# Kinematic Graphs by HC121001133815

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• pg 1
```									            Kinematic Graphs

Graphically exploring
derivatives and integrals

10/1/2012                        Dr. Sasho MacKenzie - HK 376   1
Slope
Y

(4,8)
8

(0,0)                       X
4
Slope = rise = Y = Y2 – Y1 = 8 – 0 = 8 =     2
run X      X2 – X1   4–0     4

10/1/2012                                     Dr. Sasho MacKenzie - HK 376   2
Velocity is the slope of
Displacement
Y

8
(4,8)
Displacement
(m)

(0,0)                            X
Time (s)      4

Average
Velocity = rise = D = D2 – D1 = 8 – 0 = 8 m =        2 m/s
run    t   t2 – t1   4–0      4s

10/1/2012                                     Dr. Sasho MacKenzie - HK 376   3
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

10/1/2012                           Dr. Sasho MacKenzie - HK 376   4
Instantaneous Velocity
• The average velocity over an infinitely
small time period.
• Determined using Calculus
• The derivative of displacement
• The slope of the displacement curve

10/1/2012                              Dr. Sasho MacKenzie - HK 376   5
Instantaneous Acceleration
• The average acceleration over an
infinitely small time period.
• Determined using Calculus
• The derivative of velocity
• The slope of the velocity curve

10/1/2012                           Dr. Sasho MacKenzie - HK 376   6
Average vs. Instantaneous
Y         The average velocity does not accurately
represent slope at this particular point.
8
(4,8)
Displacement
(m)

(0,0)                                    X
Time (s)           4

Average
Velocity = rise = D = D2 – D1 = 8 – 0 = 8 m = 2 m/s
run    t   t2 – t1   4–0      4s

10/1/2012                                           Dr. Sasho MacKenzie - HK 376   7
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.

10/1/2012                              Dr. Sasho MacKenzie - HK 376   8
Infinitely small time period (dt)
•Tangent line
•Instantaneous Velocity
dt
Displacement

t3

t2

t1

Time
10/1/2012                                        Dr. Sasho MacKenzie - HK 376   9
Velocity from Displacement
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.
Displacement

Velocity = 0

Velocity > 0
Velocity < 0

Time
10/1/2012                                                        Dr. Sasho MacKenzie - HK 376   10
Graph Sketching
0
Slope
Differentiation
+        Displacement           Velocity
0
0
OR

Velocity        Acceleration

0

10/1/2012                               Dr. Sasho MacKenzie - HK 376   11
Slope       Graph Sketching
0          Differentiation
+           _
Displacement           Velocity
0                   0
0
OR

Velocity       Acceleration
0

10/1/2012                                      Dr. Sasho MacKenzie - HK 376   12
Going the other way: area
under the curve
Y

4                   (2,4)

(0,0)
X
2

Area under curve = Height x Base = Y x X = 4 x 2 = 8

10/1/2012                                   Dr. Sasho MacKenzie - HK 376   13
Displacement is the Area
Under the Velocity Curve
Y

4                        (2,4)

Velocity (m/s)

(0,0)
Time (s)                   X
2

Displacement = V x t = 4 x 2 = 8 m

10/1/2012                                  Dr. Sasho MacKenzie - HK 376   14
What if velocity isn’t a straight line?
(2,4)
4
This would be
an over
estimate of
Velocity                              the area under
the curve

(0,0)
Time        2

10/1/2012                       Dr. Sasho MacKenzie - HK 376   15
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 = Vt

10/1/2012                               Dr. Sasho MacKenzie - HK 376   16
Infinitely small time periods
The area under the graph in these infinitely
small time periods are summed together.

Velocity

Time
•D = Vdt = V1t1 + V2t2 + V3t3 + ……
Instantaneous    Infinitely small
Velocity         time period
10/1/2012                                    Dr. Sasho MacKenzie - HK 376   17
Graph
Constant slope of 2               Sketching
Integration
Displacement               Velocity
0

OR

2

Velocity          Acceleration
0    t     t      t

Area under curve increases by the same amount
for each successive time period (linear increase).
10/1/2012                                            Dr. Sasho MacKenzie - HK 376   18
Graph
Exponential curve
Sketching
Integration
Displacement               Velocity
0

OR

0      t    t     t              Velocity          Acceleration

Area under curve increases by a greater amount for
each successive time period (exponential increase).
10/1/2012                                            Dr. Sasho MacKenzie - HK 376   19
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 velocity
D. Min velocity       E. Max acceleration F. Min acceleration

2

0

-1

1s             2s             3s
10/1/2012                                            Dr. Sasho MacKenzie - HK 376   20
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 velocity
D. Min velocity         E. Max acceleration      F. Min acceleration
G. Max displacement     H. Min displacement

0

10/1/2012                                                Dr. Sasho MacKenzie - HK 376   21

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