HW02-solutions - dang (ttd73) HW02 Tsoi (58020) 1 This...

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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 persons resultant displacement measured from the starting point? Correct answer: 419 . 642 m. Explanation: d 111 m 401m 120 2 6 2 m 2 9 8 m 123 Note: Figure is not drawn to scale. Basic Concepts: x = d (cos ) y = d (sin ) x tot = x 1 + x 3 + x 4 since x 2 = 0 m. y tot = y 2 + y 3 + y 4 since y 1 = 0 m. d = radicalBig ( x tot ) 2 + ( y tot ) 2 Given: d 1 = 111 . 0 m 1 = 0 d 2 = 401 . 0 m 2 = 90 d 3 = 262 . 0 m 3 = 60 . 180 . = 120 d 4 = 298 . 0 m 4 = 180 . 57 . = 123 Solution: x 1 = (111 m)(cos0 ) = 111 m y 2 = (401 m)[cos( 90 )] = 401 m x 3 = (262 m)[cos( 120 )] = 131 m y 3 = (262 m)[sin( 120 )] = 226 . 899 m x 4 = (298 m)(cos 123 ) = 162 . 302 m y 4 = (298 m)(sin123 ) = 249 . 924 m x tot = 111 m 131 m 162 . 302 m = 182 . 302 m 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 persons resultant displacement? Correct answer: 64 . 2514 . Explanation: Basic Concept: tan = y tot x tot Solution: = tan 1 parenleftbigg y tot 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 : = 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 = v 1 . Then you must compute the distance the slower ant covers in that time s 2 = v 2 t = v 2 v 1 . The slower ant is s 2 = v 2 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 This...

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