.dbdbmxxmΔ= ΔNow we express the geometrical condition that relative to the boatthe dog has moved a distanced= 2.4 m: ΔΔxxd+=which accounts for the fact that the dog moves one way and the boat moves the other. We substitute for |Δxb| from above: mmddbddbgwhich leads to 2.4 m1.92 m.1/1 (4.5/18)ddbdxmmΔ===++The dog is therefore 1.9 m closer to the shore than initially (where it was
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.