Question

can you explain 3 and 4

For Questions 1-3, Two different groups are measuring the SAME cart travelling with a constant velocity on a straight track. Table 1 and Table 2 represent the data collected by each group.
												Table 1					1. Use the standard "Kinematics Equations" for velocity and acceleration to find these values for each time step in Table 1 and Table 2 as well as their average and uncertainty.   Definitely take advantage of Excel's "Copy" and "Paste" functions so you only need to type equations once or twice. Give special attention to the cells on the edge of the data set.
Average Velocity:		Average Acceleration:
Velocity Uncertainty:	0.575694444	Acceleration Uncertainty:	0.09296875
Time (s)	Position (cm)	Velocity (cm/s)	Acceleration (cm/s/s)
0	0
1	14.875	12.05
2	24.1	13.3625	1.06875		2. Use proper rounding techniques to report the velocity and acceleration of the cart for each data table. (Note, because the data updates every time you edit this sheet, the instructor is aware that there might be a slight mismatch between your reported value and data showing at the time of submission.)
3	41.6	14.1875	0.027083333
4	52.475	13.0375	0.098863636
5	67.675	13.1	0.16625
6	78.675	11.375	0.095833333
7	90.425	12.2375	-0.2125
8	103.15	13.075	-0.414285714		Table 1: Velocity =
9	116.575	15.1375	-0.15
10	133.425	13.975	0.3275		Table 2: Velocity =
11	144.525	13.5	0.20625
12	160.425	13.15	0.141666667		Table 1: Acceleration =
13	170.825	13.075	-0.8
14	186.575	14.75	0.8875		Table 2: Acceleration =
15	200.325	12.1875	#DIV/0!
16	210.95	11.8125	1.15		3. Are your results consistent with the description given above: "Two different groups are measuring the same cart travelling with constant velocity on a straight track." Explain any numerical inconsistencies that are in the data using your understanding of "Uncertainty".
17	223.95	13.3375	1.15625
18	237.625	14.125	-0.04537037
19	252.2	13.20138889	0.8828125
Table 2
Average Velocity		Average Acceleration:
velocity uncertainty:	4.611111111	Acceleration Uncertainty:	14.77083333
Time (s)	Position (cm)	Velocity (cm/s)	Acceleration (cm/s/s)
0	0
0.1	2.02	4.325
0.2	0.865	18.325	38.75
0.3	5.685	12.075	-66.25		4. Compare and contrast the uncertainties measured by these two groups. Fundamentally what caused one group to get more precise data than another?
0.4	3.28	5.075	110
0.5	6.7	34.075	38.125
0.6	10.095	12.7	-195.625
0.7	9.24	-5.05	57.5
0.8	9.085	24.2	182.5
0.9	14.08	31.45	-60.625
1	15.375	12.075	-194.375
1.1	16.495	-7.425	-100.625
1.2	13.89	-8.05	106.875
1.3	14.885	13.95	217.5
1.4	16.68	35.45	-0.625
1.5	21.975	13.825	-138.125
1.6	19.445	7.825	54.375
1.7	23.54	24.7	43.75
1.8	24.385	16.575	-55.76388889
1.9	26.855	13.54722222	9.208333333

Answer 1

3. Explanation of numerical inconsistency in the data which is measured by two different groups of the same cart moving with a constant velocity is as follows: The data inconsistency may arise due to the change in reference frames of the two groups. For example, the first group measures the uncertainty in velocity to be 0.575... , and the other group measures it to be 4.611..... In the same way, uncertainty in acceleration of the cart comes out to be 0.0929... which is very less as compared to the uncertainty measured by the Group 2, i.e. 14.7708.....We see that the uncertainity in velocity/acceleration in case of Group 1 is less whereas, the second group found more uncertainty. That means the first group readings are more precise than the second group readings. Less is the uncertainty, more is the precision.

4. The fundamental reason behind one group getting more precise data than the other group is that change in time interval, and also the variation in total time for which the readings are taken. Group 1 took readings of acceleration and velocity after each second upto 19 seconds whereas, Group 2 took readings after every 0.1 second interval upto only 2 seconds. This variation in the time interval is the cause behind getting less precise data in case of Group 2. Readings are concentratedover a very small interval of 2 seconds, i.e. only from t=0 sec to t=2 sec with =0.1 sec. In case of Group 1, readings are spread over a time interval of 19 seconds, i.e. t=0 sec to t=19 sec, with =1 sec.