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Unformatted text preview: hÝ × ic×ÙÑÑ eÖ8ÓÑ eÛ Ó Ök# Ó ÐÙ Ø iÓÒ ×bÝ iÑC a ÐÐahaÒ § 2 Problem 5 ( a ): The total distance that both runners need to traverse is 200 m. If they are running towards each other, one at 6 . 20 m / s and one at 5 . 50 m / s, then they are reducing the distance between them at a rate of (6 . 20 m / s) + (5 . 50 m / s) = 11 . 70 m / s. The 200 m distance is thus reduced to 0 m in a time Δ t = 200 m 11 . 70 m / s = 17 . 1 s . ( b ): During this time, the distance each one has traveled is Δ x 1 = (6 . 20 m / s)(17 . 1 s) = 106 m , Δ x 2 = (5 . 50 m / s)(17 . 1 s) = 94 m . Sure enough, these add up to 200 m. § 2 Problem 11 ( a ): During the first 2.0 s the ball is moving at 2 . 0 m / s, and hence travels a distance of 4.0 m. During the last second the ball moves at 3 . 0 m / s and thus travels 3.0 m. The total distance traveled is 7.0 m, in a total of 3.0 s, so the average speed and average velocity are both 7 . 0 m 3 . 0s = 2 . 3 m / s . ( b ): If during the last second the velocity is 3 . 0 m / s then during that time the ball moves 3.0 m to the left. Thus the total displacement is Δ x = 1 . 0 m. The total odometer distance traveled is still 7.0 m. The average speed is still 2 . 3 m / s, but the average velocity is 1 . 0 m 3 . 0 s = 0 . 33 m / s . § 2 Problem 22 ( a ): First we convert everything into metric. We get a final (takeoff) speed of 77 . 3 m / s, in a distance of 93.6 m. We know the initial and final velocities and the distance, and we want to find the acceleration. We use the formula v 2 f v 2 = 2 a Δ x ⇒ a = v 2 f v 2 Δ x = (77 . 3 m / 2) 2 (0 m / s) 2 2(93 . 6 m) = 31 . 9 m / s 2 . ( b ): We need any formula that has the time, and there are several options. For example, we can use v f v = a Δ t ⇒ Δ t = v f v a = (77 . 3 m / s) (0 m / s) 31 . 9 m / s 2 = 2 . 42 s ....
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This note was uploaded on 07/12/2008 for the course PHY 121 taught by Professor Chamberlin during the Summer '08 term at ASU.
 Summer '08
 Chamberlin
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