Velocity is another vector quantity. When travelling in water, the current also has a velocity that pushes the traveller. To determine the resultant velocity, and the speed of travel, we need to add the velocity vectors!
Josh can paddle at a speed of \(5\;[\text{km}/\text{h}]\) in still water. He wishes to cross a river \(400 \; [\text{m}]\) wide that has a current of \(2 \; [\text{km}/\text{h}]\). If he steers the canoe in a direction perpendicular to the current, determine the resultant velocity and find the point on the opposite bank where the canoe touches. If he wishes to travel straight across the river, determine the direction he must head
and the time it will rake him to cross the river.
Step 1
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The current has a velocity which will push Josh downstream as he paddles across. Determine the resultant
velocity by adding the vectors.
\(\vert\vec{r}\vert=\sqrt{5^2+2^2} =\sqrt{29}\approx5.4\text{km/h} \)
\(θ=\tan^{-1}\left(\dfrac{2}{5}\right) = 21.8°\)
\(∴\vec{r}=\sqrt{29}\text{ km/h} \; [21.8° \text{off of his original direction}] \)
Step 2
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We can draw a similar triangle to represent the distance travelled. The angle will be the same as the velocity.
\(\tan{(21.8°)}=\dfrac{x}{400} \)
\(x=160\; m \)
∴ Josh touches at 160 m downstream from his starting pt. on the opposite side
Step 3
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In order to travel straight across the river, Josh will need to paddle upstream.

\(\sin{θ}=\dfrac{2}{5}\)
\(θ=23.6°\)
∴ He must head 23.6° up the stream off of the perpendicular to the current
\(\vert\vec{r}\vert=\sqrt{5^2-2^2} \)
\(|\vec{r}|=\sqrt{21}\approx4.6\text{km/h} \)
An airplane heading northwest at \(500 \; [\text{km/h}]\) encounters a wind of \(120 \; [\text{km/h}]\) from \(65°\) east of north. Determine the resultant ground velocity of the plane.
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\(\vert\vec{r}\vert=500^2+120^2-2\times500\times120\times\cos{(110°)} \)
\(\vert\vec{r}\vert=553 \; \text{km/h}\)
\(\dfrac{\sin{(110°)}}{553}=\dfrac{\sin{θ}}{500} \)
\(θ=58.2° \)
\(\vec{r}=553 \; \text{km/h} \; [\text{N56.8°W}\text{ or }\text{W33.2°N}] \)