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Does an object stop when reversing direction?

  1. Mar 1, 2012 #1
    I am teaching mechanical energy, that is, gravitational potential and kinetic energy, to middle schoolers.

    Of course, we have been talking about things like juggler’s pins, roller coasters, and snowboard half-pipes. During this, the kids and I came up with a perplexing question.

    When you toss a ball up, it has negative upward acceleration, then it has positive downward acceleration.

    Does it “stop” to change direction, or does it go directly from moving up to down without ever pausing? It seemed to me as if it would have to stop rising to start falling, but that left me with the thorny problem of how long it was stopped.

    After thinking it over, I had to conclude that it never actually stops, because I couldn’t come up with any rationale at all for determining the amount of time it would stop. Is this addressed by the laws of physics?

  2. jcsd
  3. Mar 1, 2012 #2
    An instant.

  4. Mar 1, 2012 #3
    It stops for as long as it goes any other precise velocity. If you graph the velocity it would be a straight line with slope -9.8. The y-intercept is the initial velocity that you throw the ball or juggling pin or burning torches up. At the top of the arc, is the point when the the graph crosses the x-axis and velocity is zero.
    For any velocity that you choose, it is only that velocity for one instant. Mathematically there is no duration of any velocity. If you measure the velocity, it is only the average velocity that you measure. So if you average the velocity from a tenth of a second before and after the top of the arc, it will have zero average velocity.
  5. Mar 1, 2012 #4
    First thing, negative upwards and positive downwards are equivelant notions
    Second, there is an instant where the velocity is zero, this happens at the turning point, the highest part of the trajectory.

    The trajectory can be described by
    [itex]y[t]=v_0\ t-\frac{g}{2}t^2[/itex]

    Using our calculus tricks we find the velocity to be
    [itex]v[t]=y'[t]=v_0 - g\ t[/itex]

    To find when the ball stops, or when the velocity is zero we simply solve for
    [itex]v_0=g\ t\ \rightarrow \ t=\frac{v_0}{g}[/itex]

    This is also addressed in conservation of energy;

    If you have an initial energy [itex]E_0=\frac{1}{2}m v_0^2[/itex], we are free to move anywhere as long as the total energy is still E0, [itex]\frac{1}{2}m v[t]^2 +m\ g\ y[t]=E_0[/itex]. We can then see that velocity = 0 is allowed and we can find the position at which this happens.
    [itex]E_0 = m\ g\ y\ \ \rightarrow \ y =\frac{E_0}{m\ g}[/itex]

    The ball does stop, it HAS to stop in order to change the direction of its velocity, but only for an instant!
  6. Mar 1, 2012 #5
    "Instant" is exactly what I told the students. I guess the question that I am really asking, then, is how long is an instant?
  7. Mar 1, 2012 #6
  8. Mar 1, 2012 #7
    Mathematically, an instant has a time length of zero. I would think of it another way, though.

    Ask yourself, if you throw it upwards at 10 m/s, does it reach 5 m/s at some point or does it go straight from above 5 m/s to below 5 m/s. It will be going 5 m/s for the same amount of time as it is stopped, which is an instant, as many have said.

    Sure, it's a weird thing to think about, but motion is made up of an infinite number of these instants. At least, classically speaking.
  9. Mar 1, 2012 #8
    Come back Zeno.
    All is forgiven.
  10. Mar 1, 2012 #9
    Yea, just basically adding to what others said, you can just think about it as an application of the intermediate value theorem. Namely, we are describing the motion of bodies in a continuous fashion, the I.M.T. states that if f is a real-valued continuous function on some interval [a, b] and z is a number between f(a) and f(b) then there exists some c∈[a, b] such that f(c)=z.

    You can picture a height as a function of time graph as well. Setting your coordinates such that y is never negative you can clearly see a parabola traced out in space. The vertex of this parabola, then, would be where the object is zero and turns around. Drawing a horizontal line across the parabola you can see that, because it is an even function, the height of f(x) and f(-x) lie on the same line y=f(x) and so these can be the end points of your function for which 0 lies in between. Now, because Δt is infinitesimal you can move this horizontal line up from the bottom and get the length of this line arbitrarily close to zero. That is, as the line f(x) goes to the vertex the distance of the line shrinks to zero.

    All this is just a way for you to possibly illustrate to your students the concept in different ways. It is usually easier for people to grasp things visually then analytically, thought everything I said was contained in genericusrnames post, but using pictures you can probably more easily explain it to your students (or a mix of both of course).

    Now, regarding what the physical ontology of an instant is ...... haha, that question can get tricky if given a complicated treatment. You could use it as an oppertunity to introduce Zeno's paradoxes to your students though.
  11. Mar 1, 2012 #10

    Ken G

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    Gold Member

    To follow up on this very important point, I strongly advise that students not be told that the acceleration "switches" from negative up to positive down. Instead, pick a consistent convention, like "up" is positive and the acceleration is always "negative up," which is always the exact same thing as positive down no matter which way the ball is moving. Otherwise, if you ask them what the acceleration is when the ball reaches the top of its motion (and v=0), they will invariably say the acceleration "goes through zero" while it is "switching directions." The velocity does do that, but the acceleration doesn't-- it is always the same. Newton's laws will just not make sense until the students "get" that important distinction.
  12. Mar 1, 2012 #11


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  13. Mar 6, 2012 #12
    If the graph of the velocity vs. time is an upside down pointy V, then the acceleration is undefined at the peak, correct? (infinitely many, differently angled lines can be drawn tangent to it) If so, I find that the most interesting thing.
  14. Mar 6, 2012 #13
    The velcity vs time graph is a straight line with constant gradient (-g)
    The acceleration is defined at all times and is equal to g
    The position vs time graph has a parabolic shape, at the peak of it the velocity is 0 but the velocity is still defined.

    Why do you think you can draw infinitely many lines tangent to it?

    Pic related - Position in Blue, Velocity in Purple, Acceleration in Green/Brown/Yellow
    Initial conditions are position = 0, velocity = 10

    Attached Files:

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