PHY-150 M2 Kinematics Lab Report
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School
Southern New Hampshire University *
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Course
150
Subject
Mathematics
Date
Feb 20, 2024
Type
docx
Pages
10
Uploaded by PrivateProtonSalmon11
Kinematics
Bharat Sarma
SNHU
Margaret DeAntoni (m.deantoni@snhu.edu)
Jan 21
st
, 2024
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
Insert your graph here for Distance vs Time of a Moving Object.
0
10
20
30
40
50
60
0
10
20
30
40
50
60
70
80
Distance Vs Time
Times in Seconds
DIstance in Meters
Questions for Activity 1
Question 1:
What is the average speed of the train during the time interval from 0 s to 10 s?
The Average speed = total distance /total time which is 40/10m = 4m/s
Question 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 = 50m to x = 60m.
The v
=
s
2
−
s
1
t
2
−
t
1
, = (60-50)/(30-15) = 10/15 = 0.67m/s.
Question 3:
What does the slope of the line during each time interval represent?
The time graph position in the slop represents the velocity during each time interval. Also, slop of the line represents acceleration and deceleration.
Question 4:
From time t = 35 s until t = 45 s, the train is located at the same position. What is slope of the line while the train is stationary?
The slop = 0 m/s, when train is in rest however slop is 0.
Question 5:
Calculate the average speed of the train as it moves from position x = 70m to x = 55m. What does the sign of the average velocity during this time interval represent? v
=
x
2
−
x
1
t
2
−
t
1
= (55-70)/(50-35) = -15/15 = -1m/s. Which is represent the average displacement.
Question 6:
What is the displacement of the train from time t = 0s until t = 50s?
Its shows the body changed its direction into negative sign.
Question 7:
What is the total distance traveled by the train from time t = 0s until t = 50s?
The total distance travel by train is = 70m + 15m which is 85m.
Question 8. What is the slope of the line during the time interval t = 45 to t = 50?
Slop is = (55-70/(50-45) which is -3m/s.
Question 9: What does the sign of the slope in question 8 represent in terms of the motion of the train?
The question number 8 represent the change in direction of the objects with negative sign also negative velocity.
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Question 10: What is the average velocity of the train during the interval t= 0s to t = 50s?
The Average velocity v
=
s
2
−
s
1
t
2
−
t
1
, = (55 – 0)/(50 – 0) = 55/50 = 1.1m/s.
Question 11: Does the train’s average velocity during the interval t= 0s to t = 50 s provide a complete picture of the train’s motion during this time?
I say No, because here the average velocity is positive however in some time interval, body is in rest. The body also changing direction. The train travel 85m of total distance and with being stopped at 70m.
Activity 2. Calculate the velocity of a moving object.
Activity 2. Table 1
Time (s)
Displacement (m)*
0.00
0.00
0.64
0.25
1.15
0.50
1.67
0.75
2.27
1.00
2.81
1.25
3.36
1.50
3.96
1.75
4.56
2.00
*Note that 0.25 m = 25 cm
Insert a graph of Table 1 here. Include a chart title, axes titles and units.
The table of graph is activity 2 table 1 which is Displacement(m) over Time(s).
0
0.5
1
1.5
2
2.5
Displacement Over Time
Times in Seconds
Displacement in Meters
Activity 2. Table 2
Time (s)
Velocity (m/s)
1
.46
2
.46
3
.46
4
.46
5
.46
6
.46
7
.46
8
.46
Insert a graph of Table 2 here. Include a chart title, axes titles and units.
The table is comparing velocity Over Time (m/s) wit predicted accuracies.
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
0.5
Velocity Over Time (m/s)
Velocity (m/s)
Time(s)
Velocity(m/s)
Activity 3 Graphing the motion of an Object with Constant Acceleration
Activity 3. Data Table 1.
Time (s)
Average Time (s)
Average Time
2
(s
2
)
Distance (m)
Trial 1 =
0.0
0.0
0
Trial 2 =
Trial 3 =
Trial 1 =
0.59
0.35
0.1
Trial 2 =
Trial 3 =
Trial 1 =
0.84
0.71
0.2
Trial 2 =
Trial 3 =
Trial 1 =
1.01
1.02
0.3
Trial 2 =
Trial 3 =
Trial 1 =
1.31
1.72
0.4
Trial 2 =
Trial 3 =
Trial 1 =
1.49
2.22
0.5
Trial 2 =
Trial 3 =
Trial 1 =
1.76
3.1
0.6
Trial 2 =
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Trial 3 =
Trial 1 =
1.90
3.61
0.7
Trial 2 =
Trial 3 =
Trial 1 =
2.1
4.41
0.8
Trial 2 =
Trial 3 =
*Note that 0.10 m = 10 cm
Insert your graphs of Distance vs Time (m) and Distance vs Time Squared here:
The table of graph shows that distance vs time.
0
0.5
1
1.5
2
2.5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Distance (m) vs Time(s)
Time(s)
Distance (m) The following table of graph is Distance vs time squared.
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Distance vs Time Squared
Times(s)^2 (squared)
DIstance(m)
Questions for Activity 3 Question 1: What is the shape of the graph when displacement is graphed vs. time?
It is a straight line; the line is almost Turing into linear with few outliners that maybe cause by human error.
Question 2: What is the shape of the graph when displacement is graphed against time squared?
The shape of the graph of displacement vs time however some outliners are mostly a straight line
closing towards the linear line.
Question 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?
Show they both dependent on each other; the displacement and distance with each other clam dependent.
Activity 4: Predict the time for a steel sphere to roll down an incline.
Steel Sphere
Acrylic Sphere
A
Length of Track (cm) (s)
(Step 1, use 80 cm)
80 cm
80 cm
B
Angle of Elevation (
) in Degrees ⁰
(Step 1)
6
6
C
Calculated Time from s=0 to s=80 (formula from step 2)
1.96
1.96
D
Measured Time from s=0 to s=80
(step 3 with stopwatch)
2.20
1.92
E
% Difference (step 4)
7%
2%
Question for Activity 4: What effect does the type of the sphere have on the time of the object to
travel the measured distance, explain?
The base of the types of spheres it changes the acceleration and resistances of travel of objects is measured in distance.
Activity 5: Demonstrate that a sphere rolling down the incline is moving under constant acceleration.
Questions for Activity 5:
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 discern, lower the angle of
the ramp.)
The frequency of the sound has increased because the sphere made it rolled over the rubber bands. The ball accelerating down the ramp where the rubber bands rise.
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2.
Describe your observations of the sounds made as the sphere crosses the unequally spaced rubber bands (procedure step 9)? (Use same angle as step4).
The sound was same frequency however I could not see it but feel it and the sound do not occur in a pattern. Well, I can not see the sound with my necked eye. I am under the impression that the sphere is accelerating steadily.
3.
Explain the differences you observed if any between the sounds with equal spacing and sounds with unequal spacing.
The sound is more uniform equally with equal spaced rubber bands and occurred faster as the
speed increased. The uniformed sound accords as they are unequally the spaced rubber bands. By giving the impression that ball is moving faster, it is sound like the continuing to accelerate and travel at a study speed when it occurs uneven spacing.
It was very fun lab I was having more fun doing this lab than reding and try to find the solution to the problems. I have many errors doing this lab because I did not follow the direction but at the end I did follow, and I have found the YouTube video the outside the resources. I am going to link the video. https://www.youtube.com/watch?v=fxMSBwXY82A
This link the video is in YouTube and it helps me understand how to follow and answer many questions I had in my mind.