Morgan_Adams_PHY151_LAB2
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Central Piedmont Community College *
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151
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Mathematics
Date
Feb 20, 2024
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Uploaded by ProfFireStork20
Morgan Adams
Kinematics Lab 02
Professor Marco Scipioni
06/15/2021
Constant Velocity
6. Use the Curve-Fit tool to fit the data to a linear function. What is the equation returned
from the curve-fit, and to which Equation (1-4) from the Theory section does it correspond?
The equation returned from the curve-fit was, mt+b. The equation from the theory section it corresponds to is v=v₀ + at.
7. What is the slope of the line? What does it rep-resent?
The slope of the line is 0.0987. The slope of the line represents the velocity of the cart.
8. What is the y-intercept? What does it represent? The y-intercept is 0.200. The y-intercept represents the carts starting position.
9. In a position vs. time graph, why does a straight line signify constant velocity motion? In a position vs. time graph, the velocity of the moving object is represented by the slope of the graph line. If the slope remains the same then the velocity is constant. If the line on the graph was curved that would represent a changing velocity. 10. Select the Velocity tab. Describe the graph displayed. Is it what you expected? Why or why not? The graph that is displayed is showing a straight line for velocity which is what is expected. The cart has a setting that applies a constant speed to the wheels of the cart. There are a few spots on the graph that
are deviating from what would be expected. This could be caused by the battery that is in the cart and the efficiency of the motor. The other deviations could be caused by how the cart was placed on the track when the recordings were started.
11. Use the Coordinates tool to determine the velocity of the cart. Obviously, you want to place it such that the vertical location best represents the velocity of the cart. What is this value? The velocity of the cart the coordinate’s tool has determined is 0.10 m/s.
12. In the table, use the Statistics tool to display the Maximum, Minimum, and Average values. What are
these values? How does the average velocity compare to the value in Step 7? Step 11? Table Values
Maximum: 0.12 m/s
Minimum: 0.07 m/s
Mean: 0.10 m/s
The average velocity vs the value from step 7 only has a difference of 0.0013. The difference in value from step 7 is within the uncertainty range and could be caused by several factors based on how and when the data recordings were started and how the cart was placed on the track.
The average velocity vs the value from step 11 are the same value of 0.10 m/s.
13. Go back to the graph and use the Area tool to determine the area under the curve. What is this value, and what does it represent? The area under the curve is 0.59m/s. The area under the graph represents the total displacement or how far the cart travelled.
14. Does the value returned seem reasonable based on what you did in the procedure? Why or why not? The value returned seems reasonable as that is the approximate distance of the track the cart travelled on. The starting position was found based off the y-intercept of point 0.200. According to the graph the cart ended at 0.80. So, 0.80–0.20=0.60 which is only a slight difference of 0.01.
15. Select the Acceleration tab. What rate of acceleration do you expect for the cart? What is the average value from the table? Can you determine a value from the graph? I would expect the rate of acceleration to be zero since the velocity vs. time graph showed a straight line. The average value from the table is 0.0. A value can be determined from the table, as it shows a minimum of 0.02 and a maximum of 0.2.
16. Keep the data you have collected, and repeat the procedure with the cart at maximum velocity. Describe how the position vs. time graph is different now that the velocity is higher. Answer also for the velocity vs. time graph.
The position vs. time graph with the cart at maximum velocity has a larger slope indicating that the cart was set to maximum velocity and the graph now shows the cart is reaching its final position in a shorter amount of time.
Positive Acceleration
5. Select the Position tab. On the graph, fit the data to a Quadratic function. What is the equation returned from the curve-fit, and to which Equation (1-4) from the Theory section does it correspond? The equation returned from the curve-fit is At^2 + Bt + C. The equation from the theory section it corresponds to is x=x₀ + v₀t + 1/2at²
6. Using the Slope tool determine the slope of the curve at three different points - one near the beginning of the motion, one near the midpoint, and one near the end. What are the values returned? Slope of the curve at the beginning of the motion: m = 0.839
Slope of the curve at the midpoint: m = 1.196
Slope of the curve near the end: m = 1.444
7. According the slope values above, what is happening to the velocity of the cart as it moves down the track? Is it decreasing, constant, or increasing? Is this what you expected? Why or why not? According the slope values above the carts velocity is increasing. This is what would be expected as gravity and the slope of the track is causing the cart to accelerate forward.
8. Select the Velocity tab. On the graph, fit the data to a Linear function. What is the equation returned from the curve-fit, and to which Equation (1-4) from the Theory section does it correspond? The equation returned from the curve-fit is mt+b. The equation from the theory section it corresponds to is v=v₀ + at.
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9. What is the acceleration of the cart according to this graph? The acceleration of the cart according to the graph is 1.6 m/s²
10. What is the y-intercept? What does it represent? The y-intercept is 0.787. The y-intercept represents the carts starting position.
11. Determine the acceleration of the cart from another graph or table. State the value below and explain how you obtained it.
The positive acceleration graph showed 1.6m/s². The cart picked up speed as it was traveling down the track and compared to the constant velocity graph the acceleration was zero because the cart was not accelerating as it was moving at a constant velocity. The constant velocity graph also shows a constant slope where the position vs time graph for the cart that is on a slope has a slope that is increasing over time.