Physics 1-D Motion Coordinate Systems

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```					Physics 1-D Motion

Coordinate Systems
o   A frame of reference system to define position.
o   Every coordinate system must have an origin. This is the point that all measurements are based off
of.
o If you wanted to say exactly where your seat is in the classroom, you could not just say “5 meters”,
you would have to say something like “5 meters in front of the door” . By doing this you created a
coordinate system with the door being your origin.
o To locate anything, or defines anything’s position you must first establish a coordinate system.
Examples…..

In 1-D coordinate system
Mile markers on the highway the highway can be treated as a 1-D
line so if you are stranded you tell the tow service what mile marker
you are at and what direction you are headed and they can find you.
Generally the origin (zero pt.) is the state line. All mile
measurements are based on the distance away from the state line.

In 2-D coordinate system
If you are a hiker lost in the desert, you can send a GPS distress signal for
help. This signal will send two coordinates, one for latitude (East/West
location) and one for longitude (North/South location). If only one or the
other are given the helicopter will not be able to find you b/c all that they
will know is that you are somewhere on the latitude or longitude line that
was given. If both coordinates are given they will know exactly where
you are. The origin for the latitude would be the equator, and the origin
for the longitude would be the prime meridian.

In 3-D coordinate system
An airplane’s location requires 3 coordinates because it can move East
& West and North & South, (just like the hiker in the above example)
however it can also move up and down (vertically). So in order to find
the exact location of an airplane, the plane’s latitude, longitude, and
altitude must all be given. The origin for altitude would be sea level.
Before talking about motion and how to describe motion we must first consider the values we use to
describe motion such as length and time. All quantities can be classified as being one of two types….
Scalar- a quantity that includes only magnitude NOT direction
Examples - mass, time, speed, distance
Vector- a quantity that includes both magnitude and direction
Examples- velocity, displacement, acceleration, force
Example-    35 m/s North
Magnitude Direction
Now to discuss some of these quantities…
Distance describes the total length traveled regardless of direction.
Displacement  describes the total length from starting point to ending point, another way of thinking of it is
how far you are away from the origin
Explanation-- 1 lap around the high school track around would mean you travelled a distance of 400 m,
however it would also mean that you had a displacement of zero. The displacement is zero because you
wound up in the same position that you started.

Position location on a coordinate system, for right now our coordinate system is 1-dimensional

Later on it will be 2-dimensional (x-y coordinate system)
Example…
On the below number line, you moved from a position of -2 to a position of +4, what is your displacement?
Answer here is simply +6, because we made a net movement of 6 places toward the positive direction.

-5 -4 -3 -2 -1 0 1 2 3 4 5

If you moved from a position of +5 to a position of -1, what would be your displacement?
The answer would be -6 not simply just ‘6’ because your displacement was in the negative direction.

-5 -4 -3 -2 -1 0 1 2 3 4 5

**In coordinate systems, (if dealing with the horizontal plane) movement to the left should generally be considered
negative (-) and to the right to be positive (+).

**In coordinate systems, (if dealing with the vertical plane) movement downward should generally be considered
negative (-) and upward to be positive (+).
Example….
Johnnie walked 18 kilometers north then turned around and walked 12 kilometers south.
What was Johnnie’s distance? 12 + 18 =
30 km (scalar)

What was Johnnie’s displacement?
If you make your origin at his starting point his total displacement would be 6 km North      (vector).
When combining vectors, we always ‘add’ vectors, however since South is the negative direction we get…
18 + ( -12) = + 6 ….. which is the same thing as 6 km North

18 km North and 12 km South are both
+18 km      -12 km
displacement vectors……. 6 km North would be
what we call the resultant displacement vector.
+6 km

Resultant -- the combination (or result) of 2 or more vectors…..     true for any type of vectors, not just
displacement
 Vectors in opposite directions will mathematically subtract from each other, b/c one of the vectors must be
in the negative direction
 Vectors in the same direction will mathematically add to each other, b/c both vectors will have the same sign
since they are in the same direction
 If vectors are acting in different planes (i.e. one acting East and one acting North) they CANNOT be combined
through simple addition or subtraction. This would be considered to be in 2-dimensions.... the procedure for
this will be given next chapter

Speed & Velocity
Motion is Relative…
All measured rates of motion are relative to something else.
There is no fixed standard by which to measure an objects velocity.
Different observers can observe one object moving at different speeds at the same time, depending on
their own motion
since all velocities are dependent on the motion of the observer in order to properly explain the velocity of
an object we must say 45 m/s “relative to….” That ‘relative to’ means ‘from the perspective of.
we generally don’t have to say this because more often than not most speeds are expressed relative to the
Earth
        Now that we understand the nature of motion, we can understand how to quantify it

Speed
 The rate of change of distance
 The word ‘rate’ always implies time is involved….. so we get….

   v = d/t
 Where v = speed in m/s
 d = distance in m
 t = time in s
 Speed is a scalar quantity, which means it does not have a direction
 Also notice that it is distance in the equation for speed NOT displacement
Velocity The rate of change of displacement
 Again ‘rate’ implies time…… so we get

   v = x/t
 Where v = velocity in m/s
 x = change displacement (in the x-direction) in meters = xf-xi
 t = time in s
 notice the equation for velocity (v = x/t) is , this is because speed and velocity will usually have the same
magnitude, However Speed and Velocity are not the same thing…. Although many times we will use the two
terms interchangeably because again very often their magnitudes are equal.
 Also notice that the ‘d’ in this equation represents ‘displacement’ not ‘distance’ as in the speed equation
Examples of velocity / displacement relationships…
Initial position is 18 m, velocity is -2 m/s
Time (s)       Position (m)           Displacement
per sec (m)
0              18                     -2
1              16                     -2
2              14                     -2
3              12                     -2
4              10                     -2

Initial position is 0 m, velocity is 5 m/s
Time (s)       Position (m)           Displacement
per sec (m)
0              0                      5
1              5                      5
2              10                     5
3              15                     5
4              20                     5

Ex. A high school track runner runs the 400 m dash (one full lap) in 51 seconds.

What is the runners speed?
V = 400/52 = 7.7 m/s

What is the runner’s velocity during this time?
V = 0 / 52 = 0 m/s
The change in displacement is 0 m, b/c again the runner wound up
in the same position where he started, and therefore the runners
velocity is 0 m/s.

Now answer this question about the runner above…. Is the runner running at exactly the same speed for the entire
400 m?
Answer- No, the runner is moving at the different speeds at different points in the race. He of
course has to accelerate from rest in the beginning and probably slows down at the end as his legs
are tired.
So, what we actually found was his average speed. And also for velocity we actually found the runner’s Average
Velocity over the course of the entire race.

Average Speed  the average speed of an object over some period of time
Average Velocity  the average velocity of an object over some period of time

These are the values we will usually be working with, however it is important to differentiate between these and
values for instantaneous speed and instantaneous velocity.

Instantaneous Speed / Velocity
Just as the word ‘instantaneous’ implies, these values represent the speed and velocity, respectively, at one
specific point in time. If Speed is absolutely constant then the instantaneous speed at any point will always be equal
to the average speed, however, many times this is not the case.

Like we said before the runner in the above example will have a lower instantaneous speed at a time of 48 sec
than he does at a time of 5 sec. At a time of 5 sec the runner is probably going at full speed and may have an
instantaneous speed of 8.9 m/s, as his legs are still fresh, however at a time of 48 sec. it is the end of the race and
the runners legs are tired so he will probably be going slower than his average speed, he may have an instantaneous
speed of 7.5 m/s at this point in the race.

In order to mathematically find instantaneous speed/velocity we would have to use calculus (which we will not be
doing) …..So we will always be working with some type of average speed/velocity, however it is important to be able
to understand the difference between average and instantaneous.

Acceleration
Acceleration-
the rate at which velocity changes
acceleration occurs any time there is any change in velocity
this can be a change in magnitude OR direction
example… plane circling above an airport at a constant SPEED. The plane is accelerating
because since the plane is circling, its direction is constantly changing…. This means that
there is a change in velocity and therefore must be an acceleration, even though the plane
has a constant speed.
Just as velocity explains the rate at which position changes, acceleration describes the rate at
which velocity changes.
 if an object has a positive (+) acceleration this means that the object is increasing its velocity in the
positive direction
if an object has a negative (-) acceleration this means that the object is changing its velocity in the
negative direction
equation

a = Δv/t
Δv – change in velocity in m/s
t -- time in sec
a – acceleration in m/s2
since Δv = vf – vi the equation becomes

a= (vf – vi )/t

or written in an alternative form

vf = vi + at
 Now this means that if we start from rest and we have a uniform acceleration of 5 m/s2 that our
break down of velocities over time would look like this
Time (s)    Velocity (m/s)
0           0
1           5
2           10
3           15
4           20

 Alternatively if we have an initial velocity of 10 m/s, and then accelerate uniformly with a rate of
-5 m/s2, that means that our break down of velocities over time would look like this
Time (s)      Velocity (m/s)          ** note that this chart means that as the object
0             10                      in question goes from a positive velocity to a
1            5                       negative velocity, that indicates that its velocity
2             0
is changing direction, but the acceleration is
3             -5
4             -10                     staying in the same direction

Accelerated Motion & Constant Motion
As mentioned before, an object is accelerating anytime velocity (magnitude or direction) is changing.
So if an object is not accelerating we say it has Constant Motion.
Constant motion can exist in 2 specific instances
1) Object is moving with a constant speed in a straight line
Or
2) Object is at rest

If neither of these are true for the object in question, then the object in question must be accelerating.
It is also very important to understand that if an object IS accelerating then the equation v = d/t CANNOT be used.
As that equation refers only to objects in constant motion.

If there is a problem similar to this….
Ex. A car starts from rest to a speed of 35 m/s in a time of 8 sec. What is the car’s displacement?
You cannot use v = x/t
b/c the motion is not constant.
How to solve problems like these will be shown in the below sections

Example Problems-          will be given in class
Motion Graphs
An easy way to picture motion is through the use of graphs.
Position – Time Graphs
 graph that depicts an objects position (displacement) across time
 Position on y-axis, time on the x-axis

a)                                b)
constant positive velocity                 velocity = 0

c)                                 d)
constant negative velocity,             Increasing positive velocity, constant positive acceleration
THEN constant positive velocity

****Uniform (constant) acceleration is depicted as a parabola on a position-time graph.

How to interpret these graphs.
…..How to find position at a certain time?
find the position (y-value) that corresponds to the given time (x-value) on the graph

….. How to find velocity over a certain time?
find the slope of the graph over that time period
Slope = rise/run = how much the graph goes up divided by how much it goes over

…..How to find acceleration over a certain time?
You will not have to do this on a position time graph, however you need to be able to recognize
when there is acceleration
Example….Graph of a runner’s position..
What is his position at 3 secs?
15 m.
What is his average velocity from 2 sec to
4 sec?           V= x/t = 10/2 = 5 m/s
What is his average velocity from 1 sec. to
2 sec.?        V = 0/1 = 0 m/s
What is the instantaneous velocity at time
1.5 sec.? vel. From 1 – 2 sec is constant so
instantaneous vel. Will be equal to the
average!
What is his position of the car at 4 sec.?
20 m
*** Notice on the above graph that the object changes its velocity
What is his average velocity from 0-4 sec?
several times. The velocity changes at each of the points in the
V = d/t = 20/4 = 5 m/s
graph. In a graph of a real life situation there can be no sharps
points on a position time graph b/c the velocity cannot instantaneously change from one velocity to the other
without going through everyone in between. Therefore the graph needs to be curved, and cannot have sharp points.

Graph A and Graph B are position time graphs of a constant
acceleration. In this case, how could you find the instantaneous
velocity at some specific time? There would be no way to do it…. You
could however find the average velocity over a given time interval. In the
other graphs above the velocity was constant for a time interval so any
point within that time interval would have an instantaneous velocity equal
to the average across the time interval.
Graph A                                       Now to find the instantaneous velocity at some given point, we
must first ask…. Is the velocity increasing or decreasing as time is going
by? We should notice that the displacement per time interval is increasing
(meaning as each second goes by the object is covering more and more
displacement per second). So this of course means that the velocity is
increasing. And you can tell this by looking at the graph. The steeper and
steeper the curve gets faster the object is getting. So in other words the
steepness, or slope, of the graph corresponds to the velocity. Which we
actually already discussed, so now how do we find the slope of a specific
point? We can do this by finding the slope of the tangent line to the
curve at that point.
A tangent line is a line that intersects a curve at only one point.
Every point on a curve has its own specific tangent line.
Graph B                               The tangent lines will be getting steeper and steeper as more time goes by.
As you may notice on the three colored dots that represent points on the
curve in Graph B to the left. The tangent line to the blue dot has a much
steeper slope than the tangent line to the black dot. The slope of the
black dot’s tangent line is equal to the instantaneous velocity at that point.
As is also true for the tangent lines to the red dot and the blue dot.
Velocity –Time Graphs
Another way of depicting motion is through use of a Velocity-Time Graph
Velocity on the y-axis, time on the x-axis
Can be more useful that a P-t, if the velocity is changing

o                             constant positive acceleration

o
0-10 sec. constant positive velocity, 10-20 sec. negative acceleration

How to interpret these graphs.
…..How to find velocity at a certain time?
find the velocity (y-value) that corresponds to the given time (x-value) on the graph
….. How to find acceleration over a certain time?
find the slope of the graph over that time period
Slope = rise/run = Δy/Δx
 rise and run respectively correspond to Δv and t on the graph. and if we divide the two
that would be the equation for acceleration. So it would make sense for slope to equal this
value.
…..How to find displacement over a certain time?
**to understand           find the geometric area underneath the curve. (break area into individual shapes like rectangles
why this equation
and triangles and add all the individual areas)
is being used
 doing this would make sense because for the time period 0-10 sec the area under the curve would
question.. If Mr.         be 5 x 10, which is the velocity multiplied by the time, which is the equation for displacement when
Schober just              velocity is constant (d = vt).
turned 25, what            and it would also make sense for the time period 10-20 seconds because the equation for the
was his average
area of a triangle is ½ base x height and during this time interval the velocity is changing so we
age over the last 10
years.? Think             cannot possibly just take an arbitrary velocity and plug it into d=v t, this will only work if we instead
about what you did        use d = v avg t. And to find the average velocity during a constant acceleration we would use
here and what you
are doing to find      vavg= ½ (vf + vi) this corresponds to the ½ height (1/2h) needed for area of a triangle
So it would make sense that this ‘area underneath’ technique would work to find displacement over.
Kinematics Equations
The following equations must be used for any problem in which there is accelerated motion.
In various problems you will be given different values, you must be able to figure out how to use these values in
order to solve for a missing value
The variables that will be includes in this chapter include
x – displacement (m)         vi – initial velocity   (m/s)             vf – final velocity (m/s)
a – acceleration (m/s2)                   t – time (s)

How to solve for displacement, when given vi, a, and t…
x = vit + ½ at2
(this is derived from the acceleration equation and the average velocity during constant acceleration equation)

How to solve for vf when given vi, a and d…
(this is derived from the acceleration equation and the Avg. Velocity during constant acceleration equation)

Solving for vf , when given vi , a and t…
vf = vi + at
(Which as shown before is just a version of the acceleration equation)
And also
Solving for x, when given vf, vi, and t
x= ½ (vf + vi)t
(as alluded to before this comes from the avg. velocity during constant acc. equation, and the general equation for
velocity)

Example Problems                  will be given in class
Free Fall Acceleration

An object is considered to be in Free Fall if it is only under the influence of gravity.
Meaning that the object must have a downward acceleration of magnitude 9.8 m/s2. However we refer to
the magnitude of the Acceleration of Gravity as being
2
g = 9.8 m/s
Now these types of problems can be treated just as any other acceleration problem, except that now the ‘a’ in the
equation becomes ‘g’ or rather becomes 9.8 m/s2. BUT ALSO REMEMBER that you must also recognized your
coordinate system. And in coordinate systems where you need to consider two different directions you must
recognize that the acceleration of gravity is NEGATIVE. In these type of problems, you must enter ‘g’ as -9.8 m/s2 ,
however in others where motion is only in one direction it is not necessary to do so. If this confuses you, it is always
correct to enter ‘-9.8m/ss ‘ as ‘g’ in kinematics problems, because we commonly refer to the downward direction as
being negative. However it may only sometimes be correct to enter just ‘9.8 m/s2’ without the negative.

Free Fall can be occurring even if an object is still moving upward.
For instance if you throw a ball straight up into the air, the moment it leaves your hand it is in free fall. And it
may take the ball a few moments of travelling upward before it reaches its highest point but the entire time it is
moving upwards it is still accelerating downwards. Meaning it is getting slower and slower until it eventually stops
for a moment at the top of its path, then starts accelerating downward, every second after which its velocity is
becoming larger and larger in the negative direction. In these type of problems it is very important that you
establish and stay consistent with your coordinate system.

In all of our problems we usually ignore Air Resistance because it provides too difficult of a variable to consider in
problems that are trying to focus on one topic. So in all of our problems an object in free fall will keep accelerating
towards the Earth at -9.8 m/s2 until it eventually lands or something gets in it way. However in the real world air
resistance must be considered.

Air resistance is the tendency of air to act as friction and slow objects down. As any object is moving through the
air, it is colliding with all the air molecules, and this will inevitably slow the object down. And as it so happens the
faster an object is moving through the air, the stronger the effects of air resistance are, and if an object is left to free
fall for long enough of a time the air resistance becomes so strong that it will cancel out the effects of gravity. This
effect is occurs when you reach Terminal Velocity , which is the constant velocity an object travels at when air
resistance becomes so strong that gravity no longer accelerates you, and you are stuck travelling at a constant speed.
This would occur commonly with skydivers before they open their parachute.
Relative Velocity            Revisited
All measured velocities are relative to something else.
There is no fixed standard by which to measure an objects velocity.
Different observers can observe one object moving at different
speeds at the same time, depending on their own motion

since all velocities are dependent on the motion of the observer in
order to properly explain the velocity of an object we must say 45
m/s “relative to….” That ‘relative to’ means ‘from the perspective
of.

we generally don’t have to say this because more often than not
most speeds are expressed relative to the Earth

If we give each possible observer a letter A, B, C etc. and the
symbol ‘vbc’ means ‘the velocity of b relative to c’, the general form
to find the velocity of an object relative to any relevant observer is…
Outside subscripts
vab + vbc = vac

Middle subscripts must match
notice that on the left side of the equation the two middle subscripts (b’s) match and the order of
the subscripts on the answer match the order of the outside subscripts on the left side.
However, we could also write…
vab + -vcb = vac
Because, if you want to switch the order of the subscripts you must have a negative in front of it.

Which means in general      vbc = -vcb
The use of all of these subscripts can become rather confusing, It is many times easier to picture the
situation in your head and determine what needs to be added or subtracted. However for some situations
this can be difficult, so for this reason it is a good idea to understand how these ‘rules’ for relative velocities
work.
Example…. 2 cars on the highway and a stationary observer watching from the side of the road.
Relative to the stationary observer the top car is travelling at a speed of 45 m/s and relative to the stationary
observer the bottom car is moving with a speed of 60 m/s.

What is the velocity of the top car relative to the bottom car?
We have 3 observers, top car (t) , bottom car (b), and the person (p)
We are given vtp = 45 m/s and vbp = 60 m/s
We are looking for vtb
Both of the givens have a p but the are not matched in the middle
So we write vtp + (-vbp) = vtb
45 + (-60) = -15 m/s
What is the velocity of the bottom car relative to the top car?
Vbt =??
Vbp + (-vtp) = vbt
60 + (-45) = 15 m/s
What is the velocity of the stationary person relative to the top car?
Vpt = -vtp        vtp = 45 m/s
Vpt = -45 m/s
What is the velocity of the stationary person relative to the bottom car?
Vpb = -vbp
vpb =-60 m/s

Again it is generally easier to think of these type of problems conceptually, but for more difficult problems it is
beneficial to know how to plug it into an equation.

For instance…..     if 2 cars are headed toward each other on the highway (one moving East, the other travelling
west) will one car view the other as going faster or slower than a stationary observer would?
Faster, the other car will be approaching at a greater rate than if the observer was not moving. Since
they are going fast you know that you should add the speeds.

The same would be true if the cars were headed in opposite directions and have already passed each other.

And as we saw in the earlier example if the two cars are going in the same direction, their velocities will subtract…
because the each of the cars will appear from the other car to be going slower than if the observer was at rest.

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