ch03f - Chapter 3 Motion in Two and Three Dimensions...

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Unformatted text preview: Chapter 3 Motion in Two and Three Dimensions CHAPTER 3 Motion in Two and Three Dimensions 1* · Can the magnitude of the displacement of a particle be less than the distance traveled by the particle along its path? Can its magnitude be more than the distance traveled? Explain. The magnitude of the displacement can be less but never more than the distance traveled. If the path is a semi- circle of radius R , the distance traveled is 2 π R , the magnitude of the displacement is 2 R . The maximum magnitude of the displacement occurs when the path is a straight line; then the two quantities are equal. 5* · ( a ) A man walks along a circular arc from the position x = 5 m, y = 0 to a final position x = 0, y = 5 m. What is his displacement? ( b ) A second man walks from the same initial position along the x axis to the origin and then along the y axis to y = 5 m and x = 0. What is his displacement? ( a ), ( b ) Since the initial and final positions are the same, the displacements are equal. D = (-5 i + 5 j ) m. 9* · Can a component of a vector have a magnitude greater than the magnitude of the vector? Under what circumstances can a component of a vector have a magnitude equal to the magnitude of the vector? No; A x = A cos θ ≤ A . A x = A for θ = 0, i.e., if the vector is along the component direction. 13* · A velocity vector has an x component of +5.5 m/s and a y component of -3.5 m/s. Which diagram in Figure 3-32 gives the direction of the vector? ( b ) The vector is in the fourth quadrant; θ = tan-1 (-3.5/5.5) = -32.5 o . 17* · Find the magnitude and direction of the following vectors: ( a ) A = 5 i + 3 j ; ( b ) B = 10 i- 7 j ; ( c ) C = - 2 i- 3 j + 4 k . ( a ) and ( b ) Use Equs. 3-4 and 3-5 ( c ) C + C + C = C 2 z 2 y 2 x ; the angle between C and the z axis is θ = cos-1 ( C z /C ); the angle with the x axis is φ = tan-1 ( C y /C z ) A = 5.83, θ = 31 o ; B = 12.2, θ = -35 o C = 5.39; θ = 42 o , φ = 236 o 21* · If A = 5 i- 4 j and B = -7.5 i + 6 j , write an equation relating A to B . Chapter 3 Motion in Two and Three Dimensions Note: B x = -1.5 A x and B y = -1.5 A y . Consequently, B = -1.5 A . 25* · How is it possible for a particle moving at constant speed to be accelerating? Can a particle with constant velocity be accelerating at the same time? A particle moving at constant speed in a circular path is accelerating (the direction of the velocity vector is changing). If a particle is moving at constant velocity, it is not accelerating. 29* ·· As a bungee jumper approaches the lowest point in her drop, she loses speed as she continues to move downward. Draw the velocity vectors of the jumper at times t 1 and t 2 , where ∆ t = t 2- t 1 is small. From your drawing find the direction of the change in velocity ∆ v = v 2- v 1 , and thus the direction of the acceleration vector....
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This note was uploaded on 11/25/2010 for the course PHYSICS 4A taught by Professor Eduardo during the Spring '09 term at DeAnza College.

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ch03f - Chapter 3 Motion in Two and Three Dimensions...

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