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F10P1AChap2Sol

F10P1AChap2Sol - Motion in One Dimension SOLUTIONS TO...

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1 Motion in One Dimension SOLUTIONS TO PROBLEMS Section 2.1 Average Velocity P2.3 (a) v = ! x ! t = 10 m 2 s = 5 m s (b) v = 5 m 4 s = 1.2 m s (c) v = x 2 ! x 1 t 2 ! t 1 = 5 m ! 10 m 4 s ! 2 s = ! 2.5 m s (d) v = x 2 ! x 1 t 2 ! t 1 = ! 5 m ! 5 m 7 s ! 4 s = ! 3.3 m s (e) v = x 2 ! x 1 t 2 ! t 1 = 0 ! 0 8 ! 0 = 0 m s *P2.4 (a) Let d represent the distance between A and B. Let t 1 be the time for which the walker has the higher speed in 5.00 m s = d t 1 . Let t 2 represent the longer time for the return trip in ! 3.00 m s = ! d t 2 . Then the times are t 1 = d 5.00 m s ( ) and t 2 = d 3.00 m s ( ) . The average speed is: v = Total distance Total time = d + d d 5.00 m s ( ) + d 3.00 m s ( ) = 2 d 8.00 m s ( ) d 15.0 m 2 s 2 ( ) v = 2 15.0 m 2 s 2 ( ) 8.00 m s = 3.75 m s (b) She starts and finishes at the same point A. With total displacement = 0, average velocity = 0 .
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2 Motion in One Dimension Section 2.2 Instantaneous Velocity P2.5 (a) at t i = 1.5 s , x i = 8.0 m (Point A ) at t f = 4.0 s , x f = 2.0 m (Point B ) v = x f ! x i t f ! t i = 2.0 ! 8.0 ( ) m 4 ! 1.5 ( ) s = ! 6.0 m 2.5 s = ! 2.4 m s (b) The slope of the tangent line can be found from Points C and D . t C = 1.0 s, x C = 9.5 m ( ) and t D = 3.5 s, x D = 0 ( ) , v ! " 3.8 m s . (c) The velocity is zero when x is a minimum. This is at t ! 4 s . FIG. P2.5 P2.8 (a) v = 5 ! 0 ( ) m 1 ! 0 ( ) s = 5 m s (b) v = 5 ! 10 ( ) m 4 ! 2 ( ) s = ! 2.5 m s (c) v = 5 m ! 5 m ( ) 5 s ! 4 s ( ) = 0 (d) v = 0 ! ! 5 m ( ) 8 s ! 7 s ( ) = + 5 m s FIG. P2.8 Section 2.3 Analysis Models ί૿ The Particle Under Constant Velocity P2.9
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