Physics114A_L09

# Physics114A_L09 - Lecture 9(Walker Ch 4 More 2D Kinematics...

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January 20, 2009 Physics 114A - Lecture 9 1/19 Physics 114C - Mechanics Lecture 9 (Walker: Ch. 4) More 2D Kinematics January 20, 2009 John G. Cramer Professor of Physics B451 PAB [email protected]

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January 20, 2009 Physics 114A - Lecture 9 2/19 Announcements Homework Assignments #2 and #3 are now available on Tycho. Assignment #2 is due at 11:59 PM on Thursday, January 22. Assignment #3 is due at 11:59 PM on Thursday, January 29. Register your clicker using the “Clicker” link on the Physics 114A Syllabus page. So far 147/208 students have registered their clickers. Tycho now contains currrent clicker scores for those who have registered as “Lecture Score”, giving clicker ID number and points so far (1 point for participation and 1 point for a correct answer). We will have our first Midterm Exam on Friday, January 23. It will have 60 points of multiple-choice questions based on end-of-chapter problems, a 20 point free- response question based on Tycho homework, and a 20 point free-response question testing qualitative understanding of the Chapters covered,
January 20, 2009 Physics 114A - Lecture 9 3/19 Lecture Schedule (Part 1) Physics 114A - Introduction to Mechanics Lecture: Professor John G. Cramer Textbook: Physics, Vol. 1 (UW Edition), James S. Walker Week Date L# Lecture Topic Pages Slides Reading HW Due Lab 1 5-Jan-09 1 Intro: Physics & Units 12 21 Chapter 1 No Lab 1st week 6-Jan-09 2 Position & Velocity 8 22 2-1 to 2-3 8-Jan-09 3 Velocity & Acceleration 19 25 2-4 to 2-7 9-Jan-09 4 Vectors 8 24 3-1 to 3-3 2 12-Jan-09 5 r, v & a Vectors 5 24 3-4 to 3-5 1-D Kinematics 13-Jan-09 6 Relative Motion 3 20 3-6 15-Jan-09 7 2D Motion Basics 5 20 4-1 to 4-2 16-Jan-09 8 2D Examples 13 23 4-3 to 4-5 3 19-Jan-09 H1 Martin Luther King Holiday Free Fall & Projectiles 20-Jan-09 9 More 2-D Motion - 19 - We are here.

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January 20, 2009 Physics 114A - Lecture 9 4/19 Two-Dimensional Kinematics
January 20, 2009 Physics 114A - Lecture 9 5/19 Example: A Rocket Sled A rocket sled accelerates at 50 m/s 2 for 5.0 s, coasts for 3.0 s, then deploys a braking parachute and decelerates at 3.0 m/s 2 until coming to a halt. (a) What is the maximum velocity reached by the rocket sled? v 1x = v 0x + a (t 1 – t 0 ) = a t 1 = (50 m/s 2 )(5.0 s) = 250 m/s (b) What is the total distance traveled by x 1 = x 0 + v 1 0 ) + ½a 1 0 ) 2 = ½a t 1 2 = ½(50 m/s 2 2 = 625 m x 2 1 1x 2 1 ) = 625 m + (250 m/s)(3.0 s) = 1375 m v 3x 2 = 0 = v 2x 2 +2a x = v 2 (x 3 – x 2 ), so x 3 2 + (v 2 - v 2 )/2a = 1375 m + [0 – (250 m/s) 2 ]/[2(-3.0 m/s 2 )] = 11,800 m

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January 20, 2009 Physics 114A - Lecture 9 6/19 Projectile Motion in Vector Form 0; ; 0 x y z a a g a = = - = ˆ ˆ ˆ ˆ or x y z a i a j a k gj a g + + = - = r r 2 9.81 m/s (at sea level and at 45 latitude) g = ° 0 0 ; x x y y v v v v gt = = - 0 or where v v gt v gt = + ∆ = r r r r r 0 0 0 0 ˆ ˆ ˆ ; ˆ ˆ
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## This note was uploaded on 02/24/2009 for the course PHYS 114 taught by Professor Reid during the Spring '08 term at University of Washington.

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Physics114A_L09 - Lecture 9(Walker Ch 4 More 2D Kinematics...

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