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04_P56InstructorSolution

04_P56InstructorSolution - 4.56 Model Let the earth frame...

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4.56. Model: Let the earth frame be S and a frame attached to the water be S . Frame S moves relative to S with velocity V . We define the x -axis along the direction of east and the y -axis along the direction of north for both frames. Solve: (a) The kayaker’s speed of 3.0 m/s is relative to the water. Since he’s being swept toward the east, he needs to point at angle θ west of north. In frame S , the water frame, his velocity is ˆ ˆ (3.0 m/s, west of north) ( 3.0sin m/s) (3.0cos m/s) v i j θ θ θ ′ = = − + G We can find his velocity in earth frame S from the Galilean transformation
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