PHY 101L Module Two Lab Report
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Aerospace Engineering
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Feb 20, 2024
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PHY 101L Module Two Lab Report
Name: Ashley Shaw
Date: 01/21/24
Complete this lab report by replacing the bracketed text with the relevant information.
Activity 1: Graph and Interpret Motion Data of a Moving Object
Activity 1 Table 1
Time (
x
-axis) (seconds)
Position (
y
-axis) (meters)
0
0
5
20
10
40
15
50
20
55
30
60
35
70
40
70
45
70
50
55
0
10
20
30
40
50
60
0
10
20
30
40
50
60
70
80
Motion Data (m/s)
Time (s)
Position (m)
Activity 1: Questions
1.
What is the average speed of the train during the time interval from 0 s to 10 s?
The average speed of the train from 0s to 10s is 4 m/s. 2.
Using the equation: v
=
s
2
−
s
1
t
2
−
t
1
, calculate the average speed of the train as it moves from position x
= 50 m to x
= 60 m.
The average speed of the train from 50-60m is 0.67 m/s. 3.
What does the slope of the line during each time interval represent?
While the slope is positive the train is moving in a positive direction (right) or increasing with time. At 45s, the train begins to have a negative slope implying that it has turned around (headed in the negative direction, left) or decreasing with time. The steeper the slope in either direction, the faster the object is moving. If there is no slope it means the position is not changing with time, it is stationary such as from time 35-45s, implying the position didn’t change with time. 4.
From time t
= 35 s until t
= 45 s, the train is located at the same position. What is the slope of the line while the train is stationary?
The slope of the line is 0. 5.
Calculate the average speed of the train as it moves from position x
= 70 m to x
= 55 m. What does the sign of the average velocity during this time interval represent? The average speed from 70-55m is -1.5 m/s. The negative sign is directional and indicated the train is moving in the negative direction (left). 6.
What is the displacement of the train from time t
= 0 s until t
= 50 s?
The displacement is 55m. 7.
What is the total distance traveled by train from time t
= 0 s until t
= 50 s?
The total distance traveled is 55m. 8.
What is the slope of the line during the time interval t
= 45 s to t
= 50 s?
The slope of the line is -3 m/s.
9.
What does the sign of the slope in Question 8 represent in terms of the motion of the train?
The negative slope implies that the train is moving in the negative direction (left). 10.
What is the average velocity of the train during the interval t
= 0 s to t
= 50 s?
The average velocity of the train is 1.1 m/s. 11.
Does the train’s average velocity during the interval t = 0 s to t = 50 s provide a complete picture of the train’s motion during this time?
No, there was a brief stop between 35-45s that is not shown by looking at the velocity. Activity 2: Calculate the Velocity of a Moving Object
Activity 2 Table 1
Time (s)
Displacement (m)*
0.00
0.00
0.75
0.25
1.43
0.50
2.10
0.75
2.75
1.00
3.48
1.25
4.21
1.50
4.85
1.75
5.40
2.00
*Note that 0.25 m = 25 cm
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0
1
2
3
4
5
6
0
0.5
1
1.5
2
2.5
f(x) = 0.36 x
Velocity of Car
Time (s)
Displacment (m)
Activity 2 Table 2
Time (s)
Velocity (m/s)
1
0.362
2
0.725
3
1.09
4
1.45
5
1.81
6
2.17
7
2.54
8
2.90
0
1
2
3
4
5
6
7
8
9
0
0.5
1
1.5
2
2.5
3
3.5
Activity 2
Time (s)
Velocity (m/s)
Activity 3: Graph the Motion of an Object Traveling Under Constant Acceleration
Activity 3 Table 1
Time (s)
Average Time (s)
Average Time
2
(s
2
)
Distance (m)
Trial 1 =0
0
0
0
Trial 2 =0
Trial 3 =0
Trial 1 =0.68
0.70
0.49
0.1
Trial 2 =0.70
Trial 3 =0.71
Trial 1 =1.20
1.18
1.39
0.2
Trial 2 =1.20
Trial 3 =1.15
Trial 1 =1.58
1.51
2.28
0.3
Trial 2 =1.45
Trial 3 =1.50
Trial 1 =1.60
1.64
2.69
0.4
Trial 2 =1.66
Trial 3 =1.66
Trial 1 =2.03
2.01
4.04
0.5
Trial 2 =2.03
Trial 3 =1.96
Trial 1 =2.21
2.25
5.06
0.6
Trial 2 =2.31
Trial 3 =2.23
Trial 1 =2.46
2.42
5.86
0.7
Trial 2 =2.45
Trial 3 =2.35
Trial 1 =2.33
2.38
5.66
0.8
Trial 2 =2.31
Trial 3 =2.50
*Note that 0.10 m = 10 cm
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0
1
2
3
4
5
6
7
Displacement vs. time Average Time (s)
Average Time (s^2)
Distance (m)
Time (s)
Activity 3: Questions
1.
What is the shape of the graph when displacement is graphed against time?
The shape of the graph is linear with displacement against time showing the constant velocity with the slope being the velocity. 2.
What is the shape of the graph when displacement is graphed against time squared?
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The shape of the graph is linear with a constant velocity. 3.
What do the shapes of these graphs tell you about the relationship between distance and displacement for an object traveling at a constant acceleration?
The linear graph shows the relationship of a constant velocity. I did not get a parabola which shows accelerated motion.
Activity 4: Predict the Time for a Steel Sphere to Roll Down an Incline
Activity 4 Table 1
Steel Sphere
Acrylic Sphere
A
Length of Track (cm)
(Step 1, use 80 cm)
80 cm
80 cm
B
Angle of Elevation (
) in Degrees (Step 1)
7
7
C
Calculated Time from s = 0 to s = 80 (Formula from Step 2)
3.75
3.75
D
Measured Time from s = 0 to s = 80
(Step 3 with stopwatch)
2.33
2.56
E
% Difference (Step 4)
46.7%
37.7%
Activity 4: Question
1.
What effect does the type of the sphere have on the time of the object to travel the measured distance? Explain.
The mass affects the inertia of the ball traveling down the incline, the steel sphere has a greater mass and moves faster down the incline than the lighter acrylic ball. The steel ball requires more force to move but the resistance has less of an effect on it than the acrylic ball. Activity 5: Demonstrate That a Sphere Rolling Down the Incline Is Moving Under Constant Acceleration
Activity 5: Questions
1.
Describe your observations of the sounds made as the sphere crosses the equally spaced rubber bands (procedure Step 4). (If the sounds are too fast to tell apart, lower the angle of the ramp.)
The sound of the ball traveling over the rubber bands speeds up indicating the ball is accelerating down the incline. 2.
Describe your observations of the sounds made as the sphere crosses the unequally spaced rubber bands (procedure Step 9)? (Use same angle as Step 4.)
The sound of the ball rolling over the rubber bands sounds more constant and doesn’t seem to increase with the acceleration.
3.
Explain the differences you observed, if any, between the sounds with equal spacing and sounds
with unequal spacing.
The differences indicate that the ball traveling over the equally spaced rubber bands we can tell is accelerating by the time it takes to get to the next rubber band, meaning the ball is speeding up taking
less time to travel the same distance. While the different spaced rubber bands are spaced in a way that allows for the speed to increase while maintaining the same time to cross over the next rubber band, making the sound more constant. The ball is still accelerating but the distance is decreasing.
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