Download - Kinematic Graphs
04/19/23 Dr. Sasho MacKenzie - HK 376 1
Kinematic GraphsKinematic Graphs
Graphically exploring derivatives and integrals
04/19/23 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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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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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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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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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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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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Infinitely small time period (dt)
dt
•Tangent line•Instantaneous Velocity
Displacem
ent
Time
t1
t2
t3
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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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Displacement Velocity
Velocity Acceleration
OR
0
0 0
+
0
0
_
SlopeGraph
Sketching Differentiation
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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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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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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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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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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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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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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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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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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