HW02-solutions - dang(ttd73 – HW02 – Tsoi –(58020 1...

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Unformatted text preview: dang (ttd73) – 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 111 m 401m 60 . ◦ 2 6 2 m 2 9 8 m 57 . ◦ 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: 419 . 642 m. Explanation: d θ 111 m 401m − 120 ◦ 2 6 2 m 2 9 8 m 123 ◦ 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 = 111 . 0 m θ 1 = 0 ◦ d 2 = 401 . 0 m θ 2 = − 90 ◦ d 3 = 262 . 0 m θ 3 = 60 . ◦ − 180 . ◦ = − 120 ◦ d 4 = 298 . 0 m θ 4 = 180 . ◦ − 57 . ◦ = 123 ◦ Solution: Δ x 1 = (111 m)(cos0 ◦ ) = 111 m Δ y 2 = (401 m)[cos( − 90 ◦ )] = − 401 m Δ x 3 = (262 m)[cos( − 120 ◦ )] = − 131 m Δ y 3 = (262 m)[sin( − 120 ◦ )] = − 226 . 899 m Δ x 4 = (298 m)(cos 123 ◦ ) = − 162 . 302 m Δ y 4 = (298 m)(sin123 ◦ ) = 249 . 924 m Δ x tot = 111 m − 131 m − 162 . 302 m = − 182 . 302 m Δ y tot = − 401 m − 226 . 899 m + 249 . 924 m = − 377 . 975 m and d = radicalBig ( − 182 . 302 m) 2 + ( − 377 . 975 m) 2 = 419 . 642 m 002 (part 2 of 2) 10.0 points dang (ttd73) – HW02 – Tsoi – (58020) 2 b) What is the direction (measured from due west, with counterclockwise positive) of the person’s resultant displacement? Correct answer: 64 . 2514 ◦ . 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 − 377 . 975 m − 182 . 302 m parenrightbigg = 64 . 2514 ◦ south of west. 003 10.0 points Two ants race across a table 63 cm long. One travels at 5 . 47 cm / s and the other at 3 . 99999 cm / s. When the first one crosses the finish line, how far behind is the second one? Correct answer: 16 . 9306 cm. Explanation: Let : ℓ = 63 cm , v 1 = 5 . 47 cm / s , and v 2 = 3 . 99999 cm / s . You first have to compute the time it takes the first (faster) ant to cross the finish line: t = ℓ v 1 . Then you must compute the distance the slower ant covers in that time s 2 = v 2 t = v 2 ℓ v 1 . The slower ant is ℓ − s 2 = ℓ − v 2 ℓ v 1 = (63 cm) − (3 . 99999 cm / s) (63 cm) (5 . 47 cm / s) = 16 . 9306 cm from the finish line when the faster one crosses it....
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HW02-solutions - dang(ttd73 – HW02 – Tsoi –(58020 1...

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