PHY 101L Module Two Lab Report
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PHY 101L Module Two Lab Report
Name: Sara Rigby
Date: 03/16/24
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
5
10
15
20
30
35
40
45
50
0
10
20
30
40
50
60
70
80
Activity 1 Graph 1
Time (sec)
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 is calculated using speed = distance/time (speed = 40m/10s) therefore the average speed 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.
I calculated the average speed using the calculations: v= (60-50)/(30-15) to get the answer of 6.7 m/s.
3.
What does the slope of the line during each time interval represent?
The slope of line represents the change in velocity or speed of the train. 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 zero because there is no change in the position of the train for that time.
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? I calculated the average speed using the calculations: v= (55-70)/(50-45); v=-1. The negative sign represents that the train is moving backwards compared to the initial movement or negative to positive on the plane.
6.
What is the displacement of the train from time t
= 0 s until t
= 50 s?
The displacement would be 55m, calculated by: 55m – 0 m = 55m. 7.
What is the total distance traveled by train from time t
= 0 s until t
= 50 s?
From t = 0 until t = 40s it covers a distance of 70 m
Then from t = 45s until t = 50s it covers another (70 - 55) = 15 m
So, total distance covered = 70 + 15 = 85 meters.
8.
What is the slope of the line during the time interval t
= 45 s to t
= 50 s?
The slope during this interval would be -3 caluclated by: (55-70)/(50-45)= (-15/5) = -3
9.
What does the sign of the slope in Question 8 represent in terms of the motion of the train?
The negative sign on the slop in question 8 demonstrates that the train is moving in the opposite direction from the initial movement.
10.
What is the average velocity of the train during the interval t
= 0 s to t
= 50 s?
The average velocity is 1.1m/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, it does not provide a complete picture of the trains motion because it only reflects the average, no other information is available. Activity 2: Calculate the Velocity of a Moving Object
Activity 2 Table 1
Time (s)
Displacement (m)*
0.00
0.00
0.061
0.25
1.23
0.50
1.86
0.75
2.42
1.00
3.07
1.25
3.68
1.50
4.12
1.75
4.75
2.00
*Note that 0.25 m = 25 cm
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0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0
0.5
1
1.5
2
2.5
Time vs Displacement of Battery Powered Car
Time(s)
Displacement (m)
(*0.25 m = 25cm)
Activity 2 Table 2
Time (s)
Velocity (m/s)
1
0.42
2
0.42
3
0.42
4
0.42
5
0.42
6
0.42
7
0.42
8
0.42
0
1
2
3
4
5
6
7
8
9
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
Time vs Velocity of Battery Powered Car
Velocity (m/s)
Time (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 =0.63
0.67
0.45
0.1
Trial 2 =0.66
Trial 3 =0.71
Trial 1 =0.89
0.94
0.88
0.2
Trial 2 =0.98
Trial 3 =0.94
Trial 1 =1.25
1.27
1.60
0.3
Trial 2 =1.28
Trial 3 =1.27
Trial 1 =1.36
1.35
1.81
0.4
Trial 2 =1.33
Trial 3 =1.35
Trial 1 =1.39
1.46
2.13
0.5
Trial 2 =1.51
Trial 3 =1.48
Trial 1 =1.77
1.77
3.12
0.6
Trial 2 =1.79
Trial 3 =1.74
Trial 1 =1.99
1.98
3.95
0.7
Trial 2 =2.01
Trial 3 =1.96
Trial 1 =2.05
2.05
4.20
0.8
Trial 2 =2.08
Trial 3 =2.02
*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
0.5
1
1.5
2
2.5
Distance vs Average Time of Steel Sphere Distance (m)
*(0.10 m = 10cm)
Time (s)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
Displacement vs Average Time Squared of Steel Sphere
Displacement (m)
*(0.10m = 10cm)
Time (s)
Activity 3: Questions
1.
What is the shape of the graph when displacement is graphed against time?
The graph forms a generally upwards diagonal line.
2.
What is the shape of the graph when displacement is graphed against time squared?
The graph is a steeper upward diagonal line.
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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 shapes of the graphs indicate that they share positive common trendline.
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)
8
o
8
o
C
Calculated Time from s = 0 to s = 80 (Formula from Step 2)
1.60
1.70
D
Measured Time from s = 0 to s = 80
(Step 3 with stopwatch)
1.57
1.65
E
% Difference (Step 4)
1.89%
0.03%
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.
T
he steel ball has a greater mass which creates faster acceleration due to gravity.
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 sounds of the sphere hitting the equally spaced rubber bands were also equally spaced and rhythmic. When they were equally spaced the ball also were quicker because the ball was able to gain more acceleration compared to the unevenly spaced bands.
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.)
When the bands were unevenly spaced it was not possible to tell when the sound would happen and not rhythmic at all. The ball also seemed to move slower between the bands because of a lack of acceleration between the closely placed bands.
3.
Explain the differences you observed, if any, between the sounds with equal spacing and sounds
with unequal spacing.
The sounds of the steel sphere were louder than the acrylic sphere.
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- I need help solving this problem.arrow_forwardFor the Following question Graph all 4 : [I just need all 4 graphs and please explain and make clean solution] Position vs time Velocity vs time Acceleration vs time Force vs time [For your convenience, I have solved the numerical solutions for the problem] (Please Look at the picture since it is much cleaner) Question : A 550 kilogram mass initially at rest acted upon by a force of F(t) = 50et Newtons. What are the acceleration, speed, and displacement of the mass at t = 4 second ? a =(50 e^t)/(550 ) [N/kg] v = ∫_0^t▒(50 e^t )dt/(550 )= v_0 +(50 e^t-50)/550=((e^t- 1))/11 x = ∫_0^t▒(e^t- 1)dt/(11 )= x_0 +(e^t- t - 1)/(11 ) a(4s)=(50*54.6)/550= 4.96[m/s^2 ] v(4s)=((e^4-1))/11= 4.87[m/s] x(4s)=((e^4- 4 - 1))/11= 4.51 [m]arrow_forwardK mylabmastering.pearson.com Chapter 12 - Lecture Notes.pptx: (MAE 272-01) (SP25) DY... P Pearson MyLab and Mastering Mastering Engineering Back to my courses Course Home Scores Course Homearrow_forward
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