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Chapter 02 - 2.1 a During the later 4.75-s interval the...

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2.1: a) During the later 4.75-s interval, the rocket moves a distance m 63 m 10 00 . 1 3 - × , and so the magnitude of the average velocity is . s m 197 s 75 4 m 63 m 10 00 . 1 3 = - × . b) s m 169 s 5.90 m 10 00 . 1 3 = × 2.2: a) The magnitude of the average velocity on the return flight is . s m 42 . 4 ) da s 400 , 86 ( ) da 5 . 13 ( ) m 10 5150 ( 3 = × The direction has been defined to be the – x -direction ). ˆ ( i - b) Because the bird ends up at the starting point, the average velocity for the round trip is 0 . 2.3: Although the distance could be found, the intermediate calculation can be avoided by considering that the time will be inversely proportional to the speed, and the extra time will be min. 70 1 hr km 70 hr km 105 min) 140 ( = - 2.4: The eastward run takes ) s m 5.0 m 200 ( = 40.0 s and the westward run takes ) s m 4.0 m 280 ( = 70.0 s. a) (200 m + 280 m)/(40.0 s + 70.0 s) = s m 4 . 4 to two significant figures. b) The net displacement is 80 m west, so the average velocity is ) s 110.0 m 80 ( = s m 73 . 0 in the – x -direction ). ˆ ( i - 2.5: In time t the fast runner has traveled 200 m farther than the slow runner: s 286 so , s) m (6.20 m 200 s) m 50 . 5 ( = = + t t t . Fast runner has run m. 1770 ) s m 20 . 6 ( = t Slow runner has run m. 1570 ) s m 50 . 5 ( = t
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2.6 : The s-waves travel slower, so they arrive 33 s after the p-waves. km 250 s 33 5 . 6 5 . 3 s 33 s 33 s km s km p s p s = + = + = = = + = d d d v d v d v d t vt d t t 2.7: a) The van will travel 480 m for the first 60 s and 1200 m for the next 60 s, for a total distance of 1680 m in 120 s and an average speed of . s m 0 . 14 b) The first stage of the journey takes s 30 s m 8.0 m 240 = and the second stage of the journey takes , s 12 s) m 20 m 240 ( = so the time for the 480-m trip is 42 s, for an average speed of . s m 11.4 c) The first case (part (a)); the average speed will be the numerical average only if the time intervals are the same. 2.8: From the expression for x ( t ) , x (0) = 0, x (2.00 s) = 5.60 m and x (4.00 s) = 20.8 m. a) s m 80 . 2 s 2.00 0 m 60 . 5 = - b) s m 2 . 5 s 4.00 0 m 8 . 20 = - c) s m 6 . 7 s 2.00 m 5.60 m 8 . 20 = - 2.9: a) At 0 , 0 1 1 = = x t , so Eq (2.2) gives . s m 0 . 12 s) 0 . 10 ( s) 0 . 10 )( s m 120 . 0 ( s) 0 . 10 )( s m 4 . 2 ( 3 3 2 2 2 2 av = - = = t x v b) From Eq. (2.3), the instantaneous velocity as a function of time is , ) s m 360 . 0 ( ) s m 80 . 4 ( 3 2 2 3 2 2 t t ct bt v x - = - = so i) , 0 ) 0 ( = x v ii) , s m 0 . 15 s) 0 . 5 )( s m 360 . 0 ( ) s 0 . 5 )( s m 80 . 4 ( ) s 0 . 5 ( 2 3 2 = - = x v and iii) . s m 0 . 12 s) 0 . 10 )( s m 360 . 0 ( ) s 0 . 10 )( s m 80 . 4 ( ) s 0 . 10 ( 2 3 2 = - = x v c) The car is at rest when 0 = x v . Therefore 0 ) s m 360 . 0 ( ) s m 80 . 4 ( 2 3 2 = - t t . The only time after 0 = t when the car is at rest is s 3 . 13 3 2 s m 360 . 0 s m 80 . 4 = = t
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2.10: a) IV: The curve is horizontal; this corresponds to the time when she stops. b) I: This is the time when the curve is most nearly straight and tilted upward (indicating postive velocity). c) V: Here the curve is plainly straight, tilted downward (negative velocity). d) II: The curve has a postive slope that is increasing. e) III: The curve is still tilted upward (positive slope and positive velocity), but becoming less so. 2.11: Time (s) 0 2 4 6 8 10 12 14 16 Acceleration (m/s 2 ) 0 1 2 2 3 1.5 1.5 0 a) The acceleration is not constant, but is approximately constant between the times s 4 = t and s. 8 = t 2.12: The cruising speed of the car is 60 hr km = 16.7 s m . a) 2 s 10 s m 7 . 16 s m 7 . 1 = (to two significant figures). b) 2 s 10 s m 7 . 16 0 s m 7 . 1 - = - c) No change in speed, so the acceleration is zero. d) The final speed is the same as the initial speed, so the average acceleration is
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