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04/20/23 Dr. Sasho MacKenzie - HK 376 1
Kinematic GraphsKinematic Graphs
Graphically exploring derivatives and integrals
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04/20/23 Dr. Sasho MacKenzie - HK 376 2
Slope
X
Y
(0,0)
(4,8)8
4
Slope = rise = Y = Y2 – Y1 = 8 – 0 = 8 = 2 run X X2 – X1 4 – 0 4
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04/20/23 Dr. Sasho MacKenzie - HK 376 3
Velocity is the slope of Displacement
X
Y
(0,0)
(4,8)8
4
AverageVelocity = rise = D = D2 – D1 = 8 – 0 = 8 m = 2 m/s
run t t2 – t1 4 – 0 4 s
Displacement(m)
Time (s)
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04/20/23 Dr. Sasho MacKenzie - HK 376 4
1. The displacement graph on the previous slide was a straight line, therefore the slope was 2 at every instant.
2. Which means the velocity at any instant is equal to the average velocity.
3. However if the graph was not straight the instantaneous velocity could not be determined from the average velocity
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Instantaneous Velocity
• The average velocity over an infinitely small time period.
• Determined using Calculus• The derivative of displacement• The slope of the displacement curve
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04/20/23 Dr. Sasho MacKenzie - HK 376 6
Instantaneous Acceleration
• The average acceleration over an infinitely small time period.
• Determined using Calculus• The derivative of velocity• The slope of the velocity curve
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04/20/23 Dr. Sasho MacKenzie - HK 376 7
Average vs. Instantaneous
AverageVelocity = rise = D = D2 – D1 = 8 – 0 = 8 m = 2 m/s
run t t2 – t1 4 – 0 4 s
X
Y
8
4
Displacement(m)
Time (s)(0,0)
(4,8)
The average velocity does not accurately represent slope at this particular point.
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04/20/23 Dr. Sasho MacKenzie - HK 376 8
Derivative
• The slope of the graph at a single point.• Slope of the line tangent to the curve.• The limiting value of D/ t as t
approaches zero.
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04/20/23 Dr. Sasho MacKenzie - HK 376 9
Infinitely small time period (dt)
dt
•Tangent line•Instantaneous Velocity
Displacem
ent
Time
t1
t2
t3
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04/20/23 Dr. Sasho MacKenzie - HK 376 10
Velocity from Displacement
Displacem
ent
Time
Velocity > 0
Velocity = 0
Velocity < 0
The graph below shows the vertical displacement of a golf ball starting immediately after it bounces off the floor and ending when it lands again.
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Graph Sketching Differentiation
0
+
0
Displacement Velocity
Velocity Acceleration
OR
Slope
0
0
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04/20/23 Dr. Sasho MacKenzie - HK 376 12
Displacement Velocity
Velocity Acceleration
OR
0
0 0
+
0
0
_
SlopeGraph
Sketching Differentiation
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04/20/23 Dr. Sasho MacKenzie - HK 376 13
Going the other way: area under the curve
Area under curve = Height x Base = Y x X = 4 x 2 = 8
Y
X
4
2(0,0)
(2,4)
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04/20/23 Dr. Sasho MacKenzie - HK 376 14
Displacement is the Area Under the Velocity Curve
Displacement = V x t = 4 x 2 = 8 m
Velocity (m/s)
Time (s)
Y
X
4
2(0,0)
(2,4)
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04/20/23 Dr. Sasho MacKenzie - HK 376 15
What if velocity isn’t a straight line?
This would be an over estimate of the area under the curve
(2,4)4
2(0,0)
Velocity
Time
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04/20/23 Dr. Sasho MacKenzie - HK 376 16
Integration
• Finding the area under a curve.• Uses infinitely small time periods.• All the areas under the infinitely
small time periods are then summed together.
• D = Vdt = Vt
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04/20/23 Dr. Sasho MacKenzie - HK 376 17
Infinitely small time periods
Velocity
Time
•D = Vdt = V1t1 + V2t2 + V3t3 + ……
InstantaneousVelocity
Infinitely smalltime period
The area under the graph in these infinitely small time periods are summed together.
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04/20/23 Dr. Sasho MacKenzie - HK 376 18
Graph Sketching Integration
Displacement Velocity
Velocity Acceleration
OR
0
2
0
Constant slope of 2
t t t
Area under curve increases by the same amount for each successive time period (linear increase).
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04/20/23 Dr. Sasho MacKenzie - HK 376 19
Graph Sketching Integration
Displacement Velocity
Velocity Acceleration
OR
0 t t t
Area under curve increases by a greater amount for each successive time period (exponential increase).
0
Exponential curve
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04/20/23 Dr. Sasho MacKenzie - HK 376 20
Indicate on the acceleration graph below the location of the following point(s). Place the letter on the graph.
A. Zero velocity B. Zero acceleration C. Max velocityD. Min velocity E. Max acceleration F. Min acceleration
0
1 s 2 s 3 s
-1
2
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04/20/23 Dr. Sasho MacKenzie - HK 376 21
Indicate on the velocity graph below the location of the following point(s). Place the letter on the graph.
A. Zero velocity B. Zero acceleration C. Max velocityD. Min velocity E. Max acceleration F. Min accelerationG. Max displacement H. Min displacement
0