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Need some feedback on a variation of the [[D = V°(t) + 1/2(A)(t)^2]] formula

  1. Feb 16, 2012 #1
    Hello guys, hope everything is going well. Anyways, I was thinking recently about the forumla for finding the distance an object has traveled with a constant acceleration using the formula

    D = V°(t) + 1/2 (A)(t)2

    So basically when thinking about this formula, I like to think about an object that starts from origin at rest (0 m/s), experiencing a constant force that causes a constant acceleration, which increases the velocity linearly, which increases the distance Parabolically (maybe not exactly a parabola but not linear is what I mean.)

    Ok so here is an example.

    An Object is at rest at Time(T) = 0 seconds.
    The object is experiencing a constant force (arbitrary because I wish to not introduce numbers) that causes a constant acceleration.

    Let's say the object is accelerating at 2 m/s2
    A = 2 m/s2
    V°(initial V) = 0
    V = V°+ (A)(T)

    Then the distance is equal to

    [[ D = (0)(t) + 1/2(2)(t)^2 = (1)(t)2 = t2 ]]

    So the distance increases as a parabola, t^2.



    here is a graph representing these functions according to what we just put down above. Distance = T2, V = 2(t), A = 2.

    crmv6.jpg



    so once again, a constant acceleration causes a linear velocity which causes a parabolic increase in distance.
    ___________________________________________________________________________



    Here is where it got interesting to me, instead of considering a constant force that causes a constant acceleration, which causes a linearly increasing velocity which creates a parabolic increasing distance, I was wondering what kind of effect a constantly INCREASING force, which causes a constantly INCREASING acceleration, has on velocity?

    My idea was that maybe we should just use the orignal equation, (D = V°(t) + 1/2 (A)(t)^2),
    and adjust it to help us find the velocity if we know how our acceleration is constantly increasing.



    So here is the way I have attempted to translate the formula to fit our new paradigm of a constantly increasing acceleration instead of a constantly increasing velocity.

    I basically take every variable except for time, and substitute it's derivative. The derivative of a Distance vs. Time graph is an objects instanteous velocity.

    So I replace D, in (D = V°(t) + 1/2 (A)(t)^2), with V, for Velocity.

    The deriviate of a velocity vs time graph is an objects acceleration. So I replace V°(initial Velocity) with A°(initial acceleration)

    we now have (V = A°(t) + 1/2 (A)(t)^2) [ we substituted V for D, and A° for V°]

    The derivative of a acceleration vs time graph is an objects change in acceleration. For the sake of this I am now going to refer to this objects change of acceleration as Ω, this will make it easier than typing out that I am refering to the acceleration of the objects acceleration

    We now replace A, in our original problem (D = V°(t) + 1/2 (A)(t)^2), with Ω, which gives us

    (V = A°(t) + 1/2 (Ω)(t)^2).

    So what does this say? It says for an object that is experiencing a constantly increasing force, which results in a constantly increasing acceleration, we can find the velocity by multiplying the original acceleration by the time, and adding that to the average acceleration times the time squared.

    So now let's use those same numbers from earlier, but move them up an order of magnitude if you know what I mean.

    I mean what was a constant acceleration, will now become a constant Ω, (increase in acceleration). What was a linear increase in velocity, will now become a linear increase of acceleration. Well by this logic, the Velocity will then take on the role of the distance in the orignal equation.So now the velocity will experience a parabolic type increase.


    ____________________________________________________________________________
    JUST FOR CLARITY

    Values of the original graph

    Ω(change of acceleration) -- 0

    A(acceleration) -- 2 m/s2

    V(Velocity) -- (A)*(t) = (2)(t)

    D(Distance) = [V°(t) + 1/2(A)(t2)]= --> (0)(t) + 1/2(2)(t2) = ---> t2

    _____________________________________________________________________________

    Now we are going to move each of the values up one step to the higher magnitude of derivative above it. (notice velocity = f'(x) for D vs. T graphs. Accel. = f'(x) for V vs. T charts etc)

    Values of the new graph

    Ω(change of acceleration) -- 2 m/s3

    A(acceleration) -- (Ω)(t) = --> (2)(t)

    V(Velocity) -- [A°(t) + 1/2(Ω)(t2)]= --> (0)(t) + 1/2(2)(t2) = ---> t2

    D(Distance) = I don't know what this is.. I don't even want to begin thinking about how an every increasing acceleration effects the distance traveled. Maybe it effects it exponentially?












    So once again, for clarity,( maybe over clarity) , a force is causing an object that starts at rest and from an initial acceleration of 0m/s2, to increase it's acceleration at 2 m/s3. The constant increase in accel. causes the acceleration to increase linearly. The linear increase in acceleration causes the velocity to increase Parabolically!

    Here is the graph

    2hqz5fl.jpg



    as you can see, a constantly increasing acceleration effects velocity in the same way that constantly increasing velocity effects the distance traveled?



    so is this even important? or is it blatenly obvious? Is there an easier way to come to this or did a make a terrible error?
     
  2. jcsd
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