Vector magnitude,

$\overline{)\mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathbf{v}}\mathbf{\right|}{\mathbf{=}}\sqrt{{{\mathbf{v}}_{\mathbf{x}}}^{\mathbf{2}}\mathbf{+}{{\mathbf{v}}_{\mathbf{y}}}^{\mathbf{2}}}}$

Resolving vector components,

$\overline{)\begin{array}{rcl}{\mathbf{v}}_{\mathbf{x}}& {\mathbf{=}}& \mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathbf{v}}\mathbf{\right|}\mathbf{}\mathbf{cos}\mathbf{}\mathbf{\theta}\\ {\mathbf{v}}_{\mathbf{y}}& {\mathbf{=}}& \mathbf{\left|}\stackrel{\mathbf{\rightharpoonup}}{\mathbf{v}}\mathbf{\right|}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{\theta}\end{array}}$

Let positive y-axis point to the north and positive x-axis point to the east.

**Resolving components,**

${\stackrel{\mathbf{\rightharpoonup}}{\mathbf{v}}}_{\mathbf{r}\mathbf{/}\mathbf{e}}$= 0.460 m/s (cos 0° i + sin 0° j) = 0.460 m/s i + 0 m/s j

${\stackrel{\mathbf{\rightharpoonup}}{\mathbf{v}}}_{\mathit{c}\mathbf{/}\mathbf{e}}$= 0.400 m/s {cos (- 45°) i + sin (- 45°) j } = 0.283 m/s i - 0.283 m/s j

A canoe has a velocity of 0.400 m/s southeast relative to the earth. The canoe is on a river that is flowing at 0.460 m/s east relative to the earth. (Figure 1)

Find the magnitude of the velocity **v**_{c/r} of the canoe relative to the river.

Express your answer in meters per second.

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What scientific concept do you need to know in order to solve this problem?

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