Question

question
a boat leaves the marina and sails 5 miles west, then 10 miles southeast. how far, in miles, from the marina is the boat? (round your answer to the nearest hundredth if necessary.)
provide your answer below:

Explanation:

Step1: Resolve vectors into components

Let the west - east direction be the $x$ - axis (west is negative $x$) and the north - south direction be the $y$ - axis (north is positive $y$).
The first displacement $\vec{d}_1=- 5\hat{i}+0\hat{j}$ (5 miles west).
The second displacement $\vec{d}_2 = 10\cos45^{\circ}\hat{i}-10\sin45^{\circ}\hat{j}=10\times\frac{\sqrt{2}}{2}\hat{i}-10\times\frac{\sqrt{2}}{2}\hat{j}=5\sqrt{2}\hat{i}-5\sqrt{2}\hat{j}$.

Step2: Find the net displacement vector

The net displacement $\vec{d}=\vec{d}_1+\vec{d}_2=(-5 + 5\sqrt{2})\hat{i}-5\sqrt{2}\hat{j}$.
The magnitude of the net displacement $d=\sqrt{(-5 + 5\sqrt{2})^2+(-5\sqrt{2})^2}$.
First, expand $(-5 + 5\sqrt{2})^2=(-5)^2-2\times5\times5\sqrt{2}+(5\sqrt{2})^2=25-50\sqrt{2}+50 = 75-50\sqrt{2}$.
And $(-5\sqrt{2})^2 = 50$.
Then $d=\sqrt{75-50\sqrt{2}+50}=\sqrt{125 - 50\sqrt{2}}$.
$d=\sqrt{125-50\times1.414}=\sqrt{125 - 70.7}=\sqrt{54.3}\approx7.81$.

Answer:

$7.81$