graphical analysis of motion

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GRAPHICAL ANALYSIS OF MOTION

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motion graphs come in three formats

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motion graphs come in three formatsA. Position vs. time graph

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B. Velocity vs. time graph

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C. Acceleration vs. time graph

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C. Acceleration vs. time graph ( this graph is of little use in our course)

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C. Acceleration vs. time graph ( this graph is of little use in our course)

Each graph is a function of time

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all formats use a Cartesian System for graphing 6

5position

(m) 4

3

2

1

0 0 1 2 3 4 5 6 7time (sec)

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all formats use a Cartesian System for graphing

time is always on the X axis (abscissa) independent variable

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all formats use a Cartesian System for graphing

time is always on the X axis (abscissa) independent variable

position, velocity, and acceleration are always on the Y axis (ordinate)

dependent variable

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Position vs. Time Graph6

5position

(m) 4

3

2

1

0 0 1 2 3 4 5 6 7time (sec)

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Velocity vs. Time Graph6

5velocity( m/s) 4

3

2

1

0 0 1 2 3 4 5 6 7time (sec)

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Position vs. Time Graph

This is a graph of position vs. t for an object moving with constant velocity. The velocity is the slope of the y-t curve.

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Position vs. Time Graph

Information provideda. Measured Position at a specific timeb. Estimated Position between two known pointsc. Slope of line gives average velocity between any two Cartesian pointsd. Slope of line can determine instantaneous velocity

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One-Dimensional Displacement and Velocity: Vector Quantities

Different ways of visualizing uniform velocity:

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This object’s velocity is not uniform. a. the average velocity between any two

points can be found by finding the slope of a line drawn between any two points

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This object’s velocity is not uniform. a. The average velocity between any two


b. The red line indicates a positive slope and therefore a positive velocity

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b. The red line indicates a negative slope and therefore a negative velocity

• the negative indicates direction• the object may be moving towards the

origin

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b. The red line is horizontal between these two points

• The average velocity is zero during this time period

• the object the object is moving, but its position is unchanged between points 2 & 5

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b. The red line is tangent to a point on the curved line

c. This tangent line indicates an instantaneous velocity at a particular time

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b. The red line is horizontal to a point on the curved line

c. This tangent line indicates the instantaneous velocity is zero at this particular time

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Graphical Analysis of Linear MotionOn the left we have a graph of velocity vs. time for an object with varying velocity; on the right we have the resulting position vs. time curve. The instantaneous velocity is tangent to the curve at each point.

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5position

(m) 4

3

2

1

0 0 1 2 3 4 5 6 7time (sec)

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5position

(m) 4

3

2

1

0 0 1 2 3 4 5 6 7time (sec)

Find the position of this object at times 0.0, 2.0 sec, 5.0 sec, 5.5 sec

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5position

(m) 4

3

2

1

0 0 1 2 3 4 5 6 7time (sec)

Find the slope and velocity of this object

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5position

(m) 4

3

2

1

0 0 1 2 3 4 5 6 7time (sec)

Pick two points on the line

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5position

(m) 4

3

2

1

0 0 1 2 3 4 5 6 7time (sec)

Determine the Cartesian point values of the two points

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5 (5.9, 4.9)position

(m) 4

3

2 ( 1.9,2.0)

1

0 0 1 2 3 4 5 6 7time (sec)

Determine the Cartesian point values of the two points

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Position vs. Time Graph• Finding Velocity on a position vs. time graph when object

has a constant velocity• Draw straight line between the two points being analyzed

(1.9 sec and 5.9sec)• Determine the slope of the line• Slope = rise/run = ΔY/ ΔX

= (4.9- 2.0)m/(5.9- 1.9)sec = 2.9m/4.0sec = 0.725m/s = 0.73m/s

The average velocity between times 1.9 & 5.9 sec is 0.73m/s

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Position vs. Time Graphof an object with a variable velocity

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5position

(m) 4

3

2

1

0 0 1 2 3 4 5 6 7time (sec)

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6

5position

(m) 4

3

2 Find average velocity of object between 0.0 and 4.0 sec

1

0 0 1 2 3 4 5 6 7time (sec)

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6

5position

(m) 4

3


1

0 0 1 2 3 4 5 6 7time (sec)

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6 (4.0, 5.5)5

position(m) 4

3 ( 0.0,3.0)


1

0 0 1 2 3 4 5 6 7time (sec)

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• Finding Average Velocity on a position vs. time graph• Draw straight line between the two points being analyzed • Determine the slope of the line• Slope = average velocity of the object• Slope = rise/run = ΔY/ ΔX

= (5.5- 3.0)m/(4.0- 0.0)sec = 2.5m/4.0sec = 0.625m/s = 0.63 m/s

Average velocity at between 0.0 sec and 4.0 sec is 0.63m/s

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6

5position

(m) 4

3

2 Find instantanious velocity of object at 2.0 sec

1

0 0 1 2 3 4 5 6 7time (sec)

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6

5position

(m) 4

3

2 Find instantanious velocity of object at 2.0 sec place tangent line at 2.0 sec point1

0 0 1 2 3 4 5 6 7time (sec)

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6

5position

(m) 4

3

2 Find instantanious velocity of object at 2.0 sec find two points on the tangent line1

0 0 1 2 3 4 5 6 7time (sec)

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6 (3.5, 6.2)

5position

(m) 4(3.8, 0.0)3

• Find instantanious velocity of object at 2.0 sec find two points on the tangent linelabel the points1

0 0 1 2 3 4 5 6 7time (sec)

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6 (3.5, 6.2)

5position

(m) 4(0.0, 3.8)3

• Find instantanious velocity of object at 2.0 sec find two points on the tangent linelabel the points1 determine slope of tangent line

0 0 1 2 3 4 5 6 7time (sec)

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Position vs. Time graph• Find instantaneous velocity of an object with variable

velocity• Find slope of the tangent line• tangent = slope ( 0.0s,3.8m) & (3.5s,6.2m)• find the slope of the tangent line• Slope = rise/run = ΔY/ ΔX

= (6.2-3.8)m/(3.5-0.0)sec = 2.4m/3.5sec = 0.6857m/s = 0.69 m/s

Instantaneous velocity at 2.0 sec = 0.69m/s

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6

5position

(m) 4

3

2 Find instantaneous velocity of object at 5.5 sec

1

0 0 1 2 3 4 5 6 7time (sec)

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6

5position

(m) 4

3

2 Find instantaneous velocity of object at 5.5 sec

1

0 0 1 2 3 4 5 6 7time (sec)

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6

5position

(m) 4

3

2 Find instantaneous velocity of object at 5.5 secplace a tangent line at this point1

0 0 1 2 3 4 5 6 7time (sec)

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6

5position

(m) 4 (5.0,4.0)

3 (6.2,3.0)

2 Find instantaneous velocity of object at 5.5 secdetermine two points on the tangent line1

0 0 1 2 3 4 5 6 7time (sec)

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• Find slope of the tangent line• tangent = slope ( 5.0s,4.0m) (6.2s,3.0m)• find the slope of the tangent line• Slope = rise/run = ΔY/ ΔX

= (3.0-4.0)m/(6.2-5.0)sec = -1.0m/1.2sec = -0.833m/s = -0.83 m/s

Instantaneous velocity at 5.5 sec = -0.83 m/s

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Position vs. Time graph

Instantaneous velocity at 5.5 sec = -0.83 m/s

The negative sign indicates the direction the object is moving.

The object is moving in the negative X direction

The object is not moving at a speed less than zero

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Graphical Analysis of Linear MotionOn the left we have a graph of velocity vs. time for an object with varying velocity; on the right we have the resulting position vs. time curve. The instantaneous velocity is tangent to the curve at each point.

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5velocity( m/s) 4

3

2

1

0 0 1 2 3 4 5 6 7time (sec)

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5Velocity( m/s) 4

3

2

1

0 0 1 2 3 4 5 6 7-1 time (sec)

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Velocity vs. Time Graph

• Information That can be obtained from a velocity time graph

a. Measured Velocity at a specific timeb. Estimated velocity between two known pointsc. Slope of line gives average acceleration between two

pointsd. Slope of line can give instantaneous acceleratione. Area under the curve can determine displacement of

object

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6 Find the velocity at times 1.0sec, 2.0sec, 4.5 sec, 7.0 sec

Velocity 5( m/s)

4

3

2

10 0 1 2 3 4 5 6 7-1 time (sec)

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Velocity vs. Time Graph6 Find average acceleration from

a. 0.0 sec to 2.5sec5 b. 0.0 sec to 5.0 sec

c. 2.5 sec to 5.0 sec4 d. 3.5 sec to 7.0 sec

Velocity 3( m/s)

2

1

0 0 1 2 3 4 5 6 7time (sec)

-1

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Velocity 3( m/s)

2 (2.5, 2.2)

1

(0,0) 0 0 1 2 3 4 5 6 7time (sec)

-1

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Average acceleration from 0.0 sec to 2.5 sec

aver velocity = = (2.2-0)m/2.5 sec

= + 0.88m/s2

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Velocity 3( m/s)

2 (5.0,1.6)

1

(0,0) 0 0 1 2 3 4 5 6 7time (sec)

-1

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aver acceleration = = (1.6-0)m/s/5.0 s

= + 0.320m/s2 = + 0.32 m/s2

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Velocity 3( m/s) (2.5, 2.1)

2 (5.0, 1.6)

1

0 0 1 2 3 4 5 6 7time (sec)

-1

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aver acceleration = =

= (1.6-2.10)m/s/(5.0 -2.5)sec = -0.50m/s/2.5s = - 0.20m/s2

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Velocity 3( m/s)

2

1

0 0 1 2 3 4 5 6 7time (sec)

-1

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• Average velocity from 3.5 seconds to 7.0 seconds -1.2857m/s2 = -1.3 m/s2

In the segment from 3.5 s to 7.0 s the object is slowing until it reaches zero velocity.

Its acceleration is negative After the graph crosses into the negative Y segment it gains speed in the

negative direction (going backwards) its acceleration is negative its speed is increasing its velocity is growing increasingly negative

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The area under the curve has the units velocity X time

Area = m/s X sec = meters ( length)

Area under the curve = the displacement of the object in the velocity time graph

graphical analysis of motion

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