Question

Displacement (meters)	Time (seconds)
Trial 1 (0.70 m)
0.695 m	0.373 s
0.695 m	0.374 s
0.695 m	0.373 s
0.695 m	0.373 s
Average Displacement = 0.6951 m	Average Time = 0.3733 s
Trial 2 (0.60 m)
0.598 m	0.345 s
0.595 m	0.345 s
0.595 m	0.345 s
0.598 m	0.346 s
Average Displacement = 0.5960 m	Average Time = 0.3453 s
Trial 3 (0.50 m)
0.495 m	0.314 s
0.495 m	0.315 s
0.495 m	0.314 s
0.495 m	0.314 s
Average Displacement = 0.4950 m	Average Time = 0.3143 s
Trial 4 (0.40 m)
0.395 m	0.281 s
0.3948 m	0.279 s
0.3941 m	0.280 s
0.395 m	0.279 s
Average Displacement = 0.3947 m	Average Time = 0.2798 s
Trial 5 (0.20 m)
0.200 m	0.197 s
0.1975 m	0.197 s
0.198 m	0.197 s
0.199 m	0.196 s
Average Displacement = 0.1986 m	Average Time = 0.1968 s

This is my data for Physics Measurement and Free Fall Lab. I have also graphed it. The equations I have are: y = 4.9318x^2 - 0.0061x + 0.0101 for 'Displacement vs. Time' graph and y = 4.9168x +0.0097 for 'Displacement vs Time^2' graph.

Is this okay?

Can you determine the experimental acceleration of gravity from this graph's best fit equation? Can you determine the experimental acceleration of gravity from the linear graph's best fit equation? Explain how and show work please.

How do the experimental values of "g" compare with the accepted value (g = 9.80 m/s^2) for the acceleration of gravity? Give a mathematical comparison.

Answer 1

For Displacement vs. Time graph, we have

y = (4.9318) x² - (0.0061) x + (0.0101)

For Displacement vs Time² graph, we have

y = (4.9168) x + (0.0097)

(a) We cannot determine the experimental acceleration of gravity from this graph's best fit equation.

(b) Yes, we can determine the experimental acceleration of gravity from the linear graph's best fit equation.

We know that, y = (4.9168) x + (0.0097)

Comparing an above eq. with, y = m x + b

Then, we have

m = 4.9168

The slope of such a graph will be given by -

slope = (1/2) a

a = 2 (slope)

a = [2 (4.9168)] m/s²

a = 9.8336 m/s²