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Announcements

Announcements. 39 4 students with access to smartPhysics as of this morning. Good Job! Monday ’ s intro lecture is posted on the course website (as well as a “skeleton” version of all the lectures) Tomorrow’s discussion session will have first Group Assignment!

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Announcements

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  1. Announcements • 394 students with access to smartPhysics as of this morning. Good Job! • Monday’s intro lecture is posted on the course website (as well as a “skeleton” version of all the lectures) • Tomorrow’s discussion session will have first Group Assignment! • Second Prelecture/Checkpoint due on Friday, Jan. 10th @ 10 AM • First homework (Units 1 & 2) due on Tuesday, Jan. 14th @ 11:59 pm. • smartPhysics: Progress bar and grades in %; Canvas will have the actual POINTS. • Help Lab is up and running (see web for hours). Take advantage!! • Bring a pad of paper, calculator, and pencil/pen to lecture!!

  2. Diagnostic Results & PHYS 1500 • PHYS 1500 is a 3-credit preparatory course • designed to help students be more successful in PHYS 2210   • 2 sections of PHYS 1500 scheduled concurrently with PHYS 2210 • easy to switch into 1500 if you receive such a recommendation   • PHYS 1500 doesn’t begin until second week of the semester • you can switch during the first week without missing any material

  3. Classical Mechanics Lecture 1: 1-D Kinematics Today's Concepts: a) Displacement, Velocity, Acceleration b) 1-D Kinematics with constant acceleration You should have done Prelecture1 and Checkpoint 1 already!

  4. Problem of the Day (from Exam #1, Spring 2013) A thief races toward the state line at a constant speed of 30 m/s and passes a cop who is parked 1 km from the state line. At the moment the thief passes him, the cop accelerates from rest with a constant acceleration to pursue the thief. What is the cop’s minimum acceleration to catch the thief before he reaches the state line? (Ignore any reaction time.) Thief Cop

  5. Response Cards: ACT! How many snowboarders does it take to change a lightbulb? • 1 • 3 • 5

  6. ACT! A thief races toward the state line at a constant speed of 30 m/s and passes a cop who is parked 1 km from the state line. At the moment the thief passes him, the cop accelerates from rest with a constant acceleration to pursue the thief. Which of the following statements is correct? Thief Cop The thief’s velocity does not change The cop’s velocity is increasing A and B None of the above to the right Velocity is a vector! (Must specify a direction)

  7. Prelecture Example

  8. Prelecture Example

  9. Displacement and Velocity in One Dimension Change in Displacement Time taken

  10. Displacement and Velocity in One Dimension Thev(t)vs.tplot is just theslope of thex(t)vs.tplot Definition: Speed = |v(t)| “magnitude” “absolute value” (an unsigned number)

  11. ACT! A thief races toward the state line at a constant speed of 30 m/s and passes a cop who is parked 1 km from the state line. What is the correct relationship between the thief’s displacement and velocity? Thief C A B D. I’m not sure. x x x t t t v v v t t t

  12. A thief races toward the state line at a constant speed of 30 m/s and passes a cop who is parked 1 km from the state line. slope = constant  constant speed  Instantaneous speed = average speed  Thief x t v t

  13. ACT! A thief races toward the state line at a constant speed of 30 m/s and passes a cop who is parked 1 km from the state line. Thief How much time does the cop have to catch up to the thief? Cop • About 33 seconds • About 0.33 minutes • About 3.3 seconds • I’m not sure

  14. Summary so far: • For the Thief + Cop problem, we’ve determined: • It takes Δt = 33.33 seconds for the thief to cross the state line. • The cop has this much time to catch up, starting from rest. • More generally, we’ve determined: • Velocity is a vector: need to specify magnitude (speed) and direction (in 1D, the direction can be signified by “+” or “-”) • We can easily find the position of an object moving with constant velocity:

  15. At the moment the thief passes him, the cop accelerates from rest with a constant acceleration to pursue the thief. What is the correct relationship between the cop’s displacement and velocity? Thief C A B D. I’m not sure. x x x Cop t t t v v v t t t

  16. Constant Acceleration x a t v t t

  17. Problem of the Day A thief races toward the state line at a constant speed of 30 m/s and passes a cop who is parked 1 km from the state line. At the moment the thief passes him, the cop accelerates from rest with a constant acceleration to pursue the thief. What is the cop’s minimum acceleration to catch the thief before he reaches the state line? (Ignore any reaction time.) Thief Cop A) 2.4 m/s² B) 4.0 m/s² C) 1.0 m/s² D) 1.5 m/s² E) 1.8 m/s² (back of card)

  18. Checkpoint: Rolling Down a Ramp • At t = 0 a ball, initially at rest, starts to roll down a ramp with constant acceleration. Suppose it moves 1foot between t = 0sec and t = 1 sec. • How far does it move between t = 1 sec and t = 2 sec? • A) 1 foot B) 2 feet C) 3 feet D) 4 feet E) 6 feet (back of card)

  19. CheckpointResults At t = 0 a ball, initially at rest, starts to roll down a ramp with constant acceleration. Suppose it moves 1foot between t = 0sec and t = 1 sec. How far does it move between t = 1 sec and t = 2 sec? A) 1 foot B) 2 feet C) 3 feet Less than 5% of you chose D) or E) Some of your answers: If the acceleration of the ball is neither increasing or decreasing then we know that the ball will travel the same distance from t=1 to t=2 as it did from t=0 to t=1. So the ball will travel the same distance in the second interval of time as it did in the first interval. The ball has a constant acceleration of 1 foot per second. It moved 1 foot from 0 to 1, so its moves 2 feet from 1 to 2. The way I saw it, the constant acceleration would be a horizontal line, and thus velocity would be a line. The displacement would then be a parabola, and so two squared is 4, minus the single foot from 0 to 1, and giving 3 feet.

  20. constanta(t) = a What student “C” said: From problem: At t=2 sec: Subtract:

  21. Some other answers: (B) v=v0+at: This is not an explanation (and not applicable to this problem) (D) the relationship is quadratic: Indeed, but you didn’t complete problem • The acceleration is constant. It won't go faster, it won't go slower: Constant acceleration does not mean constant speed/velocity! (C) Need a little help understanding this one. Had to googleit: Please don’t do this!!! Big Brother Google will not be there for you on exams! More importantly, asking Google for a solution (even one with an explanation) is not a good substitute for thinking about the question yourself. You are not graded on whether you get the question right or wrong (for Checkpoints): I want to know what YOU think!

  22. ACT: Displacement and Velocity in 1D Are the plots shown at the left correctly related? YES NO

  23. ACT! The velocity vs. time plot of some object is shown to the right. Which diagram below is the correct displacementvs. timeplot for the same object? A B C

  24. Let’s dig a little deeper Slope of x(t) graph ✔ Positive slope

  25. Let’s dig a little deeper Slope of x(t) graph ✔ Negative slope

  26. Let’s dig a little deeper Slope of x(t) graph ✗ Zero slope

  27. Let’s dig a little deeper ✗ ✔

  28. ACT: Acceleration & Velocity in 1D When is the object slowing down? A: Region 1 B: Region 2 C: Regions 1 & 2 1 2 Hint: “Slowing down”: v(t) and a(t) have have opposite signs “Speeding up”: v(t) and a(t) have the same sign

  29. Some Important Points • Displacement, velocity, and acceleration are separate quantities! • e.g., any one of them can be zero without either of the others being zero • In 1D motion, the sign of a quantity signifies direction (nothing else) • Speeding up and slowing down have nothing to do with sign • Speeding up and slowing down can happen in either direction • It’s best to avoid the term “deceleration”

  30. Main Points A summary of each Unit is available on the “Schedule and Downloads” page of the website (and at the end of each chapter of the book)

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