Unit 1: Linear Motion

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Unit 1 Linear Motion

• Recall from Physics I that there are four primary
  variables used in the study of kinematics
• Displacement, x (meters)
• Time, t (seconds)
• Velocity, v (m/s)
• Acceleration, a (m/s2)
• In this unit we will examine each of these in
  detail, derive the expressions for their
  interrelationships, and use them to solve systems
  that involve motion along a line.

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Position and Displacement

• The displacement of an object determines its net
  position from a reference point.
• The reference point is arbitrary and can be
  selected to simply the system or break a complex
  system into parts.

Position at time 2, or final position
Displacement
Position at time 1, or initial position
Often we may simply write x, rather than ?x
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Graphing x vs t

• It is a relatively simple matter to create a
  graph of displacement versus time for an object
  in motion along a straight line.
• To do so, the objects position (x) with respect
  to the reference point (often the starting point)
  needs to be know for each increment of time (t).
  With several ordered pairs (x,t) a graph can be
  easily constructed.

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Remember When you construct a graph it is
imperative that you label both the variable and
the unit of that quantity on each axis.
This graph represents a particle moving along a
straight line. Describe the motion of that
particle.
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Velocity

• By definition, the speed of an object describes
  how far it moves in a given quantity of time.
  Speed does not imply any direction.
• Velocity is the vector quantity for the scalar
  speed. It describes the displacement of an
  object in a given amount of time.
• Whenever you describe the change in one variable
  with respect to the change in another variable
  this is called a rate of change.

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• If we take the system we graphed before (a
  particle moving along a line), we already have
  the information we need to find the velocity.
  Lets focus only on the first two seconds for now.

What part of the graph do we look at to find
displacement? To find time? Recall that
velocity is the rate of change of displacement.
What part of the graph describes the change in
displacement versus the change in time?
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Slope

• The slope of any graph describes the rate of
  change of the independent variable to the
  dependent variable. In this case, velocity is
  the rate of change of displacement with respect
  to time, or the slope of the x vs t graph.
  Therefore
• Sometimes we may write this as v x / t,
  understanding that we mean to find the average
  velocity in a time period.

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Instantaneous vs. Average

• In the example we just looked at, from 0s to 2s,
  the particle had a displacement of 2m, resulting
  in an average velocity of 1 m/s. Notice how the
  slope of the graph (which gave us the velocity)
  is constant. That means that at every instant in
  those two seconds, this particle was traveling a
  constant 1 m/s. Now this is not terribly
  exciting, or very real. After all, most movement
  involves speeding up and slowing down.

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• To find the instantaneous velocity of an object
  that is increasing or decreasing its speed
• another way to say this would be an object
  that has a changing rate of change in
  displacement
• we will employ a calculus tool called a
  derivative.

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• The slope of the displacement function for any
  time period t2 t1 yields the average velocity.
• The derivative, or instantaneous slope, of the
  displacement function yields the instantaneous
  velocity of the object at any time t.

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Acceleration

• Acceleration is the rate of change of velocity.
  This means that acceleration is to velocity what
  velocity is to displacement. I.e., the
  derivative of a function describing the velocity
  of an object tells you the acceleration of that
  object at time t.
• Acceleration is a rate of change of a rate of
  change or a second derivative ? All this means
  is that the differential operation is performed
  twice on the same function.

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For example
For physics, we normally wouldnt be interested
in a function of x, but rather we would have a
function of time.
In this case, since the second derivative is a
constant, acceleration would be constant in
this example.
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• In math we generally have functions of the form y
  x
• In physics, our displacement function would be of
  the form x At2 Bt C, where x is the
  displacement and t represents time.
• Therefore, to find the velocity function of this
  object we would find
• the 1st derivative of displacement.
• To find the acceleration of this object we would
  have to find or the second derivative of
  the displacement function.

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Definitions and Relationships
For any position function x(t)
This is the same as saying
Velocity is the slope/derivative of the position
function
and
Acceleration is the slope/derivative of the
velocity function
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Approaching from the other direction
Given an acceleration function a(t)
This is the same as saying
Velocity is the anti-derivative of acceleration.
The average velocity over a period of time is
equal to the amount of area under the
acceleration graph.
and
Displacement is the anti-derivative of velocity.
The displacement during a period of time is
equal to the amount of area under the
velocity graph graph. Distance can be
determined from this information as well.
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Equations of Motion

• Last year I promised that in AP we would derive
  the equations of motion from the fundamental
  definitions. Well, here we are
• BUT, remember the assumption on which all these
  relationships are based
• The kinematics relationships only apply under
  conditions of constant acceleration!

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The kinematics equations

• Equations of displacement velocity

Good
Meh
Very generally
Taking into account the initial position
With a little algebra
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x vt x0

• Notice that this particular equation is missing
  the variable a (acceleration). This implies that
  the value of acceleration is zero. Be careful of
  this nuance when using this equation.
• Notice the form of the equation. Compare it to
  the standard form of a line, y mxb. Notice
  that we have an equation for displacement in
  terms of time. The velocity variable has the
  same relationship in this formula as m does to x
  in the basic mathematical expressionfurther
  proof that the derivative of displacement yields
  the velocity of an object.
• Also, the y-intercept In the physics equation
  this is x0 or the starting position of the object
  with respect to some reference point.

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And then

• An acceleration term can be added to the above
  relationship that will allow it to be used in all
  cases of motion as long as the acceleration
  remains constant.
• REMEMBER Constant acceleration is an assumption
  for all of these relationships we are deriving!!!
• When adding an acceleration term, two pieces of
  information are essential

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• The value of acceleration is given by taking the
  second derivative of the displacement function.
• The second derivative must be a constant.
• We need to consider an acceleration term that
  will yield a constant value upon taking the
  second derivative with respect to time. This
  implies the presence a t2 term since the variable
  t will still be present after a single
  differentiation.

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• Basically we need a term that will yield just the
  variable a (the value of acceleration) upon
  second differentiation. How do we take the
  differentiate a t2 term to give us a
  non-multiplied constant upon taking the 2nd
  derivative? We need a constant coefficient out
  front to cancel out the power factor from the
  differentiation.

We dont want 2a
Just a
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• So, adding a term to x vt x0 to account for
  constant acceleration we end up with

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Equations of Acceleration

• Recall Average Acceleration is the slope of a
  velocity graph. Instantaneous acceleration is
  the second derivative of the displacement
  function (also the 1st derivative of the velocity
  function).

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Notice how this equation follows the same basic
form as the displacement/velocity relationship.
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Now, my favorite
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So, given any combination of variables, x,v,a,
t, we can basically solve for any of the others,
provided that we know enough to begin with.
REMEMBER These ALWAYS
assume CONSTANT acceleration!
assumes a 0
doesnt need vfinal
doesnt need x
doesnt need t
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Interpretation of Motion Graphs

• You need to be able to
• Determine direction of movement/sign of velocity
  (when possible)
• Determine sign of acceleration
• Be able to glean info like exact position, slope
  (rate of change), and inflection (type of change)
  from a given graph.

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Describe the behavior of the particle during
each time interval
x
t
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Looks familiar, right? Is it the same? Describe
the behavior of the particle during each time
interval.
V
t
Can this graph be used to determine distance
displacement? How?
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x
Slope or Derivative
Area under graph (Integral)
v
Slope or Derivative
Area under graph (Integral)
a
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Velocity
Acceleration
Description
- -
- -
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Velocity or - ? Speeding up or slowing down?
Describe
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Velocity or - ? Are any intervals
important? Speeding up or slowing down?
Describe
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Velocity or - ? Can we know from this graph?
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Velocity or - ? What does a curved shape on a
displacement graph imply?
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What does a curve on a velocity graph imply?
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Compare x(t) and v(t)
v has a value
dx/dt positive
dx/dt negative
v zero
v has a negative value
dx/dt zero