Section04 - Physics 204A Class Notes Section 4 Motion in...

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Physics 204A Class Notes 4-1 Section 4 - Motion in More Than One Dimension Outline 1. The Definitions of Position, Displacement, Velocity and Acceleration 2. The Special Case of Projectile Motion 3. The Special Case of Uniform Circular Motion What do objects do? They move in more than one dimension. In this section we will describe motion in two dimensions by extending the same ideas of position, displacement, velocity and acceleration. The “good news” is we won’t change their definitions at all. The “bad news” is that we’ll let them become vectors. So we will trade off having to define more concepts for increasingly complex math. This is the first example of the way that physics tries to minimize the number of physical concepts and, at the same time, explain the widest possible variety of phenomena. 1. The Definitions of Position, Displacement, Velocity and Acceleration The definitions of position, displacement, velocity and acceleration all stay the same, they just become vectors. Position: The location of an object with respect to a coordinate system. Mathematically, position is now a vector r r = x ˆ i + y ˆ j . Displacement: A change in position. Now displacement is a vector as well, ! r r " r r f # r r i = (x f ˆ i + y f ˆ j ) # i ˆ i + y i ˆ j ) = (x f # x i ) ˆ i + (y f # y i ) ˆ j ! r r = ! x ˆ i + ! y ˆ j Average Velocity: The average rate of displacement. Mathematically, r v ! " r r " t . Using the displacement, r v = ! x ˆ i + ! y ˆ j ! t = ! x ! t ˆ i + ! y ! t ˆ j = v x ˆ i + v y ˆ j . The definition of speed is now a bit more complex because the velocity is a vector. Speed: The magnitude of the velocity vector. Mathematically, v ! r v = x 2 + v y 2 y x r r y x y x y f r r r r x
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Physics 204A Class Notes 4-2 Example 4.1: A hiker walks from a position of 3.00km due north of a ranger station to 1.00km east of the ranger station in 0.500h. Using an origin at the ranger station, find (a)the initial position, (b)the final position, (c)the displacement, (d)the average velocity and (e)the average speed of the hiker. Given: initial position 3.00km due north, final position 1.00km east, and t = 0.500h Find: r r i , r r f , " r r , r v , and v. (a)Using the coordinates shown at the right, r r i = 3.00 ˆ j and (b) r r f = 1.00 ˆ i . (c)Using the definition of displacement, ! r r " r r f # r r i = 1.00 ˆ i # (3.00 ˆ j ) $ " r r = 1.00 ˆ i # 3.00 ˆ j . (d)Using the definition of average velocity, r v ! " r r " t = 1.00 ˆ i # 3.00 ˆ j 0.500 $ r v = 2.00 ˆ i ! 6.00 ˆ j in km/h. (e)Using the definition of speed, v ! r v = x 2 + v y 2 = 2 + 6 2 " v = 6.32km/h . Average Acceleration: The rate of change of velocity. Mathematically, r a ! " r v " t . Using the velocity, r a = r v f ! r v i " t = (v xf ˆ i + v yf ˆ j ) ! (v xi ˆ i + v yi ˆ j ) " t = (v xf ! v xi ) ˆ i + yf ! v yi ) ˆ j " t = " v x " t ˆ i + " v y " t ˆ j = a x ˆ i + a y ˆ j . The extension from two dimensions to three dimensions you could probably guess. The definitions stay the same, but the vectors now have a z-component.
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Section04 - Physics 204A Class Notes Section 4 Motion in...

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