Fall22 Postion, Velocity, and Acceleration Lab Online EDITED 8.16.22-1
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Health and Science School *
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BSBSUS401
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Mathematics
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Jun 18, 2024
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6
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1 Analysis of Position, Velocity and Acceleration Lab Name: Nishat Shama Course/Section: PHY-1611-02F-202430
Instructor: Tyler Rucas Table 1 (10 points) t(s) x(m) v(m/s) a(m/s
2
) 0.00 0.00 0.00 0.75 1.005 .38 .75 .75 2.042 1.56 1.53 .75 3.006 3.39 2.25 .75 4.173 6.53 3.13 .75 5.127 9.86 3.85 .75 6.012 13.55 4.51 .75 7.097 18.89 5.32 .75 8.128 24.77 6.10 .75 9.005 30.41 6.75 .75 10.087 38.16 7.57 .75 1.
From the data in Table 1, and using Excel or some other graphing software, make the 3 following graphs: a.
Position vs. time b.
Velocity vs. time c.
Acceleration vs. time For the Position vs. time graph have software display BOTH the linear fit, and the quadratic fit on the graph. For the Velocity vs. time graph have the software display the linear fit on the graph. For the acceleration vs. time graph have the software display the linear fit on the graph. Make sure to turn these graphs in with the lab worksheet. (20 points)
2 y = 3.7727x - 5.6883
y = 0.3751x
2
- 0.0008x + 0.0007
-10
-5
0
5
10
15
20
25
30
35
40
45
0
2
4
6
8
10
12
Position (m)
Time (s) Position vs. Time
y = 0.7504x - 0.0022
-1
0
1
2
3
4
5
6
7
8
0
2
4
6
8
10
12
Velocity (m/s)
Time (s) Velocity vs. Time
3 Table 2 (From the ‘fits’ displayed on your graphs fill in Table 2)
(10 points)
Position vs Time Value Linear Fit m 3.7727x B - 5.6883 y = mx + b y = 3.7727x - 5.6883 Quadratic Fit A 0.3751x
2
B - 0.0008x C 0.0007 y = Ax
2 + Bx + C y = 0.3751x
2
- 0.0008x + 0.0007 (x
1
, y
1
) first point (0, 0) (x
2
, y
2
) last point (10.087,38.16) Slope 3.783 Velocity vs. Time Value M 0.7504x B - 0.0022 y = mx + b y = 0.7504x - 0.0022 (x
1
, y
1
) first point
(0, 0) (x
2
, y
2
) last point (10.087,7.57) v
avg .7504 v
avg
time 3.7727 y = -5E-17x + 0.75
0.7497
0.751275
0
2
4
6
8
10
12
Acceleration ( m/s2)
Time (s) Acceleration vs. Time
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4 Acceleration vs. Time Value M -5E-17x B .75 m/s^2 y = mx + b y = -5E-17x + 0.75 1.
What are the appropriate units for the slope of the:
(a)
Position vs Time graph? (2 points)
m/s
(b)
Velocity vs Time graph? (2 points)
m/
𝑠
2
(c)
Acceleration vs Time graph? (2 points)
m/
𝑠
3
2.
For Position vs Time data:
(a)
Did your quadratic fit of this graph provide initial position? If yes, what is its value?
(4 points)
Yes, the initial position is 0.0007 meters. (b)
Did your quadratic fit of this graph provide initial velocity? If yes, what is its value?
(4 points)
Yes, it does 0.0008 m/s. (c)
Did your quadratic fit of this graph provide acceleration? If yes, what is its value?
5 (4 points)
Yes, .07502 m/
𝑠
2
(d)
What specific physical quantity does the slope of the two middle points from the Position vs. Time graph represent?
(4 points)
It represents the average velocity. 3.
For Velocity vs Time data:
(a)
Did your linear fit of this graph provide initial position? If yes, what is its value?
(4 points)
No (b)
Did your linear fit of this graph provide initial velocity? If yes, what is its value?
(4 points)
Yes, it was - 0.0022 m/s. (c)
Did your linear fit of this graph provide acceleration? If yes, what is its value?
(4 points)
Yes, it was 0.7504 m/
𝑠
2
4.
For Acceleration vs Time data:
(a)
Did your linear fit of this graph provide initial position? If yes, what is its value?
(4 points)
Yes, it is 0.75 meters. (b)
Did your linear fit of this graph provide initial velocity? If yes, what is its value?
(4 points)
Yes, -5E-17 m/s. (c)
Did your linear fit of this graph provide acceleration? If yes, what is its value?
(4 points)
6 No
(d)
What are the SI units of the Jerk?
(4 points)
m/
𝑠
3
7. What is the general shape of each graph and why does each have that shape? (10 points)
For the position vs. time my graph is a parabola because the object is accelerating at a constant rate, making position a quadratic function of time for velocity vs. Time the graph is a linear because velocity changes at a constant rate due to constant acceleration. For the Acceleration vs. Time graph, it is a flat line because acceleration is constant.
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