec{\Delta v} = \int_{a}^b \vec{a}(t)dt##
    ##\vec{\Delta x} = \int_{a}^b \vec{v}(t)dt##

    Notice that when you do it backwards like so, you get the total change in velocity or position. This is because when you integrate, you're summing up infinitesimal changes.

    If you want the position/velocity function, you have to apply initial conditions, meaning you have to add that +C at the end of your integral and actually solve for it (generally based on some function value at t =0)

    Basically, I think you and Nathanael have it right, I just wanted to clarify something I saw as a conceptual misunderstanding.
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