hmwk02 - Su(ycs73 – HW02 – Tsoi –(58020 1 This...

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Unformatted text preview: Su (ycs73) – HW02 – Tsoi – (58020) 1 This print-out should have 22 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 person walks the path shown. The total trip consists of four straight-line paths. S N W E 108 m 347m 56 . ◦ 2 2 m 2 5 m 29 . ◦ Note: Figure is not drawn to scale. a) At the end of the walk, what is the mag- nitude of the person’s resultant displacement measured from the starting point? Correct answer: 452 . 392 m. Explanation: d θ 108 m 347m − 124 ◦ 2 2 m 2 5 m 151 ◦ Note: Figure is not drawn to scale. Basic Concepts: Δ x = d (cos θ ) Δ y = d (sin θ ) Δ x tot = Δ x 1 + Δ x 3 + Δ x 4 since Δ x 2 = 0 m. Δ y tot = Δ y 2 + Δ y 3 + Δ y 4 since Δ y 1 = 0 m. d = radicalBig (Δ x tot ) 2 + (Δ y tot ) 2 Given: d 1 = 108 . 0 m θ 1 = 0 ◦ d 2 = 347 . 0 m θ 2 = − 90 ◦ d 3 = 202 . 0 m θ 3 = 56 . ◦ − 180 . ◦ = − 124 ◦ d 4 = 250 . 0 m θ 4 = 180 . ◦ − 29 . ◦ = 151 ◦ Solution: Δ x 1 = (108 m)(cos0 ◦ ) = 108 m Δ y 2 = (347 m)[cos( − 90 ◦ )] = − 347 m Δ x 3 = (202 m)[cos( − 124 ◦ )] = − 112 . 957 m Δ y 3 = (202 m)[sin( − 124 ◦ )] = − 167 . 466 m Δ x 4 = (250 m)(cos 151 ◦ ) = − 218 . 655 m Δ y 4 = (250 m)(sin151 ◦ ) = 121 . 202 m Δ x tot = 108 m − 112 . 957 m − 218 . 655 m = − 223 . 612 m Δ y tot = − 347 m − 167 . 466 m + 121 . 202 m = − 393 . 263 m and d = radicalBig ( − 223 . 612 m) 2 + ( − 393 . 263 m) 2 = 452 . 392 m 002 (part 2 of 2) 10.0 points Su (ycs73) – HW02 – Tsoi – (58020) 2 b) What is the direction (measured from due west, with counterclockwise positive) of the person’s resultant displacement? Correct answer: 60 . 3772 ◦ . Explanation: Basic Concept: tan θ = Δ y tot Δ x tot Solution: θ = tan − 1 parenleftbigg Δ y tot Δ x tot parenrightbigg = tan − 1 parenleftbigg − 393 . 263 m − 223 . 612 m parenrightbigg = 60 . 3772 ◦ south of west. 003 10.0 points A vector of magnitude 3 CANNOT be added to a vector of magnitude 4 so that the magni- tude of the resultant is 1. 1. 2. 3. 3. 5. 4. 7. 5. 0. correct Explanation: The smallest magnitude of the resultant occurs when the vectors are anti-parallel ( R = 1); the largest occurs when they are parallel ( R = 7). Therefore all listed values are possible except R = 0. 004 (part 1 of 2) 10.0 points Consider the motion of a projectile. It is fired at t = 0. The initial velocity vector in rectangular coordinates is v x and v y , and in the polar coordinates is v and θ (see figure). x y v θ h R A B The height of the trajectory h, is given by 1. h = v x 2 g . 2. h = 2 g v x 2 . 3. h = v x 2 2 g ....
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This note was uploaded on 04/15/2009 for the course PHY 58020 taught by Professor Tsoi during the Spring '09 term at University of Texas.

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hmwk02 - Su(ycs73 – HW02 – Tsoi –(58020 1 This...

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