PHY-150 M2 Kinematics Lab Report (1)_KABBOTT (1)

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Apr 3, 2024

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Kinematics Keon Abbott November 3 rd , 2023 PHY-150-J2326

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? The average speed of the train during the time interval of 0 s to 10 s was 4 m/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. 0.67 (or 0.7) m/s. Question 3: What does the slope of the line during each time interval represent? The slope represents the train’s acceleration during each time interval. 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 is zero (0) because the position does not change while the train is stationary. 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? The answer is –3m/s when calculating for average speed. The fact that the answer is negative indicates that the train is moving either in reverse or basically in the opposite direction. Question 6: What is the displacement of the train from time t = 0s until t = 50s? The displacement of the train can be calculated as 55s – 0s = 55m. Question 7: What is the total distance traveled by the train from time t = 0s until t = 50s? The total distance is 85m . This is calculated by adding 70m with 15m. From 0-35 seconds, it is 70m. From 35-55 seconds it is 15m. Adding both gives you the answer of 85m. Question 8. What is the slope of the line during the time interval t = 45 to t = 50? The slope is calculated by (55-70) divided by (50-45). This gives you the answer of –3 . Question 9: What does the sign of the slope in question 8 represent in terms of the motion of the train? The answer has a negative sign which indicates that the train is moving in reverse or essentially in the opposite direction that it was initially in. Question 10: What is the average velocity of the train during the interval t= 0s to t = 50s?

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The average velocity can be calculated as (55-0) divided by (50-0) which gives you an answer of 1.1 m/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? This does not provide a complete picture of the train’s motion during this time because although the average velocity is positive, it has been proven that the train does move in reverse at some point. Activity 2. Calculate the velocity of a moving object. Activity 2. Table 1 Time (s) Displacement (m)* 0 0.69 1.54 2.42 3.15 3.89 4.64 5.41 6.14 Chart can be seen below on next page- *Note that 0.25 m = 25 cm

Activity 2. Table 2 Time (s) Velocity (m/s) 1 0.39 2 0.39 3 0.39 4 0.39 5 0.39 6 0.39 7 0.39 8 0.39 Graph shown below in next page:

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 Trial 2 = 0 Trial 3 = 0 Trial 1 = 0.65 0.67 0.44 0.1 Trial 2 = 0.69 Trial 3 = 0.66 Trial 1 = 0.85 0.83 0.69 0.2 Trial 2 = 0.83 Trial 3 = 0.81 Trial 1 = 0.88 0.88 0.78 0.3 Trial 2 =

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0.90 Trial 3 = 0.87 Trial 1 = 1.01 1.06 1.12 0.4 Trial 2 = 1.08 Trial 3 = 1.09 Trial 1 = 1.27 1.27 1.62 0.5 Trial 2 = 1.30 Trial 3 = 1.25 Trial 1 = 1.35 1.37 1.88 0.6 Trial 2 = 1.39 Trial 3 = 1.37 Trial 1 = 1.42 1.44 2.08 0.7 Trial 2 = 1.47 Trial 3 = 1.44 Trial 1 = 1.58 1.60 2.57 0.8 Trial 2 = 1.60 Trial 3 = 1.63 *Note that 0.10 m = 10 cm

Questions for Activity 3

Question 1: What is the shape of the graph when displacement is graphed vs. Time? The shape of the graph is a diagonal line that increases in height as it moves further right. Question 2: What is the shape of the graph when displacement is graphed against time squared? The answer is like the previous question. It is the same shape as it is a diagonal line that increases in height as it moves to the right. 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? The shape of the graph tells me with clarity that if an object is accelerating consistently and the distance increases, so will the displacement. 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 ( q ) in Degrees ⁰ (Step 1) 10 10 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) 2.12 1.91 E % Difference (step 4) 8% 3% 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?

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The resistance and acceleration differ between two of the spheres. The acrylic sphere seemed to move faster than the steel- perhaps because it had less friction. Activity 5: Demonstrate that a sphere rolling down the incline is moving under constant acceleration. Questions for Activity 5: This activity posed a challenge because I cannot hear the sounds these make naturally because I am Deaf. However, I had my father- in-law describe the sounds to me, so this observation is based on what he heard. I tried my best to have my hand close by so I could possibly feel vibrations so I could have a better understanding of the sound. 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 sound of the sphere was a clicking-like sound that increased as the rubber bands were equally spaced. The further the sphere got, the faster the clicking sound occurred. 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 of the sphere was somewhat like the clicking-like noise. The difference is that when the rubber bands were not evenly spaced, the sound did not sound cohesive and sounded uneven. 3. Explain the differences you observed if any between the sounds with equal spacing and sounds with unequal spacing. The sound was much more uniform when the rubber bands were evenly spaced, and the sound increased the further the sphere went down the ramp. This was not the case when the rubber bands were not evenly spaced. It sounded disjointed and was not the same.