Solution of Assignment (1) - 1 Solution of Assignment (1)...

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1 Solution of Assignment (1) Rectilinear Motion-1 Problem 1: A particle moves along a straight line such that its position is given by x = 105 + 100t – 5t 2 m where t is in seconds, determine: (a) The time after which the particle passes through the origin. (b) The distance traveled during this time interval. Solution: a. At the origin x = 0: x = 105 + 100t – 5t 2 105 + 100t – 5t 2 = 0 t 2 – 20t – 21 = 0 → (t – 21) (t + 1) = 0 t = 21 or t = -1 (refused) Therefore the time after which the particle passes through the origin is 21 seconds. b. D = ? Interval from t = 0 to t = 21 seconds: At ( t = 0 ): x = 105 m At (t = 21 seconds) : x = 0 t 10 100 dt dx v = = At v = 0: 100 – 10t = 0 → t = 10 seconds At t = 10 seconds: x = 105 + 100 (10) – 5 (10) 2 x = 605 m
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2 Distance traveled D = 500 + 605 = 1105 m x x = 605 m x = 0 t = 21 sec t = 0 x = 105 m t = 10 sec 605 m 500 m O +
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3 Problem 2: A particle moves along a straight line such that its position is defined by: x = 2t 3 + 3t 2 – 12t – 10 m where t is in seconds. Determine: (a) The velocity and the acceleration after 3 seconds. (b) The distance traveled during the first 2 seconds. Solution: a) 12 t 6 t 6 dt dx v 2 + = = , At t = 3 seconds : ( ) ( ) 12 3 6 3 6 v 2 + = s / m 60 v = 6 t 12 dt dv a + = = , At ( t = 3): seconds : ( ) 6 3 12 a + = 2 s / m 42 a = b) D = ? : Time interval from t = 0 to t = 2 seconds At t = 0 : x = -10 m At t = 2 seconds : x = 2 (2) 3 + 3 (2) 2 – 12 (2) – 10 x = -6m v = 6t 2 + 6t – 12 At v = 0: t 2 + t – 2 = 0 → (t + 2) (t – 1) = 0 t = -2 (refused) or t = 1 second Therefore at v = 0 →t = 1 second At t = 1 second: x = 2 + 3 – 12 -10 →
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This note was uploaded on 12/21/2010 for the course ENGINEERIN mp108 taught by Professor Elbarki during the Spring '08 term at Alexandria University.

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Solution of Assignment (1) - 1 Solution of Assignment (1)...

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