Compare the slope of the

velocity-time graph to the average of all your acceleration values. Are they

close? What does the slope of a velocity (or speed) vs. time graph mean?

Explain the answer using your data. How does the value of g that you calculated compare to the accepted value of 9.80 m/s2?

What is your percent error? Remember that the value of g can be calculated by using g

= a/sinθ and percent error can be calculated using the

following equation the equation is % error= ((measured value-accepted value)/accepted value) x 100 I tried solving this but only got this:

% error= (0.1846 – 9.8)/9.8) x 100I uploaded the data table from my experiment and i hope somone would lend me some help because ‘;m stuck!1.25

Tl T2 T3

Aug. Augof Vebcity

IM

1.33 1.15s 1.25 0.8m/s

6.9M 0.95 lills 1.33 1.ls 0.8mls

0.8M 0.93 0.985 1.01 0.99 0.8mls

0.71 0.965 0.815 0.845 35 0.2wls

061 10.8ls 0.755

0.83 35

0.2m/s

0.5M 0.65 0.665 0.65 0.6 0.8wls

0.4M/0.60, 0.65 0.653, o.lols

0.39 /0.63,-0.60 0.40s 13s 0.2m6

0.2M 0.40 0.368 0.38 1.145

o.lmls

OIM 10.403|0.33 0.28 (0.3, 03ml

Aug of Acc.

0.6m/s

0.7ml

0.9m13

097 00mis

0.06m/s

1.3m/s

0.04m/s²

0.1mls

0.15ml5

lillnls

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