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Instantaneous Velocity from Strobe Diagram

  1. Oct 11, 2015 #1
    Hi guys! My first post here. I've been frequenting Physics Forums and have found a wealth of information, and I'm hoping I can get some specific help about this concept. Thanks!

    1. The problem statement, all variables and given/known data
    "Cart A and B move along a horizontal track. The top-view strobe diagram below shows the locations of the carts and instants 1-5, separated by equal time intervals."

    Cart A: o....o....o....o....o
    Cart B: o.o..o...o....o

    (The periods represent 1cm of distance, and Cart B's position at Instant 1 is 5cm ahead of Cart A. If you want a better depiction than my horrible attempt using o's and periods, here's a link to the graphic)

    "Is there any instant at which cart A and cart B have the same instantaneous velocity? If so, identify the instant(s) and explain. If not, explain why not."

    2. Relevant equations

    I think this is more of a conceptual problem, so I don't think there are any relevant equations. However, [tex] v= v_0 + at [/tex] might be relevant.

    3. The attempt at a solution

    So, I think the answer is that there's no way of determining when the instantaneous velocity is the same (the average velocity is the same for both from instance 4-5 because there is a 4cm difference between locations of cart A and B, and [tex] v_a =∆x/∆t[/tex], so they have the same average velocity) because instantaneous velocity is just dx/dt and I can't think of a way to find that w/o a function or a graph or something. I tried using [tex] v= v_0 + at [/tex] (for cart A: v_0=0, a=0, and t=5; for cart B: v_0=0, a=1, t=5), but I don't know if that will give me the instant(s) where the instantaneous velocity is the same.

    This is probably a really basic question and I have a feeling the answer's pretty obvious, but I'm not sure. If anyone could help me out, I'd really appreciate it. Thanks!
     
  2. jcsd
  3. Oct 11, 2015 #2

    mfb

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    Do you know the intermediate value theorem?

    If not: Clearly A starts with a higher velocity. If they never have the same velocity, A always has to have a higher velocity. Is that compatible with observations?
     
  4. Oct 11, 2015 #3

    collinsmark

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    Hello toboldlygo,

    Welcome to Physics Forums! :smile:

    No, this is not a conceptual problem. There is some mathematics involved.

    Perhaps try starting with the
    [itex] x = x_0 + v_0 t + \frac{1}{2}at^2 [/itex]
    uniform acceleration formula to solve for [itex] v_0 [/itex] and [itex] a [/itex]. Use two different data points to form two equations. Check your results with additional points (not only to double check your result, but also to make sure the acceleration is actually uniform).

    Then use that result to find the instantaneous velocity (Hint: you've already hinted at how to do that in your original post.) :wink:

    [Edit: corrected some typos.]
     
  5. Oct 11, 2015 #4
    @mfb: I wasn't actually familiar with that theorem; I think that'll definitely come in handy. However, A and B in this case both have the same velocity at instance 4-5, so will it still apply in this scenario?

    EDIT: just saw collinsmark's post. Thank you so much! That makes a lot of sense.
     
  6. Oct 11, 2015 #5

    mfb

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    If they have, then you are done, because that's exactly what you are supposed to find out.

    We don't know if the acceleration is uniform (although it looks like that) and we don't need that assumption.
     
  7. Oct 11, 2015 #6
    I already used instant 4-5 when I was asked whether there were any instances when the average velocity was the same. Is it plausible to use the same instant for both instantaneous and average velocity? Isn't instantaneous velocity only equivalent to average velocity when there is no acceleration? I guess I'm still a little confused when differentiating between instantaneous and average velocities. Thanks for helping me out, though!
     
  8. Oct 12, 2015 #7

    mfb

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    [4-5] is not a single point in time, it is an interval.

    Simple argument: If A is always faster than B (for every point in time), then A will move more than B in [4-5].
    If A is faster than B at some point and slower than B at some other point, then they have to have the same instantaneous velocity at some point.
     
  9. Oct 12, 2015 #8

    collinsmark

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    Just to elaborate further on what mfb said, your answer is fine regarding the average velocity interval.

    But you still have a bit more work to do to find the instantaneous velocity. The instantaneous velocity does not occur over an interval, rather it occurs at a single instant in time. The problem (part E) is asking you to find that particular instant in time. (If there is such an instant. And if there is, at least give a general indication of where that instant is relative to the labeled instances.)
     
  10. Oct 14, 2015 #9
    Thanks so much for helping me out, guys. I definitely understand this concept better. Again, I really appreciate it!
     
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