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Can you have a negative average velocity?

  1. Sep 3, 2012 #1
    I understand that it's possible to have a negative instantaneous velocity, but I'm wondering if it's possible to have an negative average velocity.

    For example, let's say you start heading north at a constant 10 mph for an hour. The second hour, you drive south (directly toward the starting point) at 10 mph (i.e -10 mph north). For the entire 2nd hour, your instantaneous velocity is -10 mph north, but after the trip, the average comes to 0 mph since you've stopped right where you started.

    So does this mean that a negative average velocity is impossible to achieve?
     
  2. jcsd
  3. Sep 3, 2012 #2

    rcgldr

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    What if you continued heading south for a third hour, what would the average velocity be if north is considered positive and south is considered negative?
     
  4. Sep 3, 2012 #3
    Gotcha. Thanks. I guess I was assuming that any displacement from the starting point would be positive. But if north were positive and south negative, what would west be?
     
  5. Sep 3, 2012 #4
    More importantly there is no good reason any positive quantity can't be thought of as negative, as long as you keep the system consistent (a.k.a. it's fine to change + to - anywhere as long as you also change - to +)
     
  6. Sep 3, 2012 #5
    This relates to speed, not velocity. Displacement and speed are non-vector quantities, whereas position and velocity are vector quantities. It is impossible to have a negative speed just as it's impossible to have a negative length or a negative magnitude.

    As to your question, if there's an east/west, then you have to introduce a new axis on your coordinate system and define a positive and a negative on that axis.
     
  7. Sep 3, 2012 #6

    Nugatory

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    The quick answer is that west could be either positive or negative, but either way it has nothing to do with north/south; we need one number for north/south and another one for east/west.

    The longer answer:

    As long as you're only allowing movements north and south, you're confining yourself to a single straight line, and one number (for example, positive for northwards and negative for southwards) is good enough to completely specify the velocity. That's another way of saying that a line only has one dimension.

    The surface of the earth is two-dimensional, meaning that any velocity can be written as the sum of two velocities, the north-south one that we've already discussed and a second east-west one. Let's choose the positive direction for east-west motion to be eastwards (as good a convention as any, and it's irrelevant which one we pick as long as we all agree to use the same one). Now we'd say that a velocity to the northwest is the sum of a positive north-south velocity and a negative east-west one.

    If we were talking about an airplane instead of a car, we'd need to introduce a third dimension, with positive velocity corresponding to climbing and negative velocity to descending.

    This would be a good time to google about vectors and vector addition, and about the difference between "speed" and "velocity". You cannot have a negative speed or negative average speed. But it's easy to have a negative average velocity - in the example above, crash the airplane somewhere to the southwest of its starting point.
     
  8. Sep 3, 2012 #7

    Nugatory

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    The quick answer is that west could be either positive or negative, but either way it has nothing to do with north/south; we need one number for north/south and another one for east/west.

    The longer answer:

    As long as you're only allowing movements north and south, you're confining yourself to a single straight line, and one number (for example, positive for northwards and negative for southwards) is good enough to completely specify the velocity. That's another way of saying that a line only has one dimension.

    The surface of the earth is two-dimensional, meaning that any velocity can be written as the sum of two velocities, the north-south one that we've already discussed and a second east-west one. Let's choose the positive direction for east-west motion to be eastwards (as good a convention as any, and it's irrelevant which one we pick as long as we all agree to use the same one). Now we'd say that a velocity to the northwest is the sum of a positive north-south velocity and a negative east-west one.

    If we were talking about an airplane instead of a car, we'd need to introduce a third dimension, with positive velocity corresponding to climbing a negative velocity to descending.

    This would be a good time to google about vectors and vector addition, and about the difference between "speed" and "velocity". You cannot have a negative speed or negative average speed. But it's easy to have a negative average velocity - in the example above, start the airplane climbing to the northeast, then reverse course, start descending, and crash it somewhere to the southwest of its starting point.
     
  9. Sep 4, 2012 #8

    mfb

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    This is true for vectors, but some scalar values cannot be negative (or have a completely different meaning there). Think about the distance between two points, for example, or simply the magnitude of a number. Or temperature - while it can be reasonable to assign negative temperatures to some systems, "colder than 0K" does not exist.
     
  10. Sep 4, 2012 #9
    just to clarify, displacement IS a vector quantity whereas speed is not.
     
  11. Sep 4, 2012 #10

    jtbell

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    You can talk about displacement and velocity as positive or negative only when you have one-dimensional motion along a line: east-west or north-south or whatever.

    When you have two- or three-dimensional motion, you have to use vectors. The concepts of "positive" and "negative" aren't meaningful for a vector as a whole. You specify a vector either (a) using two or three components, which can each be either positive or negative; or (b) using a magnitude which is always positive, and either one or two angles for the direction.
     
  12. Sep 4, 2012 #11
    I was referring to vectors, though what I said could have been misleading so I should have explicitly said that.
     
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