PHYS 200 Lab Report 1 (Revised)
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Athabasca University, Athabasca *
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200
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Mathematics
Date
Feb 20, 2024
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docx
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Uploaded by CoachBravery11850
Athabasca University PHYS 200 Lab Report #1
Kinematics in One Dimension
Ashley Sherlow
September 12, 2023
Student Number: 3309093
Introduction
In this lab, one parameter undergoes controlled manipulation, while the other parameter is quantified as a result of the manipulation. The result in the case of this
lab will be a dataset – represented on a graph – that illustrates the average velocity of a car travelling from Calgary to Edmonton, based on a set of numbers that depict the distance travelled by the car versus time.
Procedure
During a drive from Calgary to Edmonton, a person collected data that documented the distance covered over time. The data collection began at the final green light on
the outskirts of Calgary with the odometer reset to zero. At this point, a stopwatch started, beginning to collect the time over the duration of the drive. The measured
time was used to track the distance travelled at ten minute intervals based on the odometer reading. Uncertainty in the data collected was acknowledged based on the odometer’s measurement capabilities to within 100m, therefore producing an uncertainty of δd = ± 0.1km. The uncertainty for time intervals was determined to be a timing error of about 12 seconds, therefore producing an uncertainty of δt
m = ± 0.2 min.
The data collected is presented in the table under the Data
heading.
Pictures
1
2
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Data
Original Data Collected:
Known: X
0
= 0
unc X = +/– 0.1km
unc T = +/– 0.2 min
T = intervals of 10 min
Unit Conversion (Sample Calculation):
T
h = 1/60 × t
m
= 1/60 x 10.0 min
= .0167 h
Calculation for Uncertainty
δT
h = 1/60 × δt
m
3
unc T = +/– 0.2 min
0.2min/10min = 0.02%
δT
h = 1/60 × t
m
= 1/60 x 0.02%
= +/- 0.000334 = +/- 3.34 x 10
-4
Time in
Minutes
tm±0.2(min)
Time in Hours
th±3.34 x 10-4(h)
Distance
d±0.1(km)
0.0
0.000
0
10.0
0.167
17.5
20.0
0.333
34.5
30.0
0.500
52.5
40.0
0.667
76.1
50.0
0.833
87
60.0
1.000
105
70.0
1.167
122.3
80.0
1.333
144
90.0
1.500
152.3
100.0
1.667
177
110.0
1.833
191.1
120.0
2.000
209
130.0
2.167
227.4
140.0
2.333
241
150.0
2.500
263.7
160.0
2.667
285.2
170.0
2.833
292.1
Calculation for Best-fit Linear
Slope = rise/run = Δd/Δt
= (105 km - 87 km)/(1.00h - 0.833h)
= 18 km / 0.167 h
4
= 107.8km/h = 108 km/h
Calculation for y intercept
Y = mx + b
B = y - mx
= 87 km - (108km/h x 0.833 h)
B = 0.30 km
Analysis & Discussion
According to my calculations, the instantaneous speed of the car is 108km/h, which is the same as the slope/average velocity (not greater or less than). This is because I used the slope to find the distance value in order to determine the speed and even
using another equation, the use of the correct sig fig means rounding the instantaneous speed to the same speed as the average speed/velocity/slope. If ignoring sig figs, the instantaneous speed is just 0.24km/h less than the average speed. See the Pictures
section for a graphical representation. Conclusion
In conclusion, this experiment aimed to investigate the relationship between time and distance traveled by a car. To facilitate my analysis, I converted the time from minutes to hours, then I graphically represented the data on a hand-drawn graph, plotting each point based on the values from Table L1.2. I then drew a best-fit line drawn through the data points, and calculated its slope. This line's equation, y = mx
+ b, provided insights into the relationship between time and distance.
The slope of the best-fit line is directly related to – and, in fact, representative of – the car's average speed (velocity) during the trip. To estimate the car's instantaneous speed at t = 75 minutes, I found the slope of the tangent line at that point on the graph. I found that, when rounding to the significant figure, it was actually the same as the average speed, though if not 5
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rounded, it was marginally less than the average speed. Some suggestions to enhance the accuracy of future measurements could include more frequent data points and perhaps a more precise measurement method. Questions
1.
1.001 x 10
2
m ; 9.87 x 10
2
s ; (1.000 ± 5) x 10
3
J ; 0.5 x 10
5
s ; 9.001 x 10
13
mm
2.
δy = 3(4x
4-1
)δx 3.
C = 2πr
= 2π (7.3cm)
= 45.8cm
δC = 2π(δr/r)
= 2π (0.2cm/7.3cm)
= 2π (0.03cm)
= 0.2cm
C = 45.8 ± 0.2 cm 4.
A = πr
2
= π(7.3cm)
2
= 167.4cm
6
7
5.
8
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6.
9