Accelerated Motion

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Accelerated Motion

• Chapter 3

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Chapter Objectives

• Describe accelerated motion
• Use graphs and equations to solve problems
  involving moving objects
• Describe the motion of objects in free fall.

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Section 3.1 Acceleration

• Define acceleration
• Relate velocity and acceleration to the motion of
  an object
• Create velocity-time graphs

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Uniform Motion Nonuniform Motion
Moving at a constant velocity If you close your eyes, you feel as though you are not moving at all Moving while changing velocity Can be changing the rate or the direction You feel like you are being pushed or pulled
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Changing Velocity

• Consider the following motion (particle model)
  diagram

Not moving
Constant Velocity
Increasing Velocity
Decreasing Velocity
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Changing velocity

• You can indicate change in velocity by
• the motion diagram spacing
• the magnitude (length) of the velocity vectors.
• If the object speeds up, each subsequent velocity
  vector is longer.
• If the object slows down, each vector is shorter
  than the previous one.

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Velocity-Time Graphs
Distance being covered is longer, thus the runner
is speeding up.
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Velocity-Time Graphs
Time (s) Velocity (m/s)
0 0
1 5
2 10
3 15
4 20
5 25
Area???
Slope???
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Velocity-Time Graphs

• Analyze the units
• Slope rise over run
• m ?y / ?x
• Slope m/s/ s m/s2
• m/s2 is the unit for acceleration
• Area ½ b h
• b s m/s m
• m is the unit for displacement
• The slope of a velocity-time graph is the
  ACCELERATION and the area is DISPLACEMENT.

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Slope Acceleration
Area Displacement
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Velocity Time Graphs

• How Fast something is moving at a given time?
• Average Acceleration
• Use the information on the x y axis to plug
  into the equation
• a ?v / t
• Instantaneous Acceleration
• Find the slope of the line (straight line)
• Find the slope of the tangent (curve)
• Displacement
• Find the area under the curve
• You do not know the initial or final position of
  the runner, just the displacement.

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Velocity-Time Graphs

• A
• Constant velocity
• Zero Acceleration
• Positive displacement

• B
• Constant Acceleration
• Starts from Rest
• Positive displacement

• Describe the motion of each sprinter.

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Velocity-Time Graphs

• D
• Constant Acceleration
• Positive Acceleration
• Comes to a Stop
• Zero displacement

• E
• Constant Velocity
• Zero Acceleration
• Negative displacement

• C
• Constant Acceleration
• Negative Acceleration
• Comes to a Stop
• Positive displacement

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Sample Question

• On the basis of the velocity-time graph of a car
  moving up a hill, as shown on the right,
  determine the average acceleration of the car?

A. 0.5 m/s2 B. -0.5 m/s2
C. 2 m/s2 D. -2 m/s2
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Acceleration

• The rate at which an objects velocity changes
• Variable a
• Units m/s2
• It is the change in velocity which measures the
  change in position. Thus it is measuring a
  change of a change, hence why the square time
  unit.
• When the velocity of an object changes at a
  constant rate, it has constant acceleration

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Motion Diagrams Acceleration

• In order for a motion diagram to display a full
  picture of an objects movement, it should
  contain information about acceleration by
  including average acceleration vectors.
• The vectors are average acceleration vectors
  because motion diagrams display the object at
  equal time INTERVALS (intervals always mean
  average)
• Average acceleration vectors are found by
  subtracting two consecutive velocity vectors.

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Average Acceleration Vectors
You will have
?v vf - vi vf (-vi). Then divide by the
time interval, ?t. The time interval, ?t, is 1 s.
This vector, (vf - vi)/1 s, shown in violet, is
the average acceleration during that time
interval.
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Average Acceleration Vectors

• vi velocity at the beginning of a chosen time
  interval
• vf velocity at the end of a chosen time
  interval.
• ?v change in velocity

Acceleration is equal to the change in velocity
over the time interval Since the time
interval is 1s, the acceleration is equal to the
change in velocity Anything divided by 1 is
equal to itself
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Average vs. Instantaneous Acceleration

• Average Acceleration
• Change in velocity during some measurable time
  interval divided by the time interval
• Found by plugging into the equation
• a ?v / t
• Instantaneous Acceleration
• Change in velocity at an instant of time
• Found by calculating the slope of a velocity-time
  graph at that instant

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Velocity Acceleration

• How would you describe the sprinters velocity
  and acceleration as shown on the graph?

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Velocity Acceleration

• Sprinters velocity starts at zero
• Velocity increases rapidly for the first four
  seconds until reaching about 10 m/s
• Velocity remains almost constant

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Average vs. Instantaneous Acceleration

• What is the acceleration for the first four
  seconds?
• Refers to average acceleration because there is a
  time interval
• Solve using the equation a ?v /t
• vi 0 m/s vf 11 m/s t 4s
• a (11m/s 0 m/s)/ 4s
• a 2.75 m/s2

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• What is the acceleration at 5s?
• Refers to instantaneous acceleration because it
  is looking for acceleration at an instant
• Need to find the slope of the line to solve for
  acceleration
• Slope is zero thus instantaneous acceleration is
  zero at the instant of 5s.

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Instantaneous Acceleration

• Solve for the acceleration at 1.0 s
• Draw a tangent to the curve at t 1s
• The slope of the tangent
  is equal to the
  instantaneous
  acceleration at 1s.
• a rise / run

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Instantaneous Acceleration

• The slope of the line at 1.0 s is equal to the
  acceleration at that instant .

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Positive Negative Acceleration

• These four motion diagrams represent the four
  different possible ways to move along a straight
  line with constant acceleration.

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• Object is moving in the positive direction
• Displacement is positive
• Thus, velocity is positive
• Object is getting faster
• Acceleration is positive

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• Object is moving in the positive direction
• Displacement is positive
• Thus, velocity is positive
• Object is getting slower
• Acceleration is negative

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• Object is moving in the negative direction
• Displacement is negative
• Thus, velocity is negative
• Object is getting faster
• Acceleration is negative

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• Object is moving in the negative direction
• Displacement is negative
• Thus, velocity is negative
• Object is getting slower
• Acceleration is positive

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Positive Negative Acceleration

• When the velocity vector and acceleration vector
  point in the SAME direction, the object is
  INCREASING SPEED
• When the velocity vector and acceleration vector
  point in the OPPOSITE direction, the object is
  DECREASING SPEED

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Displacement Velocity always have the same sign
Displacement Velocity Acceleration Speeding UP Or Slowing Down
UP
-
- Down
- UP
Up same Down Different
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Sample Question

• How can the instantaneous acceleration of an
  object with varying acceleration be calculated?

A. by calculating the slope of the tangent on a
distance versus time graph B. by calculating the
area under the graph on a distance versus time
graph C. by calculating the area under the graph
on a velocity versus time graph D. by calculating
the slope of the tangent on a velocity versus
time graph
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Practice v-t graph
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Segment t (s) vi (m/s) vf (m/s) ?v (m/s) avg. a (m/s2) ins. A (m/s2) Xi (m) Xf (m) ?X (m)
A             0     
B                  
C                  
D                  
E                  
Can not assume position on graph. Velocity
time graphs can only be used to figure out
displacement. You must be given an initial
position.
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3.2 Motion with Constant Acceleration

• Interpret position-time graphs for motion with
  constant acceleration
• Determine mathematical relationships among
  position, velocity, acceleration, and time
• Apply graphical and mathematical relationships to
  solve problems related to constant acceleration.

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Constant acceleration x-t Graphs

• Velocity is constantly increasing, which means
  more displacement.
• Results in a curve that is parabolic.

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Constant acceleration x-t Graphs
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Kinematics Equations

• Three equations that relate position, velocity,
  acceleration, and time.
• First two are derived from a v-t graph and the
  third is a substitution.
• Total of five different variables.
• ?x (displacement), vi (initial velocity), vf
  (final velocity), a (acceleration), and t (time).
  
• Must know any three in order to solve for the
  other two.

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First Kinematics Equation

• Remember that the slope of a v-t graph is the
  average acceleration.
• Rearranging the equation, gives us the first
  kinematics equation.

Replace tf ti with t
vf vi at
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Second Kinematics Equation
We remember that area of a v-t graph equals
displacement

• Break into two known shapes (rectangle
  triangle).
• Area Area of rectangle area of triangle
• ?x vit ½ (vf vi)t

vf vi at (substitute)
?x vit ½ at2
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Third Kinematics Equation

• First equation substituted into the second to
  cancel out the time variable.
• vf vi at t (vf vi) / a
• ?x vit ½ at2
• ?x vi((vf vi)/a) ½ a ((vf vi)/a) 2
• ?x vivf vi2 ½ a (vf2 2 vivf vi2 )/a2
• 2a ?x 2 vivf - 2 vi2 vf2 2 vivf vi2
• 2a ?x - vi2 vf2 (rearrange)

Simplify
Multiply by 2a to get rid of fraction
Combine like terms
vf2 vi2 2a ?x