Please help, thanks. A group of students, performing the same Uniformly Accelera

Question

Please help, thanks.

A group of students, performing the same Uniformly Accelerated Motion experiment that you did in lab, dropped a picket fence through a photogate and obtained the following data from the computer: Use Excel to create a spreadsheet, just like the one your group created in lab, which contains the above data, all calculations needed to obtain the average value of the acceleration, and all calculations needed to obtain the standard deviation of the average value of the acceleration. The band spacing is 0.06 m. What is the acceleration? AVERAGE ACCELERATION METHOD a = 10 3530 0.3901 m/s^2 LINEST METHOD a = 10 3926 0.0786 m/s^2

Explanation / Answer

in the given table of time vs Displacement

0 0

0.0401 0.06

0.0726 0.12

0.1005 0.18

0.1255 0.24

0.1483 0.3

0.1694 0.36

0.1891 0.42

we first calculate average velocity by calculating instantaneous velocities as (d2 - d1)/(t2 - t1)

then we calculate average acceleration by calculating instantaneous acceleraitons by (v2 - v1)/(t2 - t1)

then we find average of accelerations which comes out ot be a = -0.9444 m/s/s +- 0.6533

if we plot the distances with time on the graph and plot a polynomial fit

the equation is

y = 4.8605x^2 + 1.3022x - 3*10^-5

so, acceleration , a = 2*4.8605 = 9.721 +- 0 m/s^2 [ because R^2 value is 1]

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