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position vs displacement (was Swartz letter in AJP (work-energy theorem))



At 5:56 PM -0400 9/2/04, Hugh Logan wrote to PHYS-L:
The use of "displacement" to label the axis of ordinates seems to have
popped up out of nowhere. At the very beginning of Chap. 3, they talk
about a position vs. time graph with the axes labeled "d" and "t."
However, they state that "d is used to represent its distance from the
origin." All their examples are consistent with d being a distance --
never negative. However (delta d) is described as a displacement as
previously mentioned, and it could be negative in the case of a downward
slope.

In 1997-1998, I reviewed and corrected a number of errors, mostly
careless numerical errors, in fun@learning.physics....
one finds that the author, Dr. Mark
Sutherland, labels the axis of ordinates "displacement" rather than
"position" on an x vs. t graph. In the text, he explicitly calls x
"displacement." I suggested that he call x position and delta x
"displacement," etc.

If
forced to think about it, I always thought of a position vector as a
displacement from the origin, but it seemed inconvenient to think of a
displacement as a change of displacement, although that is what it
really is.


I have struggled with this terminology (and its inconsistent use in the
textbooks) for years; I think we even had a round of discussion on this
list about it.

Randy Knight's new book (PSE, 1e) is the first that takes a consistent
approach, IMHO. He does as you, Hugh, suggest: x is position and \Delta x
is displacement. At least this year I have a consistent story for my
students rather than apologizing for the author's own confusion in previous
texts. But I don't know if I'm 100% totally happy yet, because I also
agree with the strangeness indicated in your last paragraph.

Furthermore, while there is a name for \Delta x (i.e. displacement), there
isn't a corresponding one-word name for \Delta v.

Knight uses the three stacked graphs and labels them position, velocity,
and acceleration. The derivative of the position graph is the velocity
graph, but the area under the velocity curve (between t_i and t_f) is the
"total displacement". This is actually correct since x_f = x_i + the area
under the curve, so moving the x_i over gives \Delta x = displacement =
area under the curve; but I think it is a subtle point that is lost on most
students. They want the area under the curve to be the integral and they
want the integral to be the inverse operation of the derivative, so they
want the area under the curve to be the position, not the displacement.

Cheers,
Larry