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Re: displacement and graphs



At 08:20 AM 10/10/01 -0700, Sewall, Les P. wrote (off-list):

>The following is a presentation of this concept given to a non-calc HS
>physics course. It is a simplification, but is it incorrect? In other words,
>to use a thought expressed by others on the list, will this presentation
>hinder a more advanced understanding at a later time?
...
>A progressive diagram drawn on the board accompanies the oral
>presentation.

Good.

>I live at the east end of a dead end street. The street is straight, runs
>east-west, and is 50 meters long. I walk from one end of the street to the
>other with my dog. The dog is not on a leash and is free to explore. I walk
>a straight line to the end of the street. The dog visits the fire hydrant,
>trees a squirrel, torments the dog behind the fence by running back and
>forth along its length a couple of times, and then eventually meets meet at
>the end of the street as I turn around to head back home. If I were to put a
>puppy pedometer on my dog and measure the length of the path that he
>travelled during his excursion, I'd find that he actually travelled 250
>meters before we met at the west end of the street.
>
>Interim Measurements (relative to starting point):
>My distance: 50 m
>Dog distance: 250 m
>
>My displacement: 50 m west
>Dog displacement: 50 m west
>
>Now we go back home. I, again, walk a straight path to where I began. The
>dog is still free to roam. At the end of this leg of the trip, the puppy
>pedometer has logged an additional 150 m.
>
>Final Measurements (relative to starting point):
>My distance: 100m
>Dog distance: 400m
>My displacement: 0 m
>Dog displacement: 0 m

I am 100% happy with the foregoing. The terms are clearly defined, and
used in a way consistent with their definition. Clear communication is the
criterion, and this has been achieved.

>This presentation follows a discussion of position (location with respect to
>an observer) and is in the context of vectors versus scalars: distance is a
>scalar, displacement is a vector, speed is a scalar, velocity is a vector,
>et cetera.

I'm not 100% sure that the discussion meets the objectives suggested by
this "context". It suggests a pairing (the standard pairing), namely
displacement->distance and
velocity->speed
which is, alas, not well exemplified by the example. More on this below.

>Summary definitions are given as follows. Distance is the total length of
>the path travelled without regard to direction. Displacement is the straight
>line distance between the initial position and the final position and the
>direction the object has moved.

There are definite problems here. When we take the specific, careful term
"dog-travel-distance" and generalize it to "distance", that's when we get
into trouble. In particular, consider the question:
What is the distance from point X at one end of the street
to point Y at the other?
The answer is !not! properly formulated in terms of biped-travel-distance
or quadruped-travel-distance or anything like that. (There may be a tree
and/or a hump in the sidewalk that prevents actual travel over the actual
distance.)

>How does this presentation compare with what students will learn in a more
>advanced course?

1) In an advanced course, we will speak of arc-length. The pedometer
measures (to a reasonable approximation) arc-length. I consider
dog-travel-arc-length to be a more precise term than dog-travel-distance,
but I cannot really object to the latter because it was clearly defined and
carefully used in the story.

2) Distance (when used without qualifiers such as travel-distance) is
defined to be the norm of the separation vector, which in this context is
the displacement vector. The distance to the end of the street is 50m,
whether you walked there in a straight line or not.

Note that taking the norm does not commute with addition of vectors:
|a + b + c + d| does not equal |a| + |b| + |c| + |d|
for typical vectors a,b,c,d;
it is never more, and it is strictly less unless
-- all contributions are zero, or
-- all nonzero contributions have exactly the same direction.

Note that if a,b,c,d are the elements of a path, then |a| + |b| + |c| + |d|
gives an approximation to the arc-length, while |a + b + c + d| gives the
ordinary distance.

>What if I further defined speed as distance over time, and velocity as
>displacement over time?

Velocity is displacement (during a given time interval) divided by the time
interval, in the limit as the interval gets very small. Note the following
refinements:
-- The denominator is not the absolute time (as indicated by a wall
clock) but rather the time INTERVAL.
-- If you leave out the LIMIT, you should speak of average velocity, not
velocity per se.

v := lim (x2-x1) / (t2-t1) (equation V)

Speed is defined to be the magnitude of the velocity vector. Taking the
norm of equation V, it is clear that speed is the distance, not arc-length
but ordinary distance (during some time interval) divided by the absolute
value of the time interval, in the limit that the interval gets very small.

s := |v|
= lim |x2-x1| / |t2-t1|