P 251Laboratory Activity IIGraphical Analysis I

Graphical AnalysisThe analysis of experimental data to

determine mathematical relationships between measured

quantities

1. Constructing the Graph2. Analyzing the Graph

Constructing the Graph

Summary:Problem: To determine the relationship between mass and volume, if one exists.Measurements: Mass and volume of five different aluminum objects

Suppose that we think that there is a relationship between the mass of an object and its volume. We also think that this relationship will depend on the object's composition.

We might look for several objects that are made of the same material, such as aluminum, and measure their masses and volumes.

We could then use graphical analysis to determine if a relationship between mass and volume exists, and if it does what is its mathematical form.

Volume, cm3 Mass, g

25 67.5

67 181

90 243

142 383

160 432

DATAVolume and Mass of Aluminum Objects

Choose Volume as independent variable

Choose Mass as dependent variable

Mass versus Volume, Aluminum Objects

Independent Variable

Dependent Variable

Graph Title51 divisions

34 divisions

mas

s,

g

Volume, cm 3

Construct and Label Axes

Scales do not have to be the same for both axes.

Scales should be chosen so that the graph covers as much of the sheet as possible

Determine scales for each axis.

160cm334 divisions=4.71 cm3

division

4.71 cm3

division≈5 cm3

division

Scale for Horizontal Axis Representing Volume

Step 1: From the data table select the largest volume: 160cm3

Step 2: Count the number of divisions on the horizontal axis: 34 divisions

Step 3: Divide the largest volume by the number of divisions:

Step 4: Round UP to a convenient number:

432g51 divisions=8.47 g

division

8.47 gdivision≈10 g

division

Scale for Vertical Axis Representing Mass

Step 1: From the data table select the largest mass: 432g

Step 2: Count the number of divisions on the vertical axis:

51 divisions

Step 3: Divide the largest mass by the number of divisions:

Step 4: Round UP to a convenient number:

Volume, cm3

...1020304050

100

Plot data

0origin

Label Each Axis with its Scale

Scale : 10g

division

Scale : 5 cm3

division

Volume, cm3

...1020304050

100

Graphing First data pair

Volume=25cm3

mass=67.5g

Mass versus Volume, Aluminum Objects

1020304050

100

150

200

Volume, cm3

250

300

350

400

Mass versus Volume, Aluminum Objects

1020304050

100

150

200

Volume, cm3

250

300

350

400

Horizontal Axis Data Vertical Axis Data

0.2 16.21.3 39.31.7 47.72.2 58.23.4 83.4

Scale: Scale:Lable each axis with its scale.Plot the first data pair.

Practice 1

3.436 =0.09444 ≈0.1

0.1

43 divisions

36 divisions

83.443 =1.9395 ≈2.0

2.0

0.1 0.2 0.3 0.4 0.5 1.0

2.0

4.0

6.0

8.0

10.0

20.0

1. Determine the best scale for each axis.2. Label each axis with its scale.3. Plot the first data pair.

Practice 2

Horizontal Axis Data Vertical Axis Data

5.5 57.0512.3 125.7321.7 220.6735.3 358.0359.3 600.43

Scale: Scale:Lable each axis with its scale.Plot the first data pair.

43 divisions

36 divisions

59.336 =1.6472 ≈2.0

2.0

600.4343 =13.9634 ≈15.0

15.0

2.0 4.0 6.0 8.010.0 20.0

15.0

30.0

45.0

60.0

90.0

120.0

150.0

1. Determine the best scale for each axis.2. Label each axis with its scale.3. Plot the first data pair.

Analysis of the Graph

Y =dependent variableX= independent variable

m=ΔYΔX

b=y−intercept= value of the dependent variable

at the point that the graph intersects the vertical axis

Slope-Intercept Equation of a

Straight Line

Y =mX+b

m=slope=rate of change of dependent variable

with respect to the independent variable

mass Volume

V

Y = m X + b

M

Dependentvariable

Independentvariable

M=m⋅V+b

m=ΔYΔX

Select two points on the graph:

(67cm3,181g) & (160cm

3,432g)

m=432g−181g

160cm3−67cm

3=

251g

93cm3

=2.7 gcm3

=ΔMΔV =M2−M1

V2−V1

V1 M1 V2 M2

2.7 gcm3

To find "b" locate the point where the graph

intersects the vertical axis.

M = m V + b

M=2.7 gcm3 ⋅V+b

Mass versus Volume, Aluminum Objects

1020304050

100

150

200

Volume,

cm

3

250

300

350

400

m

The graph intersects the vertical axis at the origin. Therefore b =0

Since the graph is (1) a straight line and (2) passes through the origin, we can say the the mass and volume are directly proportional.

0Final Equation

M = 2.7gcm3 V + b

M = 2.7gcm3 V + 0

M = 2.7 gcm3 V

Time- t, seconds Distance- d, meters0.2 16.21.3 39.31.7 47.72.2 58.23.4 83.4

Practice 3

Points Chosen for Slope:(0.2s,16.2m) ; (3.4s,83.4m)

m = ΔdΔt =83.4m −16.2m

3.4s−0.2s =67.2m3.2s

m=21 ms

Intercept, b = 12m

(speed)

Equation:Y → d

X → t

m→ 21 ms

b→ 12m

d=21 ms ⋅t+12m

1. Choose two points on the graph and determine the slope.2. From the graph determine the y-intercept.3. Write the equation relating d and t.