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A question about average velocity?

  1. Sep 4, 2004 #1
    Hi, this is my first post ever published on such forums... anyway i liked this forum very much especially the classic physics category... And i'll be very happy if somebody answers my question..
    i read once that average velocity can be calculated by this formula ONLY when it's LINEAR: V average= (V1 + V2)/2 #
    now my question is:
    Why must the velocity be Linear when using this formula??
    maybe i have an explannation for it but i'm not sure if i can explain it to u...
    my explannation is maybe when velocity is linear it changes in a sequence just like these numbers: 1, 2, 3, 4, 5, 6, 7, ...
    so that we can look out the average value of velocity with the average values formula: (n1+n2)/2
    i hope ur getting this down...
    f u got confused because of this explannation, just forget it and try to answer the question above, please...
  2. jcsd
  3. Sep 4, 2004 #2


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    The average velocity is calculate by [itex]\frac{\Delta \vec{r}}{\Delta t}[/itex] it does not matter whether the object is accelarting, jerking (is there such a word or did I just make it up? - anyway jerk is the derivative of acceleration with respect to time), etc.

    Howvere adding two velocities in the wya that you are doing to obtain their average clearly requires that:

    a) the velocities are constant

    b) the object in question experinces the two veolcities for the same length of time.
  4. Sep 4, 2004 #3
    Actually balooza is correct in reasoning why (V(i) + V(f)/2) works only for velocity that can be represented by a linear line.
  5. Sep 4, 2004 #4
    Almost correct. You can calculate average speed this way only when speed changes linearly (so acceleration is constant). Then average speed equals speed at the begining (v1) + speed at the end (v2) of the movement /2.

    But you can calculate average speed as said: s/t anytime.
  6. Sep 4, 2004 #5


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    It's good that you are paying attention to the special condition for when that formula applies.

    Here's a fancier proof:

    Assume [itex]v=at[/itex].
    &\equiv\frac{\int_{t_1}^{t_2} v\ dt}{\int_{t_1}^{t_2}\ dt}=\frac{\Delta x}{\Delta t}\\
    &=\frac{\int_{t_1}^{t_2} (at)\ dt}{\int_{t_1}^{t_2}\ dt}\\
    &=\frac{a\frac{1}{2}(t_2^2 - t_1^2)}{t_2-t_1}\\

    Here are algebra- and geometry-based proofs:
  7. Sep 4, 2004 #6


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    BTW - that also works if [itex]v = v_0 + at[/itex]. :-)
  8. Sep 4, 2004 #7
    well... if it's only expplained by derivation methods, as robphy did, then i'm satisfied because i haven't studied derivation yet!!!!
    but if there is a logical (i mean more like physics not like mathematics) explannation then i'll wait for an answer... anyway thank u all very much :)
  9. Sep 4, 2004 #8


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    Okay, here's another way of looking at it. Imagine sampling the speed at different times going from t = 0 to t = T. If we sample v over N different time intervals then the [itex]n^{th}[/itex] time is [itex]t_n = \frac{n}{N}T[/itex] with integer n going from 0 to N. If the acceleration is constant then the corresponding speeds at each of those times is [itex]v_n = v_i + a t_n[/itex].

    To find the average speed, add up all the speeds and divide by N+1 (the actual number of samples!). We can simplify the algebra a bit if we note that the final velocity is [itex]v_f = v_i + aT[/itex] so that [itex]v_n = v_i + (v_f-v_i)\frac{n}{N}[/itex] and the sum of the speeds is sum = [itex](N+1)v_n + (v_f-v_i) \frac{N(N+1)}{2N}[/itex] just as you conjectured in your original post.

    Finally, divide both sides by N+1 to obtain [itex]=v_i + \frac{v_f+v_i}{2}[/itex] and your result follows! Note this only holds if the acceleration is constant.

    BTW - you wanted a nonmathematical explanation which this isn't! I'm not sure you can fully remove the math because "average" IS a mathematical concept!
    Last edited: Sep 4, 2004
  10. Sep 5, 2004 #9


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    have you looked at the algebra- and geometry-based proofs I mentioned?

    The geometry proof is often given in non-calculus-based physics textbooks.
  11. Sep 5, 2004 #10


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    For some reason I took v1 and v2 to be different functions of time (probably becasue I always write initial and final velocities as u and v).

    Anyway balloza consider this partial 'proof'.

    Imagine a velcoity time graph of the motion of the object, if the velcoity is a linear function of time, it will simply be a straight line. Clearly (well to me) is the 'midpoint' of that line which is found by (v1 + v2)/2. Also it may not be obvious but the area under the graph is Δx (this is obvious in the case of constant velcoity as the area is simply vΔt and we know that with constnat velcoit v= Δx/Δt so Δx = vΔt), so the area can be found for when velcoity is a linear function of time simply by: Δt(v1 + v2)/2 = Δx ; therefore average velocity is simply Δx/Δt = (v1 + v2)/2
  12. Sep 5, 2004 #11
    thanx guys now i got it.....
    i'm grateful!!! :)
  13. Sep 5, 2004 #12
    hello , i told u the easiest way to calculate the average velocity
    average velocity= total displacement/total time
  14. Sep 6, 2004 #13
    it's not about what would calculate average V.. my point is why can't we use the formula i have given for no linear velocities..
  15. Nov 24, 2010 #14
    I have the same question regarding jerking, and then the change of jerking based on time
    I derived this equation from the anti-derivative of the constant jerking, distance equation-

    [tex]\underline{(Final Velocity+ Average Velocity with out jerking + Initial Velcoity)}[/tex]

    (Final Velocity+ Average Velocity with out jerking + Initial Velcoity)/3 =

    (Jerking t^2)/6+ (Initial acceleration t)/2+ Initial Velocity =

    Average Velocity with Constant Jerking =

    This equation can also be used to find the average point between any two points in any parabola.
    I don't quite understand why this works. I hope that by rearranging the equation I could understand it. This is the best most helpful form I came up with.

    I also want to reform equations with a change in Jerking and so on...... or the average point of higher polynomials for better understanding.

    My friend saw something very similar in statistics- yet he couldn't explain it to me.

    Any help would be greatly appreciated!
    Last edited: Nov 24, 2010
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