Physics_Chapt4

# Physics_Chapt4 - 1 Chapter 4 1. Where the length unit is...

This preview shows pages 1–4. Sign up to view the full content.

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

View Full Document

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1 Chapter 4 1. Where the length unit is not specified (in this solution), the unit meter should be understood. (a) The position vector, according to Eq. 4-1, is ~ r =- 5 . 0 + 8 . j (in meters). (b) The magnitude is | ~ r | = x 2 + y 2 + z 2 = 9 . 4 m. (c) Many calculators have polar rectangular conversion capabilities which make this computation more efficient than what is shown below. Noting that the vector lies in the xy plane, we are using Eq. 3-6: tan- 1 8 .- 5 . =- 58 or 122 where we choose the latter possibility (122 measured counterclockwise from the + x direction) since the signs of the components imply the vector is in the second quadrant. (d) In the interest of saving space, we omit the sketch. The vector is 32 counterclockwise from the + y direction, where the + y direction is assumed to be (as is standard) +90 counterclockwise from + x , and the + z direction would therefore be out of the paper. (e) The displacement is ~ r = ~ r- ~ r where ~ r is given in part (a) and ~ r = 3 . 0 . Therefore, ~ r = 8 . 0 - 8 . j (in meters). (f) The magnitude of the displacement is | ~ r | = q 8 2 + (- 8) 2 = 11 m. (g) The angle for the displacement, using Eq. 3-6, is found from tan- 1 8 .- 8 . =- 45 or 135 where we choose the former possibility (- 45 , which means 45 measured clockwise from + x , or 315 counterclockwise from + x ) since the signs of the components imply the vector is in the fourth quadrant. 2. (a) The magnitude of ~ r is q 5 . 2 + (- 3 . 0) 2 + 2 . 2 = 6 . 2 m. (b) A sketch is shown. The coordinate values are in meters. + y + z + x ~ r 5- 3 2 3. Where the unit is not specified, the unit meter is understood. We use Eq. 4-2 and Eq. 4-3. (a) With the initial position vector as ~ r 1 and the later vector as ~ r 2 , Eq. 4-3 yields r = ((- 2)- 5) + (6- (- 6)) j + (2- 2) k =- 7 . 0 + 12 j for the displacement vector in unit-vector notation (in meters). (b) Since there is no z component (that is, the coefficient of k is zero), the displacement vector is in the xy plane. 4. We use a coordinate system with + x eastward and + y upward. We note that 123 is the angle between the initial position and later position vectors, so that the angle from + x to the later position vector is 40 + 123 = 163 . In unit- vector notation, the position vectors are ~ r 1 = 360 cos(40 ) + 360 sin(40 ) j = 276 + 231 j ~ r 2 = 790 cos(163 ) + 790 sin(163 ) j =- 755 + 231 j respectively (in meters). Consequently, we plug into Eq. 4-3 r = ((- 755)- 276) + (231- 231) j and find the displacement vector is horizontal (westward) with a length of 1 . 03 km. If unit-vector notation is not a priority in this problem, then the computation can be approached in a variety of ways particularly in view of the fact that a number of vector capable calculators are on the market which reduce this problem to a very few keystrokes (using magnitude-angle notation throughout). 5. The average velocity is given by Eq. 4-8.5....
View Full Document

## This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

### Page1 / 54

Physics_Chapt4 - 1 Chapter 4 1. Where the length unit is...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online