2-1 M2 Kinematics Lab Report

docx

School

Southern New Hampshire University *

*We aren’t endorsed by this school

Course

150

Subject

Physics

Date

Dec 6, 2023

Type

docx

Pages

8

Uploaded by JusticeDanger4006

Report
Kinematics Kyle Blan 1/23/22 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 . 0 5 10 15 20 30 35 40 45 50 0 10 20 30 40 50 60 70 80 Train Time (s) Displacement (m) 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. 0.6667 m/s Question 3: What does the slope of the line during each time interval represent?
The slope of the line represents the current speed of the train. 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 line during this period is horizontal, representing an average speed of 0 m/s. 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? -3 m/s, which represents a reverse in direction of travel. Question 6: What is the displacement of the train from time t = 0s until t = 50s? 55 meters Question 7: What is the total distance traveled by the train from time t = 0s until t = 50s? 85 meters Question 8. What is the slope of the line during the time interval t = 45 to t = 50? The slope of the line is m = -3/1 Question 9: What does the sign of the slope in question 8 represent in terms of the motion of the train? The direction of the train has been reversed. Question 10: What is the average velocity of the train during the interval t= 0s to t = 50s? The average velocity of the train during the period in question is 1.7 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? The average velocity does not provide a complete picture of the trains motion, due to the change in direction between the interval of t= 45s to t=50s. Activity 2. Calculate the velocity of a moving object.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Activity 2. Table 1 Time (s) Displaceme nt (m)* 0 0.00 0.76 0.25 1.57 0.50 2.43 0.75 3.91 1.00 4.74 1.25 5.21 1.50 6.46 1.75 7.94 2.00 *Note that 0.25 m = 25 cm 0 0.25 0.5 0.75 1 1.25 1.5 1.75 2 0 1 2 3 4 5 6 7 8 9 Model Car Displacement (m) Time (s) Activity 2. Table 2 Time (s) Velocity (m/s) 1 0.33 2 0.62 3 0.77
4 1.05 5 1.44 6 1.63 7 1.76 8 2.02 1 2 3 4 5 6 7 8 0 0.5 1 1.5 2 2.5 0.33 0.62 0.77 1.05 1.44 1.63 1.76 2.02 Model Car (Velocity) 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 Trial 2 = Trial 3 = Trial 1 = 1.36 1.35 1.82 0.1 Trial 2 =1.32 Trial 3 =1.37 Trial 1 =1.71 1.72 2.96 0.2 Trial 2 =1.74 Trial 3 =1.7 Trial 1 =2.03 2.02 4.08 0.3 Trial 2 =2.05 Trial 3 =1.97 Trial 1 =2.45 2.44 5.95 0.4 Trial 2 =2.43 Trial 3 =2.44 Trial 1 =2.67 2.68 7.18 0.5 Trial 2 =2.70 Trial 3 =2.68 Trial 1 =2.84 2.81 7.89 0.6 Trial 2 =2.79 Trial 3 =2.8 Trial 1 =2.97 2.97 8.82 0.7 Trial 2 =2.96 Trial 3 =2.99
Trial 1 =3.31 3.34 11.16 0.8 Trial 2 =3.34 Trial 3 =3.37 *Note that 0.10 m = 10 cm 0 0.2 0.4 0.6 0.8 0 2 4 6 8 10 12 Steel Ball Average Time Average Time squared Displacement (m) Time (s)
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Questions for Activity 3 Question 1: What is the shape of the graph when displacement is graphed vs. time? The resulting line is increasing. Question 2: What is the shape of the graph when displacement is graphed against time squared? The resulting line is increasing more than the time. 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 any graph that shows a constant acceleration will always be an upward trending line. 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 (Step 1) 10 10 C Calculated Time from s=0 to s=80 (formula from step 2) 4.82 4.82 D Measured Time from s=0 to s=80 (step 3 with stopwatch) 3.04 3.04 E % Difference (step 4) 145% 145% 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 less mass the sphere has, the slower it will accelerate on the track. 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.) I observed that the spheres were making adequately spaced sounds on my first trials as it accelerated across the equally spaced rubber bands. 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 sounds made became quicker as the ball accelerated down the ramp. 3. Explain the differences you observed if any between the sounds with equal spacing and sounds with unequal spacing. The main difference between the sounds heard during the equal and unequal spacing, was that the equally spaced rubber bands made the acceleration much more noticeable, due to the consistency of the sound.