Applications of Vectors - Velocity
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!
Step 1
The current has a velocity which will push Josh downstream as he paddles across. First, we can sketch a couple of diagrams representing speed:
Next, we can determine the resultant velocity using the Pythagorean theorem:
\(\vert\vec{r}\vert=\sqrt{(a)^2+(b)^2}\)
\(\vert\vec{r}\vert=\sqrt{5^2+2^2}\)
\(\vert\vec{r}\vert = \sqrt{29}\approx5.4\text{km/h}\)
Finally, we can determine the angle Josh must paddle by finding the inverse tangent of the 2 original vectors:
\(\theta =\tan^{-1}\left(\dfrac{2}{5}\right)\)
\(\theta = 21.8°\)
Therefore, we can determine Josh must paddle at a velocity of approximately \(\boldsymbol{5.4 \;\textbf{km/h} \; [21.8° \textbf{off of his original direction}]}\).
Step 2
We can draw a similar triangle to represent the distance travelled. The angle will be the same as the velocity:
Next, we can determine the distance travelled by using the tan trigonometric ratio:
\(\tan{(21.8°)}=\dfrac{x}{400}\)
\(x = (\tan{(21.8°))(400)\)
\(x \approx 160\; [\text{m}] \)
Therefore, we can determine that Josh touches at approximately \(\boldsymbol{160 \; [\textbf{m}]}\) downstream from his starting point on the opposite side.
Step 3
In order to travel straight across the river, Josh will need to paddle upstream. We can first draw a diagram to represent the distance and angle he needs to travel:
Next, we can determine the angle Josh must travel using the sin trigonometric ratio:
\(\sin{\theta}=\dfrac{2}{5}\)
\(\theta = 23.6°\)
Then, we can determine how far Josh must travel using the Pythagorean theorem:
\(a^2 = c^2 - b^2\)
In this instance, we can use \(\vert\vec{r}\) to represent \(a\). We can also identify \(c=5\) and \(b=2\):
\(\vert\vec{r}\vert=\sqrt{5^2-2^2} \)
\(|\vec{r}|=\sqrt{21}\approx4.6\text{km/h}\)
Therefore, we can determine that Josh must paddle upstream at a velocity of approximately \(\boldsymbol{4.6\textbf{km/h}}\) at an angle of \(\boldsymbol{23.6°}\) off of the perpendicular to the current.
As we have \(2\) velocities and an angle, we can use the Law of Cosines to determine the resultant ground velocity:
\(c^2 = a^2 + b^2 - 2ab\cos C\)
In this instance, we can classify \(a = 500\), \(b=120\), and \(\angle C = 110^{\circ}\). We can also use \(|\vec{r}|\) to represent \(c\).
Next, we can substitute these values into the above formula as such:
\(\vert\vec{r}\vert=500^2+120^2-2\times500\times120\times\cos{(110°)}\)
\(\vert\vec{r}\vert=553 \; \text{km/h}\)
Then, since we have 2 velocities and a corresponding angle, we can use the Law of Sines to determine the other angle:
\(\cfrac{a}{\sin A} = \cfrac{b}{\sin B}\)
We can susbtitute the appropriate values into the above formula as such:
\(\cfrac{\sin (110°)}{553}=\cfrac{\sin \theta}{500}\)
\(\theta = 58.2°\)
Therefore, we can determine the resultant ground velocity of the plane is \(\boldsymbol{553 \; \text{km/h} \; [\text{N56.8°W}}]\).
We can also express it as \(\boldsymbol{553 \; \text{km/h} \; [\text{W33.2°N}]}\).