Plans for the Week and Assignments: 1. Monday(08/30): Introduction to Mechanics, the study of motion. Displacement, speed, velocity, and acceleration. Average vs. instantaneous quantities, mathematically and graphically. HW: Read p.24-34, then solve prob.3, 7, 11, and 16 on pages 50-51. 2. Tuesday(08/31): Applications of the Kinematics equations, and freely falling objects. Analysis of Motion using Calculus. HW: Read p.33-43, then solve prob.22, 33, 40, and 47 on pages 51-53. 3. Wednesday(09/01): LAB experiment on Accelerated Motion. HW: Process Lab data and read p.44-48, then solve prob.53, 55, and 56 on page 54. 4. Thursday(09/02): Post-Lab Discussion. HW: Complete lab report and write Abstract (due Tuesday). 5. Friday(09/03): REVIEW I for Test on Ch.2. HW: Complete Handout and Study for Test. Monday(09/06): Labor Day - No School. 6. Tuesday(09/07): REVIEW II for Test on Ch.2. HW: Complete Handout and Study for Test. 7. Wednesday(09/8): TEST on Ch.2 - Motion in One Dimension. HW: Go to web site for notes on Ch.3 - Vectors. Very Important: If you have any questions or miss a class, see me before school (8:00 - 8:30 AM), during Lunch, or after school. Best to send an email to rpersin@fau.edu.
WEBSITE NOTES: Ch.2 - Motion in One-Dimension. 1. In Physics, the study of motion is known as Mechanics. This is divided into Kinematics, the study of HOW things move, and Dynamics, which is concerned with WHY they move. 2. Galileo (1564-1642) was the first to study motion and developed Kinematics. He performed his experiments Pisa, Italy, frequently dropping objects from the Leaning Tower and/or rolling spheres along level surfaces and down ramps. Isaac Newton (1642-1727), a theoretical physicist, formulated Dynamics by deriving his 3 Laws of Motion and the Law of Universal Gravitation. We will study his work in depth when we get to Chap. 4. 3. Galileo's study of motion produced four motion formulas: (I) Δx = vavg·Δt , with Δx = xf - xi , and vavg= (vi + vf)/2 , (II) vf = vi + a·Δt , (III) vf2 = vi2 + 2a·(xf - xi) , and (IV) xf - xi = vi·t + ½a·(Δt)2 . 4. Instantaneous velocity can be computed by finding the slope of a tangent line at any point on the graph of displacement as a function of time, or v = Δx/Δt. 5. Instantaneous acceleration can be computed by finding the slope of a tangent line at any point on the graph of velocity as a function of time, or a = Δv/Δt. 6. Velocity and acceleration can also be evaluated using another method involving limits and the first derivative. Here, v = dx/dt and a = dv/dt. 7. Integral calculus can be used to derive velocity from acceleration, and displacement from velocity. In terms of Calculus, v = ∫a dt + C. Similarly, x = ∫v dt + C. 8. Displacement, velocity, and acceleration can all be demonstrated graphically. For example, straight lines on a position-time graph indicate constant velocity, determined by the slope. 9. The area under a velocity-time graph is displacement while the slope of the graph indicates acceleration. The area under an acceleration-time graph is the velocity. 10. The acceleration due to gravity is -9.8 m/s2 for Earth at sea level. The negative direction is downward. 11. This is known as free-fall acceleration and is the same for all objects regardless of mass near the surface of the Earth. 12. To solve a motion problem in Physics use these steps: (i) read the problem and identify the given variables (ii) determine what you are asked to solve for (iii) find the correct motion formula to use (iv) use algebra to isolate the unknown (v) substitute-in the given information and simplify. Answers to the Homework: Pages 50-51: #3 ((a) 5 m/s, (b) 1.2 m/s, (c) -2.5 m/s, (d) -3.3 m/s,(e) 0 ) , #7 ((a)-2.4 m/s, (b) -3.8 m/s, (c) 4s) , #11 (1.34x104 m/s2), #16 (13.0 m/s, (b) 10.0 m/s and 16.0 m/s, 6.00 m/s2) Pages 51-53: #22 ((a)-9.75 m/s2, (b)-9.94 m/s2 , (c) -10.2 m/s23), #33 ((a) 4.98x10-9 s), (b) 1.20x1015 m/s2), #40 ((a) -4.90 m, -19.6 m, -44.1 m, (b) -9.80 m/s, -19.6 m/s, -29.4 m/s), #47 ((a) 29.4 m/s, (b) 44.1 m). Page 54: #53 ( (a) a = Jt + ai, v = ½ Jt2 + ait + vi, x = 1/6 Jt3 + ½ ait2 + vit + xi , (b) a2 = ai2 + 2J(v - vi), #55 ( (a) a = 108 m/s3 t + 3.00x105 m/s2 then x = -1.67x107 m/s3 t3 + 1.50x105 m/s2 t2, (b) .003 s, (c) 450 m/s, (d) .900 m) #56 ( .222 s)
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