Bundor-PHY150-M2-KinematicsLabReport

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Southern New Hampshire University *

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Physics

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Apr 25, 2024

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PHY-150 Intro Physics: Module Two Assignment C Daniel T Bundor Southern New Hampshire University PHY-150-H7663 Intro Physics: Mechanics 24EW4 Professor Ralph Spraker March 30, 2024 PHY-150 Intro Physics: Module Two Assignment
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 Questions for Activity 1 Question 1: What is the average speed of the train during the time interval from 0 s to 10 s? Average Speed = Total Distance / Total Time = 40 meters / 10 seconds = 4 meters per second 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. V=60m-50m 30s-15s V=10m 15s V=10m 15s
V=0.667 m/s Question 3: What does the slope of the line during each time interval represent? The slope of the line on a position-time graph represents the velocity of the object being tracked. 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 slope of the line while the train is stationary (from t=35t=35 s to t=45t=45 s) is zero. This is because the position of the train remains constant during this time interval, meaning there is no change in position with respect to time. 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? Total Time=t2−t1=50s−40s=10s Now, plug these values into the formula for average speed: Average Speed=15 m10 s=1.5 m/s The sign of the average velocity during this time interval represents the direction of motion. Since the velocity is positive (1.5 m/s), it indicates that the train is moving towards the positive direction, which means it's moving to the left on the position- time graph. Question 6: What is the displacement of the train from time t = 0s until t = 50s? Δx=x2−x1 Δx=55 m−0 Δx=55 m So, the displacement of the train from t=0s until t=50s is 55 meters. Question 7: What is the total distance traveled by the train from time t = 0s until t = 50s? Total Distance= 20−0 + 40−20 + 50−40 + 55−50 + 60−55 + 70−60 + 70−70 + 70− 70 + 55−70 Total Distance=20+20+10+5+5+10+0+0+15 Total Distance=85 meters So, the total distance traveled by the train from t=0t=0 s until t=50t=50 s is 8585 meters. Question 8. What is the slope of the line during the time interval t = 45 to t = 50?
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