Answer:371.564 mi
Explanation:
Given
Airplane flies northwest for 250 mi and then travels west 150 mi
That is first it travels 250cos45 in - ve x direction and simultaneously 250sin45 in y direction
after that it travels 150 mi in -ve x direction
So its position vector is given by
[tex]r=-250cos45\hat{i}-150\hat{i}+250sin45\hat{j}[/tex]
[tex]r=-\left ( 250cos45+150\right )\hat{i}+250sin45\hat{j}[/tex]
so magnitude of displacement is
[tex]|r|=\sqrt{\left ( \frac{250}{\sqrt{2}}+150\right )^2+\left ( \frac{250}{\sqrt{2}}\right )^2}[/tex]
|r|=371.564 mi
