PSC 151
Laboratory Activity 3
Graphical Analysis IIA
Nonlinear Graphs 1
and
The Acceleration Due to Gravity
Graphical Analysis Exercise
Determining 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
2.1
4.8
8.8
11.5
17
DATA
Circumference, cm
7.5
15.4
28.3
36.2
53.1
Plot a graph of Circumference versus diameter.
CALCULATIONS AND OBSERVATIONS:
1. Is your graph a straight line?
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  Y  C  36.2cm  15.4cm 
X d 11.5cm  4.8cm
20.8cm  3.1
6.7cm
Slope has NO units
What is the equation relating Circumference and diameter?
Y = mX+b
C 3.1 d + 0
C 3.1d
Compare slope = 3.1 to p (3.14)
% error 
experimental value - accepted value
accepted value
3.1  3.14
.04  100%

% error 
 100% 3.14
3.14
 100%
 1.3%
Non-Linear Graphs
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.
Data
time-t, s
0
1
2
3
4
5
6
7
8
9
10
distance-d, m
5
9.9
24.6
49.1
83.4
127.5
181.4
245.1
318.6
401.9
495
We then plot a graph of distance versus time.
Distance versus Time
500
Not a straight line
but is a uniform curve
450
400
distance, m
350
300
250
200
150
100
50
0
0
1
2
3
4
5
6
time, s
7
8
9
10
11
Compare graph to
graphs of other
functions of the
independent variable
7
y =x
Y X
6
Y  mX  b
5
4
Y
3
2
1
0
0
1
2
3
4
X
5
6
7
40
y =x2
YX
Y  mX  b
2
30
Y
2
20
10
0
0
1
2
3
4
X
5
6
7
y = x1/2
2.5
2
1.5
Y X
Y
Y  m X b
1
0.5
0
0
2
4
X
6
8
1.25
y = 1/x
1
Y
X
1
Y  m 1 b
X
0.75
Y
0.5
0.25
0
0
1
2
3
4
X
5
6
7
1.25
y = 1/x2
1
Y 2
X
Y  m 12  b
X
1
0.75
Y
0.5
0.25
0
0
1
2
3
4
X
5
6
7
1.1
y = 1/x1/2
1
1
Y
X
0.9
1
Ym
b
X
0.8
Y
0.7
0.6
0.5
0.4
0
1
2
3
4
X
5
6
7
7
40
Y X
y =x
Y  X2
6
Y  mX  b
5
y = x1/2
2.5
y =x2
2
Y  mX  b
30
2
Y X
1.5
4
Y

20
Y
3
2
10
Y
Y  m X b
1
0.5
1
0
0
1
2
3
4
5
6
0
0
7
0
X
1
2
3
4
5
6
0
7
Y
0.5
8
1.1
1.25
y = 1/x
Y  12
X
Y  m 12  b
X
6
X
y = 1/x2
0.75
4
X
1.25
1
2
Y 1
X
1
0.75
Y
Y 1
X
1
0.9
Y  m 1 b
X
y = 1/x1/2
Y  m 1 b
X
0.8
Y
0.7
0.5
0.6
0.25
0.25
0.5
0
0
1
2
3
4
X
5
6
7
0.4
0
0
0
1
2
3
4
X
5
6
1
2
3
4
7
X
5
6
7
Plot a new graph where time squared
is the independent variable:
Distance, d versus Time Squared,
2
t
Revised Data Table
time-t,s time2-t2, s2 distance-d,m
0
5
0
1
9.9
1
2
24.6
4
3
49.1
9
4
83.4
16
5
127.5
25
6
181.4
36
7
245.1
49
8
318.6
64
9
401.9
81
10
495
100


Distance versus Time Squared
500
450
400
distance, m
350
300
250
200
150
100
50
0
0
10 20 30 40 50 60 70 80 90 100 110
time squared
Analysis of Graph
Y  mX b
distance
time
squared
d
2
t
2
d  mt  b
Slope Calculation :
m = Y  d2
X t
Points Chosen :
2
2
(4s , 24.6m) and (64s ,318.6m)
24.6m  294m
m = 318.6m
2
2
2
64s  4s
60s
m  4.9 m2
s
2
m
d  4.9 2 t  b
s
With units of m/s2 the slope represents the acceleration of the
object.
Distance versus Time Squared
500
450
400
distance, m
350
300
250
200
intercept,b
150
It will be difficult to
determine the intercept from
the graph!
100
50
0
0
10 20 30 40 50 60 70 80 90 100 110
time squared
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.
2
t  0  b  5m
2
m
 4.9 2  t  b
s
2
m
b  d  4.9 2  t
s
2. Start with the partial equation: d
Solve for “b”:
Choose any data pair and substitute the values of “d” and “t2”
into the equation for “b”:
(25s2, 127.5m)
2
m
b  127.5m  4.9 2  25s
s
b  5m
Final Equation
2
m
d  4.9 2 t  5m
s
Graphical Analysis of
“Free-Fall” Motion
Determining the Acceleration Due to Gravity
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.
Motion at a constant speed can be described by the equation:
d  v  t  di
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.
Comparing the equation above with the slope-intercept equation of
a straight line, Y = mX + b,
d  v  t  di
Y  mX  b
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.
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:
2
1
d  2 a t  d i
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.
Comparing the equation above with the slope-intercept equation of
a straight line, Y = mX + b,
2
1
d  a t  d i
2
Y
m X 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.
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 timesquared.
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.
v, velocity = m, slope
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.
a, acceleration = 2m, 2 x slope
The Acceleration Due to Gravity
If distance versus time-squared is a straight line graph 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.
If the length of the incline is L and the height of its raised end is h:
L
g a h
Experimental Procedure
The metal ball will be released from rest at designated positions
(1-5) along the inclined track and the time to travel to the bottom
measured with a stopwatch.
Data Table 1
Length, L = 176cm, Height, h = 2cm
Position #
Distance-d, cm
Distance-d, m
1
13.8
0.14
2
48.8
0.49
3
83.8
0.84
4
118.5
1.19
5
153.5
1.54
Time-t, s
Time Squared-t 2 , s2
Plot graphs of:
distance versus time
and
distance versus time squared