PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to...
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![Page 1: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/1.jpg)
PSC 151Laboratory Activity 3
Graphical Analysis IIA Nonlinear Graphs 1
andThe Acceleration Due to Gravity
![Page 2: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/2.jpg)
Graphical Analysis ExerciseDetermining the Relationship between
Circumference and Diameter
Procedure:
1. Measure the Circumference and diameter of five circular objects.
2. Analyze data using graphical analysis.
diameter, cm Circumference, cmDATA
53.1
2.14.88.811.517
7.515.428.336.2
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DIAMETER, cm
CIRCUMFERENCE vs. DIAMETER
2468
10
20
30
40
50
60
70
80
Plot a graph of Circumference versus diameter.
![Page 4: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/4.jpg)
1. Is your graph a straight line?
CALCULATIONS AND OBSERVATIONS:
YES
2. Does the graph pass through the origin? YES…b = 0
3. Are circumference and diameter directly proportional? YES
4. Calculate the slope; Points Used:
(4.8cm,15.4cm) & (11.5cm,36.2cm)
slope =m=ΔYΔX =ΔC
Δd =36.2cm −15.4cm11.5cm−4.8cm =
20.8cm6.7cm =3.1
Slope has NO units
![Page 5: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/5.jpg)
Y = m⋅X+b
What is the equation relating Circumference and diameter?
C d3.1⋅ 0+
C=3.1⋅dCompare slope = 3.1 to 314)
%error=experimental value-accepted value
accepted value×100%
% error =3.1 −3.14
3.14×100% = .04
3.14× 100% =1.3%
![Page 6: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/6.jpg)
Non-Linear Graphs
![Page 7: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/7.jpg)
What procedure do we follow if our graph is not a straight line?
Consider an experiment designed to investigate the motion of an object.
We want to determine the relationship between the object’s distance traveled and time.
We measure its distance each second for 10s.
Here is the resulting data.
![Page 8: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/8.jpg)
time-t, s distance-d, m0 51 9.92 24.63 49.14 83.45 127.56 181.47 245.18 318.69 401.910 495
Data
We then plot a graph of distance versus time.
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0
50
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150
200
250
300
350
400
450
500
0 1 2 3 4 5 6 7 8 9 10 11
time, s
Distance versus Time
Not a straight line but is a uniform curve
![Page 10: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/10.jpg)
Compare graph to graphs of other functions of the
independent variable
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0
1
2
3
4
5
6
7
Y
0 1 2 3 4 5 6 7
X
y=x
Y =mX+b
Y ∝ X
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0
10
20
30
40
Y
0 1 2 3 4 5 6 7
X
y=x2
Y ∝ X2
Y =mX2 +b
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0
0.5
1
1.5
2
2.5
Y
0 2 4 6 8
X
y = x1/2
Y ∝ X
Y =m X +b
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0
0.25
0.5
0.75
1
1.25
Y
0 1 2 3 4 5 6 7
X
y = 1/x
Y ∝ 1X
Y =m 1X +b
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0
0.25
0.5
0.75
1
1.25
Y
0 1 2 3 4 5 6 7
X
y = 1/x 2
Y ∝ 1X2
Y =m 1X2 +b
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0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
Y
0 1 2 3 4 5 6 7
X
y = 1/x 1/2
Y ∝ 1X
Y =m 1X +b
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0
1
2
3
4
5
6
7
Y
0 1 2 3 4 5 6 7
X
y=x
Y =mX+b
Y ∝ X
0
10
20
30
40
Y
0 1 2 3 4 5 6 7
X
y=x2
Y ∝ X2
Y =mX2 +b
0
0.5
1
1.5
2
2.5
Y
0 2 4 6 8
X
y = x1/2
Y ∝ X
Y =m X +b
0
0.25
0.5
0.75
1
1.25
Y
0 1 2 3 4 5 6 7
X
y = 1/x
Y ∝ 1X
Y =m 1X +b
0
0.25
0.5
0.75
1
1.25
Y
0 1 2 3 4 5 6 7
X
y = 1/x 2
Y ∝ 1X2
Y =m 1X2 +b
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
Y
0 1 2 3 4 5 6 7
X
y = 1/x 1/2
Y ∝ 1X
Y =m 1X +b
¸
0
50
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200
250
300
350
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450
500
0 1 2 3 4 5 6 7 8 9 10 11
time, s
Distance versus Time
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Plot a new graph where time squared is the independent variable:
Distance, d versus Time Squared, t2
![Page 19: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/19.jpg)
time-t,s distance-d,m0 51 9.92 24.63 49.14 83.45 127.56 181.47 245.18 318.69 401.910 495
Revised Data Table
0149
162536496481
100
time2-t2, s2
![Page 20: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/20.jpg)
![Page 21: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/21.jpg)
Y=m⋅X +b
Analysis of Graph
distance timesquared
d t2
d=m⋅t2 +b
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Slope Calculation :m = ΔY
ΔX = ΔdΔt2
Points Chosen :
(4s2 , 24.6m) and (64s2 ,318.6m)
m = 318.6m −24.6m64s2 −4s2 =294m
60s2
m =4.9 ms2
With units of m/s2 the slope represents the acceleration of the object.
d=4.9 ms2 ⋅t2 +b
![Page 23: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/23.jpg)
intercept,bIt will be difficult to
determine the intercept from the graph!
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Two Other Methods for Determining the Intercept
1. The intercept is the value of the dependent variable where the graph intersects the vertical axis. At this point the value of the independent variable is zero.
Look at the data table to determine the value of d where t2 equals zero. t2 =0⇒ b=5m2. Start with the partial equation:
Solve for “b”:
d =4.9 ms2 ⋅t2 +b
b =d−4.9 ms2 ⋅t2
Choose any data pair and substitute the values of “d” and “t2” into the equation for “b”: (25s2, 127.5m)
b =127.5m−4.9 ms2 ⋅25s2 b =5m
![Page 25: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/25.jpg)
Final Equation
d=4.9ms2 ⋅t2 +5m
![Page 26: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/26.jpg)
Graphical Analysis of “Free-Fall” Motion
Determining the Acceleration Due to Gravity
![Page 27: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/27.jpg)
Purpose:In this lab, you will determine the correct description of free-fall motion and to measure the value of the
acceleration due to gravity, g.
Introduction: The Greek natural philosopher Aristotle was one of the first to attempt a “natural” description of an object undergoing free-fall motion. Aristotle believed that objects moved according to their composition of four elements, earth, water, air, and fire. Each of these elements had a natural position with earth at the bottom, then water, then air, and fire at the top. If a rock, composed primarily of earth, was held in the air and then released its composition would cause it to return to the earth. Accordingly, Aristotle thought that objects fell with a constant speed which was proportional to the object's weight, that is, a heavier object would fall faster than a lighter one.
![Page 28: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/28.jpg)
Motion at a constant speed can be described by the equation:
Comparing the equation above with the slope-intercept equation of a straight line, Y = mX + b,
where d is the distance fallen, v is the speed, t is the time the object has been falling, and d1 is the initial distance from the origin.
we see that a graph of distance fallen versus time should be a straight line with di as the y-intercept, and the slope of the line would give the speed, v, at which the object was falling.
d =v⋅t + di
d = v⋅ t + di
Y =m⋅X + b
![Page 29: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/29.jpg)
where d is the distance fallen, a is the acceleration, t is the time the object has been falling, and di is the initial distance from the origin.
In the late 16th and early 17th centuries Galileo challenged much of the work of Aristotle. Working with objects rolling down inclined planes he demonstrated that objects fall with a constant acceleration that is independent of their weight. According to Galileo objects fell with a speed that changed uniformly and at the same rate for all objects.
Motion at a constant acceleration, starting from rest, can be described by the equation:
d =12 a⋅t2 + di
![Page 30: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/30.jpg)
Comparing the equation above with the slope-intercept equation of a straight line, Y = mX + b,
we see that a graph of distance fallen versus time squared should be a straight line with d1 as the y-intercept and the slope of the line would equal one-half of the acceleration at which the object was falling.
Y m X +b
d=12 a⋅t2 + di
![Page 31: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/31.jpg)
To find the true nature of Free-Fall:
Let a ball roll down an incline,
Measure the distance traveled after certain times,
Plot graphs of distance versus time and distance versus time-squared.
If distance versus time is a straight line then Free-Fall is at a constant velocity and the slope of the graph measures that velocity.
If distance versus time-squared is a straight line then Free-Fall is at a constant acceleration and the slope of the graph measures one-half of that acceleration.
v, velocity = m, slope
a, acceleration = 2m, 2 x slope
![Page 32: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/32.jpg)
The Acceleration Due to Gravity
If distance versus time-squared is a straight line then Free-Fall is at a constant acceleration and the slope of the graph measures one-half of that acceleration.
The acceleration, a, found from the slope of the d vs t2 graph is related to but not equal to the acceleration due to gravity, g.
To find the actual value of g we must account for the effect of the incline.
g =a×Lh
L
h
CBR
CBL
![Page 33: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/33.jpg)
Experimental Procedure
![Page 34: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/34.jpg)
CBR / TI-83 Set-UPConnect the CBR to the TI-83
Press: APPS
Press “4” CBL/CBR
Press “Enter”
Press “2” Data Logger
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Data Logger Set-Up
Probe
# SAMPLES
INTRVL (SEC)
UNITS
PLOT
DIRECTNS
GO...
Sonic
Enter: 20
Enter: .04
Select: m
Select: REAL TIME
Select: ON
Press "Enter"
Press: Enter
![Page 36: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/36.jpg)
CBR/CBL Set-Upcontinued
Press “2” CBR
After CBR-CBL link has been tested:Press: “Enter”
After “Status OK”:Press: “Enter”
When you are ready to begin taking data”Press: “Enter”
After data collection is complete the TI-83 will plot a graph of distance versus time.
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0
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0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6
Time, s
Use this key to advance cursor
* * **
**
*
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0
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0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6
Time, s
(0.4, 0.676)
(0.6, 0.847)
(0.8, 1.086)
(1, 1.393)
(1.2, 1.768)
![Page 39: PSC 151 Laboratory Activity 3 Graphical Analysis IIA Nonlinear Graphs 1 and The Acceleration Due to Gravity.](https://reader036.fdocuments.in/reader036/viewer/2022062308/56649f0a5503460f94c1e552/html5/thumbnails/39.jpg)
Length, L = 162cm
Time-t, s Time Squared-t2, s2 Distance-d, m
Height, h = 27.7cm
Data Table 1
Plot graphs of:distance versus time
and distance versus time squared