This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: kodeih (njk359) – HW 3 – Coker – (56625) 1 This printout should have 24 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. This homework for Chapter 4 covers the last set of material that will be tested on Quiz 1, to be held on September 15 at 7 PM in GAR 0.102. 001 10.0 points The correct, general definitions of velocity and acceleration, in terms of position vector vectorr , are vectorv = dvectorr dt , and vectora = dvectorv dt . During a very short time interval, the velocity of an object changes from vectorv i to vectorv f , as shown. vectorv i vectorv f What is the approximate direction of the acceleration during this time interval? 1. correct 2. 3. 4. Explanation: vectora Δ t = vectorv f vectorv i . vectorv i vectorv f vectora Δ t 002 (part 1 of 2) 10.0 points A particle has vectorr (0) = (4 m)ˆ and vectorv (0) = (2 m / s)ˆ ı . If its acceleration is constant and given by vectora = (2 m / s 2 ) (ˆ ı +ˆ ), at what time t does the particle first cross the x axis? Correct answer: 2 s. Explanation: vectorr ( t ) = vectorr (0) + vectorv (0) t + 1 2 vectora t 2 , so vectorr ( t ) = (4 m) ˆ + (2 m / s) ˆ ıt +( 1 m / s 2 ) (ˆ ı +ˆ ) t 2 = [(2 m / s) t (1 m / s 2 ) t 2 ]ˆ ı +[4 m (1 m / s 2 ) t 2 ] ˆ vectorr ( t ) will not have a y component when 4 m (1 m / s 2 ) t 2 = 0 (1 m / s 2 ) t 2 = 4 m t = 2 s . 003 (part 2 of 2) 10.0 points At what time t is the particle moving parallel to the y axis; that is, in the ˆ direction? Correct answer: 1 s. Explanation: vectorr ( t ) = vectorr (0) + vectorv (0) t + 1 2 vectorat 2 vectorv ( t ) = dvectorr d t = vectorv (0) + vectorat , so kodeih (njk359) – HW 3 – Coker – (56625) 2 vectorv ( t ) = (2 m / s)ˆ ı (2 m / s 2 ) (ˆ ı +ˆ ) t = [(2 m / s) (2 m / s 2 ) t ]ˆ ı (2 m / s 2 ) t ˆ vectorv ( t ) will not have an x component when 2 m / s (2 m / s 2 ) t = 0 t = 1 s . At this time vectorv ( t ) = (2 m / s)ˆ parallel to the yaxis. 004 (part 1 of 2) 10.0 points A cannon fires a 0 . 654 kg shell with initial velocity v i = 8 . 1 m / s in the direction θ = 46 ◦ above the horizontal. Δ x Δ h 8 . 1 m / s 4 6 ◦ Δ y y The shell’s trajectory curves downward be cause of gravity, so at the time t = 0 . 561 s the shell is below the straight line by some verti cal distance Δ h . Your task is to calculate the distance Δ h in the absence of air resistance. On what does Δ h depend (besides g )? 1. It is a constant and does not depend on the flight time t or the initial velocity v i or the initial angle θ . 2. It depends on everything: the flight time t , the initial angle θ , and the initial velocity v i ....
View
Full
Document
This note was uploaded on 09/25/2010 for the course PHY 56640 taught by Professor Coker during the Fall '10 term at University of Texas.
 Fall '10
 COKER

Click to edit the document details