HW-03-solutions

HW-03-solutions - kuruvila (lk5992) HW-03 gokce (P111S11) 1...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: kuruvila (lk5992) HW-03 gokce (P111S11) 1 This print-out should have 16 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 2) 10.0 points A particle has an initial horizontal velocity of 2 . 8 m / s and an initial upward velocity of 3 . 8 m / s. It is then given a horizontal accelera- tion of 2 . 5 m / s 2 and a downward acceleration of 2 m / s 2 . What is its speed after 3 s? Correct answer: 10 . 5323 m / s. Explanation: Basic Concepts: The direction of the motion depends only on the horizontal and vertical components of the velocity at any moment. Solution: For the horizontal motion, v x = v x + a x t = 2 . 8 m / s + ( 2 . 5 m / s 2 ) (3 s) = 10 . 3 m / s For the vertical motion, v y = v y- a y t = 3 . 8 m / s- ( 2 m / s 2 ) (3 s) =- 2 . 2 m / s A resultant negative velocity v y would indi- cate downward motion. A right triangle is formed by the compo- nents, so v = radicalBig v 2 x + v 2 y = radicalBig (10 . 3 m / s) 2 + (- 2 . 2 m / s) 2 = 10 . 5323 m / s 002 (part 2 of 2) 10.0 points What is the direction of its velocity at this time with respect to the horizontal? Answer between- 180 and +180 . Correct answer:- 12 . 0568 . Explanation: The vertical component v y is the side oppo- site the angle and the horizontal component v x is the side adjacent to the angle, so tan = v y v x = arctan v y v x = arctan- 2 . 2 m / s 10 . 3 m / s = 10 . 5323 m / s The angle must be in degrees and a positive angle indicates upward motion while a nega- tive angle indicates downward motion. 003 (part 1 of 3) 10.0 points A particle moves in the xy plane with constant acceleration. At time zero, the particle is at x = 2 . 5 m, y = 5 . 5 m, and has velocity vectorv o = (1 m / s) + (- 5 . 5 m / s) . The acceleration is given by vectora = (7 m / s 2 ) + (5 m / s 2 ) . What is the x component of velocity after 8 . 5 s? Correct answer: 60 . 5 m / s. Explanation: Let : a x = 7 m / s 2 , v xo = 1 m / s , and t = 8 . 5 s . After 8 . 5 s, vectorv x = vectorv xo + vectora x t = (1 m / s) + (7 m / s 2 ) (8 . 5 s) = (60 . 5 m / s) . 004 (part 2 of 3) 10.0 points What is the y component of velocity after 8 . 5 s? Correct answer: 37 m / s. kuruvila (lk5992) HW-03 gokce (P111S11) 2 Explanation: Let : a y = 5 m / s 2 and v yo =- 5 . 5 m / s . vectorv y = vectorv yo + vectora y t = (- 5 . 5 m / s) + (5 m / s 2 ) (8 . 5 s) = (37 m / s) . 005 (part 3 of 3) 10.0 points What is the magnitude of the displacement from the origin ( x = 0 m, y = 0 m) after 8 . 5 s? Correct answer: 298 . 422 m. Explanation: Let : d o = (2 . 5 m , 5 . 5 m) , v o = (1 m / s ,- 5 . 5 m / s) , and a = (7 m / s 2 , 5 m / s 2 ) ....
View Full Document

This note was uploaded on 02/12/2011 for the course PHYS 111 taught by Professor Moro during the Spring '08 term at NJIT.

Page1 / 7

HW-03-solutions - kuruvila (lk5992) HW-03 gokce (P111S11) 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online