HW 3-solutions - morris(gmm643 – HW 3 – Coker –(56625...

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Unformatted text preview: morris (gmm643) – HW 3 – Coker – (56625) 1 This print-out should have 24 questions. Multiple-choice 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 morris (gmm643) – 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 y-axis. 004 (part 1 of 2) 10.0 points A cannon fires a 0 . 595 kg shell with initial velocity v i = 8 . 2 m / s in the direction θ = 40 ◦ above the horizontal. Δ x Δ h 8 . 2 m / s 4 ◦ Δ y y The shell’s trajectory curves downward be- cause of gravity, so at the time t = 0 . 224 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 depends only on the initial angle θ , and does not depend on the flight time t or the initial velocity v i . 2. It depends on some data not given in the problem. 3. It depends only on the flight time t , and does not depend on the initial velocity v i or the initial angle θ . correct 4. It is a constant and does not depend on the flight time t or the initial velocity v i or the initial angle θ ....
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HW 3-solutions - morris(gmm643 – HW 3 – Coker –(56625...

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