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

This preview shows pages 1–3. Sign up to view the full content.

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 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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. The sketch is shown. 33* · A particle moving at 4.0 m/s in the positive x direction is given an acceleration of 3.0 m/s 2 in the positive y direction for 2.0 s. The final speed of the particle is ____. ( a ) -2.0 m/s ( b ) 7.2 m/s ( c ) 6.0 m/s ( d ) 10 m/s ( e ) None of the above 1. Find the final velocity vector 2. Find the magnitude of the velocity vector v = v x i + a y t j = 4.0 m/s i + 6.0 m/s j v = 7.2 m/s ( b ) 37* ·· A particle moves in the xy plane with constant acceleration. At time zero, the particle is at
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern