hmwk02 - Su(ycs73 HW02 Tsoi(58020 This print-out should...

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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 347 m 56 . 0 202 m 250 m 29 . 0 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 347 m 124 202 m 250 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 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 . 0 180 . 0 = 124 d 4 = 250 . 0 m θ 4 = 180 . 0 29 . 0 = 151 Solution: Δ x 1 = (108 m)(cos 0 ) = 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)(sin 151 ) = 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
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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 0 x and v 0 y , and in the polar coordinates is v 0 and θ (see figure). x y v 0 θ h R A B The height of the trajectory h, is given by 1. h = v 0 x 2 g . 2. h = 2 g v 0 x 2 . 3. h = v 0 x 2 2 g . 4. h = v 0 y 2 2 g . correct 5. h = v 0 2 2 g . 6. h = 2 g v 0 y 2 . 7. h = 2 g v 0 . 8. h = v 0 2 g . 9. h = 2 g v 0 2 . 10. h = v 0 y 2 g . Explanation: Basic Concepts: Constant acceleration: x x 0 = v 0 t + 1 2 a t 2 (1) v = v 0 + a t . (2) Solution: Consider the motion in y direc- tion, v y 2 = v 0 y 2 + 2 a s (3) At the maximum height h, 0 = v 0 y 2 2 g h (4) or h = v 0 y 2 2 g . (4) 005 (part 2 of 2) 10.0 points
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Su (ycs73) – HW02 – Tsoi – (58020) 3 Consider two cases with the initial angles θ 1 and θ 2 where the initial speed v 0 is fixed.
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