Question

In the lab you obtained a velocity vs. time graph when a person was moving on a straight line with constant velocity away from the motion sensor. You highlighted a region of data containing 5 points with the following values of velocity: 0.38, 0.26, 0.29, 0.34, 0.34 m/s

IMPORTANT NOTE: When you report your final result in the lab report, the number of the significant figures on the error should NOT be the same as the number of significant figures of the mean. For example notation: v=0.356 ± 0.00623 m/s, despite having the same number of significant figures (3) in the mean value and in the error is incorrect. You should decide how many significant figures to keep for the mean and then the number of the decimal places in the error should be the same. For example a correct way of reporting the above result would be: v=0.356 ± 0.006 m/s or, alternatively: v=0.36 ± 0.01 m/s . So let's practice this lab notation here.

Report your result for the mean velocity to 2 significant figures and the corresponding error.

Value of velocity = _____ ± _____

Answer 1

Mean=(.38+.26+.29+.34+.34)/5

=.322m/s

Error in 1=|.38-.322|=0.058

Error in 2=|.26-.322|=0.062

Error in 3=|.29-.322|=.032

Error in 4=|.34-.322|=.018

Error in 5=|.34-.322|=.018

Avg value=(Error in 1+Error in 2+Error in 3+Error in 4+Error in 1+Error in 5)/5

=0.0376

Avg value=0.322+-0.038m/s