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a) What quantity name is represented by the slope of your v vs t best-fit line? Hint: Look at the slope units for a clue. Remember that a quantity is a variable and not a unit. For example position is a quantity name, whereas cm or meters are units for the position quantity.
b) What is the quantity name represented by the vertical axis intercept on the velocity vs. time graph? Is it the value that you expected based on how you were directed to release the cart? If not, what should you have done differently?
c) Show your calculation for the method 2 velocity. Explain how it compared to the reading from your velocity vs. time graph.
Transcribed Image Text:Position vs Time
450
400
350
....**
300
250
200
150
100
50
0.0
-50
5.0
10.0
15.0
20.0
25.0
-100
Time(sec)
Series1
Linear (Series 1)
Plot the Velocity vs. Time graph from the values in your formal table. Draw best-fit
line(s) that best fit the data.
Velocity vs Time
25.00
20.00
..........
15.00
10.00
5.00
0.00
0.0
5.0
10.0
15.0
20.0
25.0
Time (sec)
Series1
Linear (Series 1)
....
Best fit slope of the graph is 1.09 with intercept 1.41
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(s/w) Aapojan
Transcribed Image Text:Position
Time
Velocity
(m/s)
Time
(m)
(+/-) Im
(sec)
(+/-)0.1sec
(sec)
2
1.5
1.38
1.5
10
2.8
3.64
2.8
22
4.0
5.50
4.0
38
5.4
7.04
5.4
60
6.7
8.96
6.7
86
8.0
10.75
8.0
118
9.4
12.55
9.4
154
10.7
14.39
10.7
184
11.8
15.59
11.8
216
12.9
16.74
12.9
246
14.0
17.57
14.0
276
15.0
18.40
15.0
308
16.1
338
17.2
19.13
16.1
18.3
19.65
17.2
370
400
19.4
20.22
18.3
Analysis-Graphs: Be sure to use the graph side of the paper.
Attach the position vs time graph for the Race Cart. Be sure it shows the 8 tangent lines.
Show a sample calculation for the slope of one tangent line.
Slope at Time = 4.0 sec, position = 22.0 m and Time =15.0sec, position =276 m
Slope = rise / run = Ax(position) / At(time) = (276-22)m / (15-4)s = 23.09 m/s
+c(15=23.09 x 4 +C)=-77.36
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