PES 1110 Fall 2013, Spendier
Lecture 2/Page 1
First HW assigned today - due next Wednesday
Today: first HW assigned (chapter 1 and parts of chapter 2) due next Wednesday at the
beginning of class. If you cannot make class hand it in to me earlier or e-mail me an
electronic version due at class time.
What did we do last time: (chapter 1)
- S.I. units:
Length
meter
m
mass
kilogram
kg
time
second
s
- Changing units: Conversion factor = a ratio of units that is equal to unity
- Scientific Notation x 1011
- Metric Unit Prefixes kilo (k), mega (M)
- Significant figures told you don't give me more than 3
- Dimensional analysis
We can check for error in an equation or expression by checking the dimensions.
Quantities on the opposite sides of an equal sign must have the same dimensions.
Quantities of different dimensions can be multiplied but not added together.
e.g., a proposed equation of motion, relating distance traveled (x) to the acceleration (a)
and elapsed time (t).
x=
1 2
at
2
Dimensionally, this looks like
[L] = [L] ⋅ [T ]⋅ [T ]
[T ]
[L] = [L]
2
At least, the equation is dimensionally correct; it may still be wrong on other grounds, of
course.
Motion along a straight line: Chapter 2
Our number one topic of this term is moving objects - Motion - that is going to take two
thirds of the class.
Mechanics - Study of how and why objects move.
So why and how objects move is called mechanics. This is the overall broad study of
motion. Here and in chapters 2 and 4, we are going to be doing kinematics.
Kinematics - Motion without regard to how it is caused.
Any problem were you can discuss, predict something about motion without needing to
know something about the cause of that motion. Trust me by the end of chapter 4 you
PES 1110 Fall 2013, Spendier
Lecture 2/Page 2
will have a very good idea about what this actually entails. Problem of how long does a
car take to get to a certain point. You don't need to know what a car actually makes it
speed up or slows down.
Particle Model
We will (for a very long time) be studying the mathematical representation of the motion
of a point mass. A point mass is an idealized object having no extent, nor volume, nor
internal structure - suffices to treat moving objects as particles (little dots) with a single
value of position.
When you think about this of course it is completely unrealistic - you have a car going
down the road - you have the front of the car -you have the middle of the car - you have
the back of the car. We don't want to be that realistic - yet. We just pick one point on the
car, lets say the front, the middle or the back, it does not matter it is up to you and we
represent the entirety of the car as just being a little dot. So it is only a single value of
position that we have to give in our problems for now.
One Dimensional Motion
The motion is constrained to lie along a single straight line: back & forth, up & down,
left & right, etc. The coordinate system consists of a single line, with components on
either side of an origin.
So what do we need to know in order to describe motion? Not in an every day way but
more in a scientific and precise way.
Variables of Motion
To describe motion completely, we need to know:
- Where the object is located at every time = Position
- How fast and in what direction the object is going at every time = Velocity
- Whether the object is speeding up or slowing down at every time = Acceleration
- how long – time: position, velocity, and acceleration all functions of time
We will be looking for a mathematical means of describing the relationships among these
variables—an equation of motion.
Position
How far and what direction from an origin.
Where an object is located and also what direction it is "from the origin" (is arbitrary and
can be picked – we will try to make a smart choice). In order to locate something we need
a reference point. For this we use a number line, i.e. for a problem where something id
going to the right
On the other hand if you are having something going straight up in your homework what should your picture look like? Going straight up.
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Lecture 2/Page 3
It is always useful to remember that what we actually mean here is the position HOW
FAR from zero. A better representation for the first picture is this picture with a little
arrow going from zero to the current location
There are two possible directions, so how do we indicate direction here? We will use
positive or negative values and that will be true for all three physical quantities I just told
you about, position, velocity, and acceleration.
For 1D Motion, direction is indicated by giving positive or negative values for physics
quantities. The usual convention is that
Right/up = positive
Left/down = negative
Displacement
If things move they change their position and that is where displacement comes in. Now
displacement , if you have never gone over this in physics, as little strange abbreviation is
the change in the position we don't abbreviate it by "d" we abbreviate it by "delta x"
Moving objects change their position, so we introduce displacement.
Displacement = change in position = ∆x (Delta x)
x for position, delta meaning "change in", I know it is a triangle, but is also the Greek
name for delta
Your books notation is the following:
Initial Position = x1, (where an object starts is its initial position)
Final Position = x2 (where an object stops is its final position)
So graphically the displacement, the change in position, again it is an arrow, it goes from
the initial to the final position, from x1 to x2. So remember what we mean by x1 is how far
from zero. Positions are always measured from zero. Displacements are from initial to
final.
So if you include these little arrows on the picture it becomes really obvious from the
initial and the final position -
how do I calculated the displacement?
displacement = final position - initial position
∆x = x2 - x1
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The displacement can be positive or negative because it is a vector quantity - the minus
and plus sign give direction. We will discuss vectors in more detail next week when we
study 2D motion.
Example:
x1 = 1m
x2 = -2 m
∆x = x2 - x1 = -2 m - 1 m = -3 m
x
-2
-1
0
1
Distance:
Distance is a positive number which tells you how far an object has traveled.
Distance, d = always positive number which gives how far an object has traveled.
For our example above:
d=3m
Displacement Exercise: Displacement is not the same as distance traveled!
An eagle is flying 4m above a lake when it spies a fish that is 35 cm below the surface. If
the eagle dives straight down and grabs the fish, what the eagle’s displacement? Use the
typical convention that up is positive.
Let's draw a picture to illustrate good problem solving
Make sure you write down the units - can we compare cm and m -NO - we need to
convert
 1m 
∆x = x2 - x1 = -35 cm - 4 m = -35 cm (1) - 4 m = -35 cm 
- 4 m
 100 cm 
= -0.35 m - 4 m = - 4.35 m
∆x = - 4.35 m
What is the negative sign telling me here? - 4.35 m? The eagle went downwards. Good.
Now let's try this one:
PES 1110 Fall 2013, Spendier
Lecture 2/Page 5
An eagle is flying 4m above a lake when it spies a fish that is 35 cm below the surface.
The eagle dives straight down, grabs the fish, and then flies straight back up to where it
started. For the entire trip, what the eagle’s displacement ∆x and distance d traveled? Use
the typical convention that up is positive.
∆xtotal = ∆xdown + ∆xup
∆xdown = - 4.35 m
∆xup = + 4.35 m
∆xtotal = ∆xdown + ∆xup = - 4.35 m + 4.35 m = 0 m
distance = ddown + dup = |∆xdown| + |∆xup| = 4.35 m + 4.35 m = 8.7 m
This is when displacement and distance become different - when you have these
combination problems - two motions combined.
The displacement is zero, because displacement does not care what happens in between you started at 4 m you ended at 4 m - your displacement is zero. For any round trip the
displacement is zero,.
Distance on the other hand is like what your car's odometer gives you, it is always
positive, it is actually measuring how far did you go here - makes it positive 4.25 m and
when you came back up it is still positive.
Average velocity and average speed:
Displacement/distance alone do not give us all the information about the motion. We
would like to know how long it too to cover a given distance. Hence we will now talk
about speed and velocity. The mean different things. We are very precise - normal
everyday people get to use these term interchangeably - we do not!
Speed is simply how fast
So to find the average speed of a moving object, you take the distance it has traveled you see distance, and you divide by time, the elapsed time - how long it took for this trip
total distance d
 length   m 
=
units: 
first compound unit
=
elapsed time ∆t
 time   s 
and ∆t = t2 − t1 (elapsed time)
t1.....initial time
t2.....final time
Average speed: savg =
Velocity is how fast and direction of motion
Velocity is a bit different, it is not just how fast, it also includes the direction you are
traveling in. So here which ones told you about direction, the fact that the eagle was
going downwards? Displacement! So the average velocity is defined to be the
displacement over elapsed time
Average velocity: vavg =
displacment ∆x x2 (t2 ) − x1 (t1 )
m
=
=
units:  
elapsed time ∆t
t2 − t1
s
PES 1110 Fall 2013, Spendier
Lecture 2/Page 6
Example:
A Car goes to the right for 5 s at 10m/s, immediately turns around, and goes to the left at
25m/s for 2 s. What is the cars average velocity?
vavg,right = +10 m/s
vavg,left = - 25 m/s
we know:
average speed to the right vavg,right = +10 m/s and left vavg,left = - 25 m/s
elapsed time to right ∆tright = 5s and left ∆tleft = 2s
total displacment ∆xright + ∆xleft
v avg =
=
total elapsedtime ∆tright + ∆tleft
don't know:
∆xright and ∆xleft
But from equation we see that
∆x
v avg ( ∆t ) =
( ∆t )
∆t
∆x = vavg ( ∆t )
Hence
∆xright = vavg , right ( ∆tright ) = +10 m / s(5s ) = +50 m
and
∆xleft = vavg ,left ( ∆tleft ) = −25m / s(2 s ) = −50 m
and therefore
∆xright + ∆xleft 50 m − 50 m 0 m
vavg =
=
=
= 0m / s
5s + 2 s
7s
∆tright + ∆tleft
What would be the average speed?
total distance 100 m
savg =
=
= 14.3m / s
elapsed time
7s
We see that average velocity does not cut it – we had this car that went to the right and
then it came back and went to the left, and its displacement, its delta x was zero so it
missed all of this information about what happened in between. So while the average
velocity is a lovely little quantity, it is not quite good enough physics because it only tells
you what happens on average.
What we need is a way to find an instantaneous velocity, which is just v
Instantaneous velocity v how fast and the direction of motion for one instant of time.
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For example:
At t = 0.03 s how fast was the object going and in which directions was it going?
The way we are going to do this, because not everyone is on the same page in calculus
yet, is that we are going to stress the idea of making graphs.
Motion Graphs
Physicists like to make graphs to describe motion
What is a position versus time graph?
What is a velocity versus time graph?
Horizontal motion
Uniform Motion - constant velocity
Equal spacing between dots because with constant velocity the object travels the same
distance during equal elapsed times.
So what does the position versus time graph look like here? Straight line – yes
This distance corresponds to the elapsed time ∆t = t2-t1 and on the other side this distance
corresponds to the displacement ∆x = x2-x1.
∆x rise
v avg =
=
∆t run
Here, when something does not speed up or slow down, the average and the
instantaneous velocity are always the same.
Changing velocity
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Lecture 2/Page 8
When we have changing velocity (velocity is not constant) the position versus time is
now a curve.
How come? When something accelerates what does it tell you about the distance traveled
with time? It increases or decreases depending on where the object is speeding up or
slowing down.
Example: speeding up
How can I get the instantaneous velocity from this graph? The instantaneous velocity at
one instant of time is the slope of the tangent line.
Now we pick some tangent line at a given time. Look at a little region around time t
greatly magnify the curve – what does it look like? It looks like a straight line
To make this 100% correct, what magnification do we have to take here – an infinite
amount of magnification. In order to make your magnification infinite you got to make
your time interval as closely to zero as mathematically even possible. So that is what a
derivative actually is – it is this magnification process made infinite
Instantaneous velocity
∆x dx
v = lim
=
∆t → 0 ∆ t
dt
is the time derivative of position
Speeding up or slowing down:
PES 1110 Fall 2013, Spendier
speeding up - slope at a given time is increasing
slowing down - slope at a given time is decreasing
Example:
A particle's position is described by
x(t) = t2 -2 t +8
a) find the velocity v(t)
b) When does the particle turn around?
a)
v(t ) =
dx d 2
= ( t − 2 t + 8 ) = 2t − 2
dt dt
b) particle turns around when velocity is zero:
v(t) = 0 = 2t - 2
t=1s
The particle turns around at 1 s
Lecture 2/Page 9