Multiple Representations and Interactions with Motion and Force Examples
Dan Phelps <wyeast@uniserve.com>
An Initial Note
The point is that one can first mathematically demonstrate that there is this relationship between a
change in position and a change in time and then use measurements and analysis to determine the
value of this relationship and how it may be changing with time. Please do a Google (or other search)
using “Rob MacDuff” to follow this up.
For example, rather than beginning by measuring and analyzing the motion of an object vs. time, one
can begin with the statement “A change in position and a change in time go together.” And there is a
relationship between the change in position, x, and the corresponding change in time, t. This
relationship can then be developed using an identity, multiplication by a “fancy form of one” (t / t),
and then rearranging the result:
x = x = (x) (t / t) = (x / t) ( t ) = ( v ) ( t )
And the ratio (x / t), which we call the velocity, is the relationship between the change in position
and the change in time. – If you are not comfortable with using “x” for position please use “s”.
AND PLEASE, DO NOT use the letter “d” under any circumstance in spite of what things from the
BC Ministry of Education says. This is because both the words “distance” and “displacement” begin
with the letter “d”. And position is neither “distance” or “displacement.
Multiple Representations and Steps to Require in All Problem Work
Multiple Representations each show a different aspect of “what is happening”. And different
representations are processed in different parts of the brain. The point of using and requiring that
students use all of these representations is so that they will come to know them and gain confidence in
their ability to solve problems. Also students will then come to appreciate that something other than an
algebraic solution can provide a more simple and faster solution to a question.
Require that students provide all of the following in their work. At the beginning of each term state
that you will stop marking if a step is left out. (At the very least, deduct marks for each missing item.)
Following is a simple example.
a) Prose text presenting and/or describing the situation – INDICATE ASSUMPTIONS AND/OR
DEFINITIONS TO BE USED!
Example: Initially ( use t0 = 0 s ), a person is standing still, 0.5 metres east of the origin (define east
as the positive direction). Then they move east at 0.5 m/s for 3.0 seconds. (Assume the person’s
velocity can change very quickly.) Next the person turns 180 degrees in place in 2.0 seconds. Finally
the person moves west at 1.0 m/s for 3 seconds. What is the final position of the person?
b) Labelled sketch or sketches -- For the reader to complete
c) A Motion Map
A time is written next to each circled position point. If an object is at rest for a period, two or more
concentric circles are used
d) If forces and/or energy is involved show the corresponding objects involved and the interactions
between the objects. (More will be said below in the section with the label “Interactions”.)
e) A Graph or Graphs with units
f) An equation or equations
Position, x = x0 + Δx1 + Δx2+ Δx3
A given displacement Δxn= (vn)(Δtn)
Next the equation(s) should be solved algebraically. Only then should numerical values with units
be substituted into the final algebraic solution. And finally the numerical work should be simplified
and a final result (or results) should be stated with units.
Initial position, x0 = 0.5 m
Displacements: Δx1= (v1)(Δt1) = ( 0.5 m/s)(3.0 s) = + 1.5 m
Δx1= (v2)(Δt2) = ( 0.0 m/s)(2.0 s) =
0.0 m
Δx1= (v3)(Δt3) = ( - 1.0 m/s)(3.0 s) = - 3.0 m
The result is x = x0 + Δx1 + Δx2+ Δx3 = + 0.5 m + 1.5 m + 0.0 m - 3.0 m = - 1.0 m
The values used here were chosen so that the initial position was different from zero and so that both
at least one displacement and position value were negative.
This may be the time to help students see that areas under velocity vs time graphs provide
displacements.
What to say when … ?
Here are some examples: Suppose we chose to the right as the positive direction. And now lets
consider: (a) an object moving right with the velocity increasing, (b) an object moving left and (c) an
object initially moving right with the velocity decreasing so that the direction of motion changes.
Hopefully you will introduce you students to the possible combinations of positive, negative and zero
values for initial position, initial velocity and acceleration with the corresponding motion maps,
velocity vs. time graphs and position vs. time graphs.
There is choice of simply choosing a positive direction and the referring to the sign of the velocity
and the sign of the change in velocity with a change time or of describing the direction of motion in
words and referring to the speed of the object. Let us use motion maps to represent the motion for (a)
and (b).
0
-|---------------------------> Position, x(m)
a)
.
.
.
.
0s
1s
2s
3s
We could say the velocity is positive and increasing or the velocity is positive and the acceleration is
positive. Alternatively we could say the object is moving to the right and speed is increasing.
b)
.
.
4s 3s
.
.
2s
1s
Here we could say the velocity is negative and decreasing or the velocity is negative and the
acceleration is again positive. Alternatively we could say the object is moving to the left and speed is
decreasing.
c) Here it is probably easier to consider a velocity vs. time graph.
An example such as an object initially moving up a hill to the
right (in the positive direction). A very important thing to say is
that at time, t = 2 s, the direction of the motion changes, not that
the object momentarily stops. Otherwise, any student who is
still thinking “a = v/t” will then say that the acceleration is zero
at t = 2.0 s.
And please do not use the word “deceleration”! So directions
of motion, etc. are clear, simply say the initial velocity is positive
and the acceleration is negative. Alternatively, say the object is
initially moving right and slowing down, motion reverses
direction and the object moves left and speeds up
Using a Velocity vs. Time to Solve a Problem
I chose this example to stress that a strictly algebraic solution to a problem is often NOT the most
effective way to proceed. Here we will use the idea that areas under a velocity vs. time graph provide
the necessary displacements.
On August 16, 2009, Usain Bolt ran the 100 metre dash. Suppose he started from rest and accelerated
uniformly for 2.50 seconds when he reached a speed of 12.0 m/s and then ran at constant speed until he
completed the race. Choose the direction he ran as the positive one. How long did it take him to run
the complete 100 metre race?
Let us look at this using a velocity vs. time graph.
The sum of the two displacements, Δx1 and Δx2 is 100.0 metres so Δx2 = 100.0 m - Δx1
The displacement, Δx1 corresponds to the triangular area of the velocity vs. time graph
from t = 0 to t = 2.5 s
So Δx1 = (1/2) (v1 – 0) (t1 – 0) = (1/2)(12.0 m/s – 0)(2.50 s – 0) = 15.0 m
We can now work on Δt2 = t2 – t1 = t2 – 2.50 s. The velocity for that entire time interval is a constant
v2 = 12.0 m/s. And Δx2 = 100.0 m - Δx1 = 100.0 m - 15.0 m = 85.0 m. This displacement also
corresponds to the area of the rectangle on the right with height, (v2 – 0) and a width of (t2 – t1).
Since, Δx2 = (v2 - 0)(t2 – t1) we can find (t2 – t1) = (Δx2) / (v2 - 0) = (85.0 m) / (12.0 m/s) = 7.08 s.
Now we can answer “How long did it take him to run the complete 100 metre race” which is
(t1 – 0) + ( t2 – t1 ) = 2.50 s + 7.08 s = 9.58 s
And happily, the set of values chosen for this problem gives a result that agrees with:
http://www.washingtonpost.com/wp-dyn/content/article/2009/08/16/AR2009081601161.html
“Bolt ran 100 meters in 9.58 seconds” in Berlin on that day setting a new men’s world record time.
Interactions
First let us focus on forces as an interaction between two objects. It is important to read these in
words even though they may be abbreviated using symbols. So students MUST be taught to read and
say “The force of ‘A’ on ‘B” whenever they see --->
FA on B
The forces we are most likely to be considering are gravitational, g, electrical, e, magnetic, m, or
contact, c. The other very important thing to remember is that if object “A” interacts with object “B”
then the force of “B” on “A” is exactly equal in magnitude and opposite in direction as the force of
“A” on “B”. This is just Newton’s third law.
One must also include information about the duration of the interaction between two objects in any
description and/or diagram. Some interactions are temporary and can be shown using a dotted line
with points at each end between the pairs of letters standing in for the two objects. A gravitational
force will continuously act between two objects and an electrical or magnetic force may do so also.
Show any continuous interactions with a solid line with a point at each end.
The following example may help with this idea. Consider initially holding an object such as a block
of wood at rest in your hand. Then throw the block in the air and let it fall to the surface of the earth
and come to rest. While in your hand the block is interacting with your hand via a contact force and
with the bulk of the earth via a continuous gravitational force. Once free of your hand we need only
consider the continuous gravitational force interacting between the block and the bulk of the earth.
Finally, once the block comes to rest on the surface of the earth we need to consider the contact force
providing the interaction between the block and that surface. Again the continuous gravitational force
interacting between the block and the bulk of the earth is still there.
In the following diagram these letters are used: block, B; Hand, H; bulk of the earth, E; and surface
of the earth, S.
Another View of Friction Forces
A very useful way to view friction is to simply combine it with other forces from a surface on the
object involved.
First consider an object, such as a block of wood, resting on a plane inclined surface that is at a gentle
enough angle that the object does not slide. Traditionally one considers a “normal force”
perpendicular to the surface and a “friction force” parallel to the surface. Alternately, one only needs
to consider a single force of the surface on the object and a force of the bulk of the earth on the object.
Since the object is not moving, the motion of the object is not changing and this means that the sum of
forces on the object is zero. Since the gravitational force from the bulk of the earth is vertically down,
the force of the surface on the object is vertically up.
If the inclined surface is “steep” enough, the object will start to slide and probably accelerate. This
can be described by noticing that force of the surface on the object is slightly smaller in magnitude
than the gravitational force of the earth on the object and that its direction has shifted somewhat from
the vertical toward the direction of a normal to the surface. The vector sum of these two forces is now
pointed “downward” parallel to the surface.
Here are the vector diagrams for these two examples: Earth, E; Surface, S; and Object, O
Note: The maximum angle between the normal direction and the net surface force due to “friction”
parallel to the surface and the normal force is Φmax = tan-1 ( Ff static / FN ) = tan-1( μs ). Once the angle
between the surface and the horizontal exceeds this value, the block will begin to slide and the angle
between the normal direction and the net surface force becomes
Φ = tan-1 ( Ff kinetic / FN ) = tan-1( μk ).
A Second Friction Example
This may be a good way to introduce students to the idea of “friction” particularly if they only think
of it acting only when one object is sliding over the surface of another object in contact with it.
Consider this example. A block of wood is resting on a horizontal surface, an elastic band connects
from a hook on the right side of the block to your finger. You gently stretch the elastic band enough
that students can see that it has been extended but not enough to cause the block to slide. Students will
probably agree that the bulk of the earth is “pulling down” on the block and that the elastic band is
“pulling horizontally to the right” on the block. And hopefully, they are comfortable with the idea that
since the block is not moving the vector sum of forces on the block must be zero. And if this is so,
then there must be “some other” force acting on the block.
In the following force diagram, “B” stands for block, “E” stands for earth and “El” stands for elastic
and “S” stands for surface.
A brief discussion of Energy and Interactions
When we consider energy and the corresponding interactions we do need to look at the various forces
involved, and the work done on or by each object in terms of displacements and forces. We also need
to include “thermal energy”. For example, suppose that a block of wood is pushed horizontally across
a horizontal surface. The temperature of the object will increase and the thermal energy of the object
will also increase. What happens once the block is still? The temperature of the block returns to an
equilibrium with the surrounding environment. This corresponds to a transfer of thermal energy from
the block to the surrounding environment.
Finally 2 Pages from Carl Wieman
I am attaching a 2 page paper written by Carl Wieman for your consideration. Carl Wieman
developed the Science Education Initiative at UBC. Please look at the items under the section titled
“Generic components of expertise in all fields of science and engineering” on the first page. Also please
look at the comments about what little of these are considered in “back of the chapter” problems.
In the time set aside for you to design a problem, please try to include as many of Wieman’s points as
you can.