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Lecture11 - PHYSICS 149 Lecture 11 Chapter 4 4.3 Free Fall...

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PHYSICS 149: Lecture 11 • Chapter 4 – 4.3 Free Fall – 4.4 Motion of Projectiles Lecture 11 Purdue University, Physics 149 1

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Midterm Exam 1 • Wednesday, February 23, 18:30 – 19:30 • Place: PHY 333 • Chapters 1 - 4 • The exam is closed book. • The exam is a multiple-choice test. • There will be ~15 multiple-choice problems. – maximum possible score will be 150 points. • Note that total possible score for the course is 1,000 points (see the course syllabus) • The difficulty level is about the same as the level of textbook problems. • You may make a single crib sheet – you may write on both sides of an 8.5” × 11.0” sheet Lecture 11 2 Purdue University, Physics 149
Midterm Exam 1 • Do not forget to bring – Purdue ID Card, – Crib Sheet, – #2 Pencil (with an eraser), –C a l c u l a t o r • Adaptive learners should contact Prof. Neumeister ASAP. Academic Dishonesty: – Do not cheat! Cheaters will be given an F grade in the course and will be reported to the Dean of Students. • Review session: – During next recitation session – Come and see your TA in the help center (private mini-reviews) Lecture 10 3 Purdue University, Physics 149

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Lecture 11 Purdue University, Physics 149 4 After winning a baseball game, one player drops a glove, while another tosses a glove into the air. How do the accelerations of the two gloves compare after release? A) The one going down is larger B) They are the same C) The one going up is larger ILQ 1
Lecture 11 Purdue University, Physics 149 5 Fred throws a ball 30 mph vertically upward. Which of the following statements are true about the ball s velocity at the very top of its trajectory. (Let up be the positive direction) A) v < 0 B) v = 0 C) v > 0 ILQ 2

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Lecture 11 Purdue University, Physics 149 6 Constant Acceleration •x = x 0 + v 0 t + 1/2 at 2 •v = v 0 + at 2 = v 0 2 + 2a(x-x 0 ) Δ x = v 0 t + 1/2 at 2 Δ v = at v 2 = v 0 2 + 2a Δ x x v a t t t
Bottom Line • Be able to handle these equations at any time! If acceleration a is constant, x f = x i + v i (t f –t i ) + ½ a (t f –t i ) 2 (x f –x i ) = ½ (v f +v i ) (t f –t i ) v f = v i + a (t f –t i ) v f 2 –v i 2 = 2 a (x f –x i ) Note: If a is not const, you should not use above equations. Lecture 11 7 Purdue University, Physics 149

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Graphs for Motions with Const Acceleration Í const a = 0 Í const a > 0 Í const a < 0 Lecture 11 8 Purdue University, Physics 149
Kinematics Example A car is traveling 30 m/s and applies its breaks to stop. Assuming constant acceleration of -6 m/s 2 , how long does it take for the car to stop, and how far does it travel before stopping?

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This note was uploaded on 04/23/2011 for the course PHYS 149 taught by Professor Staff during the Spring '08 term at Purdue.

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Lecture11 - PHYSICS 149 Lecture 11 Chapter 4 4.3 Free Fall...

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