Mueller.Maxwell_PHY-150Lab1

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Southern New Hampshire University *

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Mechanical Engineering

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Apr 3, 2024

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Kinematics Maxwell Mueller 1/21/2022 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 Insert your graph here for Distance vs Time of a Moving Object. Questions for Activity 1 Question 1: What is the average speed of the train during the time interval from 0 s to 10 s? 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. .66666 m/s Question 3: What does the slope of the line during each time interval represent? Acceleration 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? 0 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? -3m/s, The negative average velocity means the train is moving toward it’s original position during that interval. Question 6: What is the displacement of the train from time t = 0s until t = 50s? 55 m Question 7: What is the total distance traveled by the train from time t = 0s until t = 50s?

85 m Question 8. What is the slope of the line during the time interval t = 45 to t = 50? -3 Question 9: What does the sign of the slope in question 8 represent in terms of the motion of the train? The negative sign means that the train is moving towards its original position. Question 10: What is the average velocity of the train during the interval t= 0s to t = 50s? 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? Looking at only the average velocity during this time would not account for the period in which was stationary or in which the train was moving back toward its original position. The average velocity for the interval is positive and that would. be misleading as to the actual motion of the train during this interval. Activity 2. Calculate the velocity of a moving object. Activity 2. Table 1 Time (s) Displacement (m)* 0 0.77 1.55 2.51 3.26 4.00 4.76 5.50 6.24 *Note that 0.25 m = 25 cm Insert a graph of Table 1 here. Include a chart title, axes titles and units.

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0 1 2 3 4 5 6 0 0.5 1 1.5 2 2.5 Displacement over Time Time (s) Displacement (m) Activity 2. Table 2 Time (s) Velocity (m/s) 1 .3246m/s 2 .324 m/s 3 .3246 m/s 4 .3246 m/s 5 .3246 m/s 6 .3246 m/s 7 .3246 m/s 8 .3246 m/s Insert a graph of Table 2 here. Include a chart title, axes titles and units.

0 1 2 3 4 5 6 7 8 9 0.2 0.22 0.24 0.26 0.28 0.3 0.32 0.34 Velocity over Time Time (s) Velocity (m/s) 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 =.63 .62 .38 0.1 Trial 2 =.60 Trial 3 =.62 Trial 1 =.83 .81 .66 0.2 Trial 2 =.80 Trial 3 =.80 Trial 1 =.91 .91 .83 0.3 Trial 2 =.93 Trial 3 =.90 Trial 1 =1.01 1.01 1.02 0.4 Trial 2 =1.02 Trial 3 =1.00 Trial 1 =1.21 1.22 1.49 0.5 Trial 2 =1.24 Trial 3 =1.20 Trial 1 =1.32 1.32 1.74 0.6 Trial 2 =1.35 Trial 3 =1.30 Trial 1 =1.38 1.37 1.88 0.7 Trial 2 =1.40 Trial 3 =1.34 Trial 1 =1.51 1.51 2.28 0.8 Trial 2 =1.54 Trial 3 =1.49 *Note that 0.10 m = 10 cm

Insert your graphs of Distance vs Time (m) and Distance vs Time Squared here: 0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Distance vs Time Time (s) Distance (m) 0 0.5 1 1.5 2 2.5 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 Distance vs Time Squared Time (s^2) Distance (m)

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Questions for Activity 3 Question 1: What is the shape of the graph when displacement is graphed vs. time? The graph is diagonally moving upward on the y-axis as it moves right on the x-axis. Question 2: What is the shape of the graph when displacement is graphed against time squared? The graph is diagonally moving upward on the y-axis as it moves right on the x-axis. 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? If the object is traveling in only one direction while moving away from its original position, distance and displacement are equal. 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 (  ) in Degrees ⁰ (Step 1) 7 7 C Calculated Time from s=0 to s=80 (formula from step 2) 1.37 1.37 D Measured Time from s=0 to s=80 (step 3 with stopwatch) 1.58 1.55 E % Difference (step 4) 14% 12% 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? The type of sphere did not appear to have any effect in my experiment. I believe that gravity acts on both these objects equally and any change in time between the two is either due to different coefficients of friction or densities. Activity 5: Demonstrate that a sphere rolling down the incline is moving under constant acceleration. Questions for Activity 5:

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 frequency in which the noise of the ball going over the bands seemed to increase as the ball moved further down the track. 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 change in frequency seemed to be less drastic as the bands were spaced unequally. 3. Explain the differences you observed if any between the sounds with equal spacing and sounds with unequal spacing. As the acceleration of the ball is constant, the sounds of the ball passing the equally spaced rubber bands appeared to be increasing in frequency based on the ball’s increasing velocity. In other words, the frequency is increasing because the ball is taking less time to cover the space between the bands. When the bands are unequally placed, this offsets the increase in velocity as the ball must cover (now) varying distances to reach the next band and make the sound.