Louis Thomas Detoro Lab 1_ Constant Velocity Motion in 1D

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Arizona State University *

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Apr 3, 2024

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Title of the Experiment: Constant Velocity Motion in One Dimension Student’s name: Louis Thomas Detoro Section SLN: PHY 122 - 14231 TA’s Name: Ayush Kumar Singh, Yash Patil Week of the experiment: 1/16/24 OBJECTIVE (3 points) : This experiment focuses on Newton's 2nd law, the law of motion. This lab is designed to help students understand the concepts of position, distance, displacement, and velocity using
1 dimensional motion with constant velocity. EXPERIMENTAL DATA (3 points): PART 1: Object moving away from the motion sensor. Table 1 Run # Time interval, ( s ) Position ( m ) [Initial & End] Distance ( m ) Displacem ent ( m ) Speed ( m/s ) Velocity ( m/s ) 1a t 1 = 0 s t 2 = 5.8438 s x 1 = 0.0754 x 2 = 1.9274 1.852 m + 1.852 m 0.317 m/s + 0.317 m/s 1b t 1 = 1.5 s t 2 = 1.5313 s x 1 = 0.1404 x 2 = 0.1534 0.013 m + 0.013 m 0.419 m/s + 0.419 m/s 1c t 1 = 5 s t 2 = 5.0313 s x 1 = 1.5887 x 2 = 1.6015 0.013 m + 0.013 m 0.419 m/s + 0.419 m/s Table 2 Run # Slope ( m/s ) Y-intercept ( m ) Name of Physics quantity (i.e. position, distance, etc.) 1 m = 0.4133 m/s b = -0.4773 m Average Velocity 2 m = 0.6191 m/s b = -0.5810 m Average Velocity Using equation 5, substitute in your slope and y-intercept to create an equation that describes the motion of the cart: Run 1: x(t) = 0.4133 m/s* t - 0.04773 m Run 2: x(t) = 0.6191 m/s* t - 0.5810 m Table 3. (Use Logger Pro for statistical data to calculate avg. velocity and uncertainty) Run # Average Velocity, ( m/s ) Standard deviation ( m ) 1 V avg = 0.4116 m/s 0.02237 m 2 V avg = 0.6123 m/s 0.06496 m
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PART 2: Object moving toward the motion sensor. Table 4
Run # Slope ( m/s ) Y-intercept ( m ) Name of Physics quantity (i.e. position, distance, speed, etc.) 3 x = -0.8233 m/s b = 2.772 m Velocity 4 x = -1.227 m/s b = 2.606 m Velocity Using equation 5, substitute in your slope and y-intercept to create an equation that describes the motion of the cart: Run 3: x(t) = - 0.8233 t + 2.772 Run 4: x(t) = - 1.227 t + 2.606 Table 5. (Use Logger Pro for statistical data to calculate avg. velocity and uncertainty) Run # Average Velocity, ( m/s ) Standard deviation ( m/s ) 3 V avg = -0.8118 m/s 0.09717 m/s 4 V avg = -1.162 m/s 0.2935 m/s
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PART 3. Matching position vs. time graphs Run 5: Include in the Discussion the methods used to reproduce the graph.
DATA ANALYSIS ( 10 points ): the section includes equations, calculations and error analysis if required. Be sure all equations are present! PART 1 & 2: Show the equation with the plugged in numbers you used in run 1 & 3 to calculate the cart’s: 1a) t 1 = 0, x 1 = 0.0754 t 2 = 5.8438 , x 2 = 1.9274 1b) t 1 = 1.5, x 1 = 0.1404 t 2 = 1.5313, x 2 = 0.1534 1c) t 1 = 5, x 1 = 1.5887 t 2 = 5.0313, x 2 = 1.6015 Distance : x 2 - x 1 = Distance 1a) 1.9274 - 0.0754 = 1.852
1b) 0.1534 - 0.1404 = 0.013 1c) 1.6015 - 1.5887 = 0.013 Displacement: Distance = Displacement 1a) 1.9274 - 0.0754 = 1.852 1b) 0.1534 - 0.1404 = 0.013 1c) 1.6015 - 1.5887 = 0.013 Average speed: V avg = Displacement / Δt Δt = t 2 - t 1 1a) Δt = 5.8438 - 0 = 5.8438 1b) Δt = 1.5313 - 1.5 = 0.031 1c) Δt = 5.0313 - 5 = 0.031 1a) 1.852 / 5.8438 = 0.317 1b) 0.013 / 0.031 = 0.419 1c) 0.013 / 0.031 = 0.419 Average velocity: V avg = Displacement / Δt Δt = t 2 - t 1 1a) Δt = 5.8438 - 0 = 5.8438 1b) Δt = 1.5313 - 1.5 = 0.031 1c) Δt = 5.0313 - 5 = 0.031 1a) 1.852 / 5.8438 = 0.317 1b) 0.013 / 0.031 = 0.419 1c) 0.013 / 0.031 = 0.419 For run 2 and 4, calculate the percent difference between the average velocity from the slope of the x(t) graph and mean velocity from v(t) graph. % Difference = |v avg - v mean | / ((v avg - v mean )/2) Run 2: average velocity = 0.6191 Run 4: average velocity = -1.227 Run 2: % Difference = | 0.6191 - 0.6123| / (( 0.6191 - 0.6123)/2) = 2 * 100 = 200 % Run 4: % Difference = | 1.227 - 1.126| / (( 1.227 - 1.126)/2) = 2 * 100 = 200 %
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PART 4: Run 6: Show calculation (use eqn. 5) used to make a prediction for the final position of the cart after 2 s from the moment it crosses the 80 cm mark on the track. x(t) = x 0 + (v * (t - t 0 )) Run 6: V = -20 cm/s X 0 = 80 cm t = 2 s t 0 = 0 s x(2) = 80 cm + (-20 cm/s * (2 s - 0 s)) x(2) = 40 cm RESULTS ( 3 points ): Table 6. PARTS 1 & 2 (Report results below with 4 significant figures) Average velocity ( units ) Run # Position vs. Time graph Velocity vs. Time graph % Difference 1 2 3 4 Table 7. PARTS 1 & 2 Equations of motion Run # Substituted Equations of Motion ( equation 5 ) 1 2 3 4 Table 8. PART 4 final position of cart after 2 s Predicted position: Experimental position:
DISCUSSION AND CONCLUSION (10 points): The objective of the experiment is to demonstrate motion in one dimension, and how to find the relationship between position and time when velocity is constant. During this experiment we used a test to determine that the average velocity and speed with a constant velocity are equal or the same when on their own from the initial and final values of position and time. From the results of this experiment we have proven that the average velocity can be calculated from 3 different sections or intervals of the same run, yielding values that are close enough to be considered equal to one another. These results were done multiple times with different limitations producing the same results. During Run 1, the movement is observed in the positive x-axis direction. Using the data gathered in the experiment we calculate the Distance, Displacement, Speed, and Velocity: 1a) t 1 = 0, x 1 = 0.0754 t 2 = 5.8438 , x 2 = 1.9274 1b) t 1 = 1.5, x 1 = 0.1404 t 2 = 1.5313, x 2 = 0.1534 1c) t 1 = 5, x 1 = 1.5887 t 2 = 5.0313, x 2 = 1.6015 Distance : x 2 - x 1 = Distance 1a) 1.9274 - 0.0754 = 1.852 1b) 0.1534 - 0.1404 = 0.013 1c) 1.6015 - 1.5887 = 0.013 Displacement: Distance = Displacement 1a) 1.9274 - 0.0754 = 1.852 1b) 0.1534 - 0.1404 = 0.013 1c) 1.6015 - 1.5887 = 0.013 Average speed: V avg = Displacement / Δt Δt = t 2 - t 1 1a) Δt = 5.8438 - 0 = 5.8438
1b) Δt = 1.5313 - 1.5 = 0.031 1c) Δt = 5.0313 - 5 = 0.031 1a) 1.852 / 5.8438 = 0.317 1b) 0.013 / 0.031 = 0.419 1c) 0.013 / 0.031 = 0.419 Average velocity: V avg = Displacement / Δt Δt = t 2 - t 1 1a) Δt = 5.8438 - 0 = 5.8438 1b) Δt = 1.5313 - 1.5 = 0.031 1c) Δt = 5.0313 - 5 = 0.031 1a) 1.852 / 5.8438 = 0.317 1b) 0.013 / 0.031 = 0.419 1c) 0.013 / 0.031 = 0.419 This process was repeated with 5 other trials all with similar procedures as run 1. Based on the data that were collected and or calculated throughout all 6 experiments, we can conclude that the average velocity and speed with a constant velocity are equal or the same when on their own from the initial and final values of position and time.
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