Physics 114A Introduction to Mechanics (without calculus)

Transcription

1 Physics 114A Introduction to Mechanics (without calculus) A course about learning basic physics concepts and applying them to solve real-world, quantitative, mechanical problems Lecture 6 Review of Vectors x, v, a and Relative Motion January 15, 2008 Lecturer: Prof. Paul Boynton

2 Reminder of vector representations POSITION: r(t) = x(t) i + y(t) j VELOCITY: v(t) = v x (t) i + v y (t) j v x (t) = dx v y (t) = dy dt dt ACCELERATION: a (t) = a x (t) i + a y (t) j a x (t) = dv x dt a y (t) = dv y dt

3 This space-time diagram shows the time variation of position (in a single spatial dimension). The slope at any point x(t) corresponds to the instantaneous (scalar) velocity component, v x, of this 1-D motion at that time.

4 This velocity-time diagram shows the time variation of velocity (in a single spatial dimension). The slope at any point v x (t) corresponds to the instantaneous (scalar) acceleration component, a x, of this 1-D motion at that time. v x

5 This diagram is very different from the previous two. It is plot illustrating the trajectory of a point in a 2-D space. At two points on this trajectory, velocity vectors are superposed to show the direction and magnitude of the velocity in these two space coordinates (note that these axes are not in units of velocity). Also shown is a av. This depiction of a mixture of vectors is common, but can be confusing

6 At point 1, the a and v indicate the speed is deceasing, but the direction of v is not changing (consistent with the shape of the trajectory). At 2 and 3, the speed is not changing, but the direction of v is changing. What is the clue that speed is constant? Point 4 is like 1, but speed is increasing at this point. You should be able to diagnose the motion from inspection of such diagrams.

7 v i!v v i v f a v f!v!v a = lim -!t 0!t

8 In the previous slide, the top panel illustrates the nature of the average acceleration vector, a av, for motion in a circle at constant speed. The average is depicted for motion though a quarter of the rotation cycle. The lower panel shows that the direction of a av. Does not change even when evaluated for a much smaller time interval. In fact, for ucm* a = a av. A little thought should reveal that by the construction shown here, the magnitude of!v/!t is independent from!t. Even so, the instantaneous acceleration, a,continually points toward the center of rotation for ucm. *uniform circular motion

9 Example problem from Ch. 3, Relative Motion 45. The pilot of an airplane wishes to fly due north from Seattle to Vancouver, but there is a 65 km/h wind blowing toward the east. (a) In what direction should the pilot head her plane if its speed relative to the air is 340 km/h? (b) Draw a vector diagram that illustrates your result in part (a). (c) Given that the distance between Seattle and Vancouver airports is about 200 km, what is the flight time assuming constant velocity relative to the ground of magnitude given by your analysis in part (a)? (d) If the pilot decreases the air speed of the plane, but still wants to head due north, should the angle found in part (a) be increased or decreased? (Problem somewhat modified.) Solution proceeds directly from equation 3-7 rewritten with moving air replacing moving train (see later discussion). Don t choose the plug-andchug approach. Try to formulate a solution using the concepts introduced in the text plus logic and intuition. This will best develop your problemsolving skills. Most problems can be solved many ways, what follows is one approach. First, all the relative motion problems in your text ask you to consider two frames of reference: one stationary (your rest frame), and one moving frame. Motion in that moving frame is described as observed by someone at rest in that moving frame (their rest frame) and you must then determine how it appears in your rest frame, or vice versa.

10 First, examine the problem in the ground frame of reference, then proceed to the rest frame of the air. Vectors are noted in bold face type. Spatial configuration N y.. v pg x Seattle Vancouver plane Velocity of plane with respect to ground in ground frame of reference Relevant motion v y v pg vag E Velocity of air with respect to ground v x In rest frame of air V pa (V pa ) x " v y (V pa ) y v x In the rest frame of the air, (V pa ) x must be equal to -v ag if the plane is not to move east with the wind. Also, (V pa ) y must be equal to the plane s ground speed v pg, as there is no air motion along the y axis. Hence, from diagram at the left, V pa = v pg - v ag or v pg = V pa + v ag [which is exactly eq. 3-7 (or 3-8) in your text!]

11 So we proceed to the solution: (a) Use the inverse sine function to find the heading,!: # v! = sin "1 ag & % ( = sin "1 # 65 km/h & $ ' $ % 340 km/h ' ( = 11 west of north v pa (b) The previous diagrams depict the vectors. (c) From the previous page, v pg = v pa cos" = 340 km/h (0.98) = 334 km/h. and t = 200km/(334km/hr) = 0.6 = 36 minutes (d) She would have to increase the heading,!, according to the diagram.