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Unformatted text preview: Chapter 3 Kinematics in Two Dimensions 37 Chapter 3 KINEMATICS IN TWO DIMENSIONS PREVIEW Twodimensional motion includes objects which are moving in two directions at the same time, such as a projectile , which has both horizontal and vertical motion. These two motions of a projectile are completely independent of one another, and can be described by constant velocity in the horizontal direction, and free fall in the vertical direction. Since the twodimensional motion described in this chapter involves only constant accelerations, we may use the kinematic equations . The content contained in sections 1, 2, 3, and 5 of chapter 3 of the textbook is included on the AP Physics B exam. QUICK REFERENCE Important Terms projectile any object that is projected by a force and continues to move by its own inertia range of a projectile the horizontal distance between the launch point of a projectile and where it returns to its launch height trajectory the path followed by a projectile Equations and Symbols Horizontal direction: x a v v t a t v x t v v x t a v v x ox x x ox x x o x ox x 2 2 1 ) ( 2 1 2 2 2 Vertical direction: y a v v t a t v y t v v y t a v v y oy y y oy y y o y oy y 2 2 1 ) ( 2 1 2 2 2 For a projectile near the surface of the earth: a x = 0, v x is constant, and a y = g = 10 m/s 2 . Chapter 3 Kinematics in Two Dimensions 38 Ten Homework Problems Chapter 3 Problems 12, 13, 16, 22, 25, 28, 39, 43, 64, 71 DISCUSSION OF SELECTED SECTIONS 3.2 Equations of Kinematics in Two Dimensions Chapter 2 dealt with displacement, velocity, and acceleration in one dimension . But if an object moves in the horizontal and vertical direction at the same time, we say that the object is is moving in two dimensions . We subscript any quantity which is horizontal with an x (such as v x and a x ), and we subscript any quantity which is vertical with a y (such as v y and a y . ) Example 1 A helicopter moves in such a way that its position at any time is described by the horizontal and vertical equations x = 5 t + 12 t 2 and y = 10 + 2 t + 6 t 2 , where x and y are in meters and t is in seconds. (a) What is the initial position of the helicopter at time t = 0? (b) What are the x and y components of the helicopter’s acceleration at 3 seconds? (c) What is the speed of the helicopter at 4 seconds? Solution: (a) For the initial position, we simply substitute zero for time: x = 5(0) + 12(0) 2 and y = 10 + 2(0) + 6(0) 2 yielding x = 0 and y = 10 m at t = 0. (b) Notice that both equations are of the familiar form 2 2 1 at t v s s o . This means that the acceleration in the equation for x must be 24 m/s 2 (that is, ½ (24) t 2 ), and the acceleration in the equation for y must be 12 m/s 2 . Thus, a x = 24 m/s 2 , and a y = 12 m/s 2 ....
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This note was uploaded on 03/30/2012 for the course PHYSICS 201 taught by Professor Rollino during the Fall '11 term at Rutgers.
 Fall '11
 rollino
 Physics

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