Question

Enter the data from the table below into Microsoft Excel and determine the linear correlation coefficient, R2 for time versus position.
Then, calculate y/t and determine the linear correlation coefficient, R2 for time versus y/t. Write in the calculated values in the table below (use two decimal places).

Time, t (s)	Position, y (cm)	Velocity, y/t (cm/s)
0/60	0.00
1/60	2.41
2/60	5.09
3/60	8.08
4/60	11.30
5/60	14.79
6/60	18.56
7/60	22.52
8/60	26.96
9/60	31.56

1. Determine the linear correlation coefficients, R2, for both the y vs. t data and the y/t vs. t data. (Both to four decimal places.)

   For y vs. t, R2 = ____________ For y/t vs. t, R2 = ____________

2. Using the linearized relationship, y/t vs. t, report the properly rounded best estimate for ?. (?"## = 980.35 ??/?0.)

g =

%diff=

Answer 1

Please look at the formula bar next to the big bold sigma notaion for the formula entered in the highlighted cell. Drag down to apply the formula to the remaining cells in the column.

We can use the CORREL() function in excell to find the linear correlation coefficients.

Note that since velocity is y/t, we do not have a velocity value for y = 0 and t = 0 because it will give divided by zero error. So, we have one data less for y/t vs t correlation. Note the change in the formula.

So,

for y vs t, R2 = 0.9951

for y/t vs t, R2 = 0.9999

Now, to obtain the linearized relationship, we can plot a scatter plot of y/t vs t and fit a straight line and obtain the equation of the linear fit.

Select the data in the velocity and time columns -> insert scatter plot -> right click on any data point -> insert linear fit/trend line -> check the box to chow equation

The value of g is twice the slope of the linearized relationsheep between Velocity and Time

So, g = 2 X 490.6571 = 981.3142

rounded to two decimal places,

g = 981.31 cm/s²

% difference = 100 X (981.31 - 980.35) / 980.35 = 0.0979 %

If you do not want to plot a scatter plot and want to obtain the linearized relationsheep from the linear correlation coefficient, you can use the linear regression model for Velocity (lets call it v) vs Time (t):

where,

where, R₂ is the linear correlation coefficient, s_v and s_t are the standard deviations in v and t respectively. And,

where, v and t with a bar are the averages of v and t

The difference in the % difference here and that obtained earlier is because of error introduced due to rounding in the earliner case.

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