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Unformatted text preview: Chapter 3 Kinematics in Two Dimensions 37 Chapter 3 KINEMATICS IN TWO DIMENSIONS PREVIEW Two-dimensional 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 two-dimensional 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