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Teaching change in velocity

  1. Aug 15, 2017 #1
    Good afternoon all,
    For those of you that teach physics, if an object has an initial velocity of - 5m/s and a final velocity of -20 m/s, do you consider the velocity to be increasing or decreasing in this case?

    I view this to be strictly defined mathematically, that is that the acceleration is negative, therefor I describe this as a decreasing velocity and an increasing speed in the negative direction.

    Both the HS physics teacher I student taught under, and the physics teachers where I teach now, regard this as an increasing velocity. I continue to go in circles with them trying to reconcile this. Essentially, they interchange speed and velocity. Both of these gentleman have been teaching physics for 20+ years, so I am trying to take a humble approach here and see if there is something I am missing. Am I missing something?
    How do you teach it?
    Thanks.
     
  2. jcsd
  3. Aug 15, 2017 #2
    Velocity is made up of two components magnitude (speed) and direction.
    So, in the simplest case, if the magnitude changes but the direction is the same, then I would think it proper to say that the velocity has "increased" or "decreased".
    Otherwise, I would just say there was a change in velocity.
     
  4. Aug 15, 2017 #3
    Thanks for the reply .Scott. So are you saying that either would be fine?
    To me, if I say that the velocity is "increasing," then by definition, I am saying that he acceleration is positive. In this case the acceleration is negative.
     
  5. Aug 15, 2017 #4

    robphy

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    This is an issue involving vectors, magnitudes-of-vectors, and components-of-vectors.
    Of course, velocities are vectors, components-of-velocity are signed scalars, and magnitudes-of-vectors ("speeds") are non-negative scalars.
    For clarity [especially when needed], one should use full descriptions with a good notation.

    To say "an object has an initial velocity of - 5m/s and a final velocity of -20 m/s" implicitly references a coordinate axis:
    so... ##v_{x,i}=-5{\rm\ m/s}## and ##v_{x,f}=-20{\rm\ m/s}##.

    Thus, ##\Delta v_x=v_{x,f}-v_{x,i}=(-20{\rm\ m/s})-(-5{\rm\ m/s})=-15{\rm\ m/s}##.
    • "The x-component of velocity has decreased" since ##v_{x,f} \equiv (v_{x,i}+\Delta v_x) <v_{x,i}##.
    • "The magnitude of the velocity ("speed") has increased" since ##|v_{x,f}|>|v_{x,i}|##.
    • "The x-component of the average-acceleration is negative" since ##a_{x,avg}=\frac{\Delta v_x}{\Delta t}<0##.
    • Often, one says that "an object accelerates ["speeds up", "has increasing magnitude-of-velocity"]" when ##\vec a\cdot \vec v>0##,
      which, in the 1-d case, means "when the change-in-velocity[vector] is in the same direction as the initial-velocity[vector]".
      (Note: the rate of change of square-speed ##\frac{d}{dt} |\vec v|^2=\frac{d}{dt} (\vec v\cdot\vec v)=2 \vec a\cdot \vec v##.)
    So, in this example, "the object accelerates ["speeds up"]" (since ##\frac{\Delta v_x}{\Delta t} v_{x,i}>0##)
    although "its x-component of average-acceleration is negative" (since ##\frac{\Delta v_x}{\Delta t}<0##).
     
  6. Aug 15, 2017 #5
    Thanks for the reply. I would also say an object accelerates when it is slowing down or the magnitude of the velocity is decreasing, not just speeding up and increasing, respectively. Its just a negative acceleration. Im not sure i got your response about how you would regard this change, do you regard the -20 m/s as a higher or lower velocity than -5 m/s? Sorry for the crude notation; im on a tablet.
     
  7. Aug 15, 2017 #6

    robphy

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    When the object "slows down" ("has a decreasing magnitude-of-velocity"), one often says "the object is decelerating".

    Concerning "an object has an initial velocity of - 5m/s and a final velocity of -20 m/s",
    I would say that
    • the x-component-of-velocity has decreased:
      ##(-20{\rm\ m/s}\ \hat x)## has a "lower x-component-of-velocity" than ##(-5{\rm\ m/s}\ \hat x)##
    • the magnitude-of-velocity ("speed") has increased:
      ##(-20{\rm\ m/s}\ \hat x)## has a "higher magnitude-of-velocity" than ##(-5{\rm\ m/s}\ \hat x)##
    • Strictly speaking, vectors [on their own] can't be sorted or ranked... they need additional structure.
      So, "higher velocity" doesn't really make sense... and I won't use it.
      One should really specify whether one wants to rank "x-components" or rank "magnitudes".
      Admittedly, in special cases, where everything is positive, one can get by with loose talk...
      ...but there are situations, as you have raised, where one can get into trouble trying to continue with loose talk.
    Drawing a position-vs-time graph, a velocity-vs-time graph, and/or a number-line might help supplement this discussion.
     
  8. Aug 15, 2017 #7

    jtbell

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    I would say that -20 m/s is "lower" than -5 m/s, in a purely numerical sense, whereas the magnitude of -20 m/s is larger than the magnitude of -5 m/s, and this has a physical meaning because the first is "faster" than the second.

    However, the first statement depends on your coordinate system. Suppose that it assumes that "+" is towards the west and "-" is towards the east. Now flip the coordinate system so "+" is east and "-" is west. Now the velocities are +20 m/s and +5 m/s. The second velocity is now "lower," in a purely numerical sense, but the magnitudes still have the same relationship. The first one is still "faster" than the second.
     
  9. Aug 15, 2017 #8
    I agree. So look at this graph. I read this to say that the velocity is decreasing from 4.0 s to 14.0 s.
    upload_2017-8-15_20-37-6.png

    Quite frankly, I don't see any other way to interpret it unless you start incorporating speed into the discussion, but that isn't necessary because all of the information about the speed is contained in the velocity.
     
  10. Aug 15, 2017 #9

    robphy

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    If this is 1-D motion, then you have plotted [say] the x-component of the velocity... and it's okay to just call it "velocity" (since it would be a 1-D vector).
    And, yes, the [x-component of] velocity is decreasing from 4.0s to 14.0s.

    (If the problem were a projectile problem in 2-D or 3-D, I would have a problem with the graph.)
     
  11. Aug 15, 2017 #10
    Yes, I am just talking about 1-Dimensional motion.
     
  12. Aug 15, 2017 #11
    I think this perfectly illustrates why most people are so confused by physics. We can't even agree on the way to put it into words among ourselves. Why make it more complicated than it needs to be? There is a change in velocity toward the negative direction. From 4s to 9s the change in velocity is in the opposite direction as the current velocity, so it is slowing down. From 9s to 14s (and maybe beyond) the change in velocity is in the same direction as the current velocity and the object is speeding up. When talking about 1D motion, sign only denotes direction, it is completely arbitrary which is which. It'd be like saying north is greater than south and moving from north to south is decreasing.

    So to the original question, I would say the velocity is increasing magnitude in the negative direction.
     
    Last edited: Aug 15, 2017
  13. Aug 15, 2017 #12
    I would also agree that the magnitude of the velocity, or the speed, is increasing in the negative direction.
     
  14. Aug 16, 2017 #13
    Consider the case of a centrifuge. There is constant speed and constant acceleration. The object is neither slowing down nor speeding up.
    The same acceleration vector can cause either speeding up or slowing down. An acceleration of -1 m/s doesn't mean your slowing down.
    There is a bit of a semantics issue. If you mix the terms "speed" and "acceleration" in the same discussion, you may be really talking about "absolute acceleration". But when you go to the math, you are always dealing with velocity and acceleration vectors.
     
  15. Aug 17, 2017 #14
    I don't argue over definitions. The scalar number that describes the velocity is decreasing. The magnitude is increasing.

    But if one selects a coordinate system that points in the opposite direction, both are increasing.

    Never argue over things that depend on definitions or choices of coordinate system.

    In all my courses, I teach that students need to choose and clearly describe the coordinate system they have chosen to solve a problem. Usually this just means drawing coordinate axes with arrows and x and y labels somewhere on their diagram. Solutions without defined coordinate systems are ALWAYS WRONG. Solutions are judged with respect to their correctness relative to the coordinate system used.

    I try to avoid ambiguity and encourage the same from students. Saying a velocity going from -5 to -20 m/s is increasing is ambiguous because one has not specified whether one is talking about the scalar number or the magnitude of the scalar number. Asking a multiple choice type question that depends on the student to resolve this ambiguity is bad form.
     
  16. Aug 27, 2017 #15
    It is an increasing velocity in the negative direction. The negative sign indicates the direction and should be not used as part of the magnitude.
     
  17. Aug 27, 2017 #16

    David Lewis

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    The direction of a vector is specified by both the line of action and its sense. The + or - prefix only indicates the sense.
     
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