F10P1AChap3Sol

# F10P1AChap3Sol - Motion in Two Dimensions SOLUTIONS TO...

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1 Motion in Two Dimensions SOLUTIONS TO PROBLEMS Section 3.1 The Position, Velocity, and Acceleration Vectors P3.1 x m ( ) 0 ! 3 000 ! 1 270 ! 4 270 m y m ( ) ! 3 600 0 1 270 ! 2 330 m (a) Net displacement = x 2 + y 2 = 4.87 km at 28.6 ° S of W FIG. P3.1 (b) Average speed = 20.0 m s ( ) 180 s ( ) + 25.0 m s ( ) 120 s ( ) + 30.0 m s ( ) 60.0 s ( ) 180 s + 120 s + 60.0 s = 23.3 m s (c) Average velocity = 4.87 ! 10 3 m 360 s = 13.5 m s along r R

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2 Motion in Two Dimensions Section 3.2 Two-Dimensional Motion with Constant Acceleration P3.3 r v i = 4.00 ˆ i + 1.00 ˆ j ( ) m s and r v 20.0 ( ) = 20.0 ˆ i ! 5.00 ˆ j ( ) m s (a) a x = ! v x ! t = 20.0 " 4.00 20.0 m s 2 = 0.800 m s 2 a y = ! v y ! t = " 5.00 " 1.00 20.0 m s 2 = " 0.300 m s 2 (b) ! = tan " 1 " 0.300 0.800 # \$ % & ' ( = " 20.6 ° = 339 ° from + x axis (c) At t = 25.0 s x f = x i + v xi t + 1 2 a x t 2 = 10.0 + 4.00 25.0 ( ) + 1 2 0.800 ( ) 25.0 ( ) 2 = 360 m y f = y i + v yi t + 1 2 a y t 2 = ! 4.00 + 1.00 25.0 ( ) + 1 2 ! 0.300 ( ) 25.0 ( ) 2 = ! 72.7 m v xf = v xi + a x t = 4 + 0.8 25 ( ) = 24 m s v yf = v yi + a y t = 1 ! 0.3 25 ( ) = ! 6.5 m s " = tan ! 1 v y v x # \$ % & ' ( = tan ! 1 ! 6.50 24.0 # \$ % & ' ( = ! 15.2 ° P3.5 r a = 3.00 ˆ j m s 2 ; r v i = 5.00 ˆ i m s ; r r i = 0 ˆ i + 0 ˆ j (a) r r f = r r i + r v i t + 1 2 r a t 2 = 5.00 t ˆ i + 1 2 3.00 t 2 ˆ j ! " # \$ % & m r v f = r v i + r a t = 5.00 ˆ i + 3.00 t ˆ j ( ) m s (b) t = 2.00 s , r r f = 5.00 2.00 ( ) ˆ i + 1 2 3.00 ( ) 2.00 ( ) 2 ˆ j = 10.0 ˆ i + 6.00 ˆ j ( ) m so x f = 10.0 m , y f = 6.00 m r v f = 5.00 ˆ i + 3.00 2.00 ( ) ˆ j = 5.00 ˆ i + 6.00 ˆ j ( ) m s v f = r v f = v xf 2 + v yf 2 = 5.00 ( ) 2 + 6.00 ( ) 2 = 7.81 m s
Chapter 3 3 P3.6 (a) For the x -component of the motion we have x f = x i + v xi t + 1 2 a x t 2 . 0.01 m = 0 + 1.80 ! 10 7 m s ( ) t + 1 2 8 ! 10 14 m s 2 ( ) t 2 4 ! 10 14 m s 2 ( ) t 2 + 1.80 ! 10 7 m s ( ) t " 10 " 2 m = 0 t = " 1.80 ! 10 7 m s ± 1.8 ! 10 7 m s ( ) 2 " 4 4 ! 10 14 m s 2 ( ) " 10 " 2 m ( ) 2 4 ! 10 14 m s 2 ( ) = " 1.8 ! 10 7 ± 1.84 ! 10 7 m s 8 ! 10 14 m s 2 We choose the + sign to represent the physical situation t = 4.39 ! 10 5 m s 8 ! 10 14 m s 2 = 5.49 ! 10 " 10 s . Here y f = y i + v yi t + 1 2 a y t 2 = 0 + 0 + 1 2 1.6 ! 10 15 m s 2 ( ) 5.49 ! 10 " 10 s ( ) 2 = 2.41 ! 10 " 4 m . So, r r f = 10.0 ˆ i + 0.241 ˆ j ( ) mm . (b) r v f = r v i + r a t = 1.80 ! 10 7 m s ˆ i + 8 ! 10 14 m s 2 ˆ i + 1.6 ! 10 15 m s 2 ˆ j ( ) 5.49 ! 10 " 10 s ( ) = 1.80 ! 10 7 m s ( ) ˆ i + 4.39 ! 10 5 m s ( ) ˆ i + 8.78 ! 10 5 m s ( ) ˆ j = 1.84 ! 10 7 m s ( ) ˆ i + 8.78 ! 10 5 m s ( ) ˆ j (c) r v f = 1.84 ! 10 7 m s ( ) 2 + 8.78 !