I need help soon as possible. My question says: Refer to Figure 3 in the Lab Man

Question

I need help soon as possible. My question says:

Refer to Figure 3 in the Lab Manual, which shows five quadratic fits to five runs of data. In the fit on the far left, the three fit coefficients are:

A = 0.392

B = -0.669

C = 0.527

From these numbers, what was the acceleration of the system?

Hint: When something is moving under constant acceleration, its kinematic equation of motion is: x = x0 + v0t + (1/2)at^2. The computer is fitting a function of the form: x = At^2 + Bt + C. So what is the relationship between big "A" (the coefficient from the fit) and little "a" (the acceleration)? These are not the same number.

I do not understand this question. Can anyone please explain the steps and how can I figure the answer to be correct soon as possible?

Click on the horizontal and vertical axis of the graph and manually rescale the graph so that the data fills the graph area. To add a fit lineleft click any data pair on the graph. Then right click and choose add trendline. Choose the following: linear fit, display equation on chart. Check to see if there are any obvious discrepancies in the graph. Make sure you did not enter data incorrectly and if necessary double check quadratic fits for consistency of data selection. You can now use the slope and intercept from the linear fit of (a vs. mh) to determine the acceleration due to gravity. If you are proficient with Excel don't hesitate to use the slope and intercept function to calculate the slope and intercept so that you wont have to hard type in the values from the fit equation on the graph. Use your slope and intercept to compute "little g" and the coefficient of kinetic friction UK. zer on a KIA rSmoothing — of the EA x 4Gla to 0.85 uadratic A + BK + C ouadratic quadratic A + BX + C a +B A + BK + C 0.80 + C Run #2 k A = 1.35 An 1.82 A = 2.39 0.75 Ruum #4 B = -16.1 B = 53.3 Run 6 0.70 HRun 48 RMSE = 3.4110-4 RMSE = 7.56x 10-4 0.65 Rum #10 0.60 Quadratic 0.55 A = 0.392 050 a 0.527 RISE = 8.5 3X 10 0.40 0.35 0.30 0.25 Figure 3: Five Capstone runs, with five fit quadratics. 45

Explanation / Answer

Now, the fucntion from the graph is

x = At2 + Bt + C

The kinematic equation of motion is

x = x0 + v0t + (1/2)at2

Now, the values of x are the same.

Thus,

At2 + Bt + C = x0 + v0t + (1/2)at2

Comparing the coefficient of t2,

A = 1/2a

a = 2A

Substitute 0.392 for A in the above equation,

a = 2A

= 2 (0.392)

= 0.784 m/s2

Therefore, the acceleration of the system is 0.784 m/s2.

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