$0\mathrm{.}55\frac{m}{s}0\mathrm{.}48\frac{m}{s}$vec$(v{)}_{\frac{C}{E}}$vec$(v{)}_{\frac{R}{E}}$vec$(v{)}_{\frac{C}{R}}$vec$(v{)}_{\frac{C}{E}}=$vec$(v{)}_{\frac{C}{R}}+$vec$(v{)}_{\frac{R}{E}}$vec$(v{)}_{\frac{C}{R}}=$vec$(v{)}_{\frac{C}{E}}-$vec$(v{)}_{\frac{R}{E}}$vec$(v{)}_{\frac{C}{R}}-0\mathrm{.}48\frac{m}{s}+\frac{0\mathrm{.}55\frac{m}{s}}{\sqrt[\phantom{2}]{2}}\frac{0\mathrm{.}55\frac{m}{s}}{\sqrt[\phantom{2}]{2}}0\mathrm{.}40\frac{m}{s}2(CH3)$

$(?)$ Help

Question $7$

$7of12$

A canoe has a velocity $of0\mathrm{.}55\frac{m}{s}$ southeast relative $to$ the earth. The canoe $ison$ a river that $is$ flowing $0\mathrm{.}48\frac{m}{s}$ east relative $to$ the earth.

For related problem$-$solving tips and $strtegies,$ you may want $to$ view a Video Tutor Solution $of$ Flying $in$ a crosswind

Important: $It$ you use this answer $in$ later parts, use the full unrounded value $in$ your calculations.

Review

I Constants

IDENTIFY: Apply the relative velocity relation.

SET $UP$: The relative velocities are vec$(v{)}_{\frac{C}{E}},$ the canoe relative $to$ the earth, vec$(v{)}_{\frac{R}{E}}$ the velocity $of$ the river relative $to$ the earth and vec$(v{)}_{\frac{C}{R}},$ the velocity $of$ the canoe relative $to$ the river.

EXECUTE: vec$(v{)}_{\frac{C}{E}}=$vec$(v{)}_{\frac{C}{R}}+$vec$(v{)}_{\frac{R}{E}}$ and therefore vec$(v{)}_{\frac{C}{R}}=$vec$(v{)}_{\frac{C}{E}}-$vec$(v{)}_{\frac{R}{E}}.$ The velocity components $of$ vec$(v{)}_{\frac{C}{R}}$ are $-0\mathrm{.}48\frac{m}{s}+\frac{0\mathrm{.}55\frac{m}{s}}{\sqrt[\phantom{2}]{2}},$ east and $\frac{0\mathrm{.}55\frac{m}{s}}{\sqrt[\phantom{2}]{2}},$ south, for a velocity relative $to$ the river $of0\mathrm{.}40\frac{m}{s}$

Part $B$

Find the direction $of$ the velocity $of$ the canoe relative $to$ the river.

Express your answer $in$ degrees.

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