Question

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.

Blocking Times (s)
0.05206
0.08969
0.12076
0.14763
0.17203
0.19438
0.21502

The band spacing is 0.06 m.

(a) Use Excel to do the following. (You will not submit this spreadsheet. However, the results will be needed later in this problem.)

(i) Calculate the accelerations.

(ii) Calculate the average of the accelerations.

(iii) Create a graph of velocity versus time.

(iv) Use the trendline option to determine the slope.

(v) Report your results below.

(b) What is the average acceleration? (Give your answer to four significant digits.)
a_avg =

(c) What is the average acceleration a_average as determined from the slope? (Give your answer to four significant digits.)
a_avg =

(d) What is the percent difference between the values of the acceleration you obtained in (b) and (c) above?
%

I am struggling as to why I keep getting the wrong answer.

Answer 1

a. From the given data,   
given band gap is 0.06 m
so for consecutive timings we add up bandgaps
time distance (cumulative banndgap distance)
0.05206   0.06
0.08969   0.12
0.12076   0.18
0.14763   0.24
0.17203   0.3
0.19438   0.36
0.21502   0.42

i. Now, for acceleration
from the graph (from excel), the equation of motion becomes

y = 4.8686t^2 + 0.9094t - 0.006
hence
y" = 2*4.8686 = 9.7372 m/s/s
  
2. as the acceleration is constant
average acceleration is 9.7372 m/s
  
3. the following is plot of velocity vs time
y' = 9.7372*t + 0.9094

4. from the trendline, slope is 9.7372

v. b) average acc = 9.7372 m/s/s
c) from slopes
average velocity = 2.301521 m/s
average acc = 9.25381566 m/s/s
d) %diff = (9.7372 - 9.25381566)*100/9.7372 = 4.964305 %