Question

A car travels 3km due north, then 5km northeast. Determine the resultant displacement

Answer 1

Answer on Question #65513, Math / Analytic Geometry

Question:

A car travels 3km due north, then 5km northeast.

Determine the resultant displacement.

Solution:

In Cartesian coordinates the first part of car's travel is vector $\vec{A} = (0;3)$ , and the second part is vector $\vec{B} = (5\cdot \cos \alpha ;5\cdot \sin \alpha) = \left(\frac{5}{\sqrt{2}};\frac{5}{\sqrt{2}}\right)$ .

The resultant displacement vector $\vec{C} = \vec{A} +\vec{B} = (0;3) + \left(\frac{5}{\sqrt{2}};\frac{5}{\sqrt{2}}\right) = \left(\frac{5}{\sqrt{2}};3 + \frac{5}{\sqrt{2}}\right).$

Its length $\left|\vec{C}\right| = \sqrt{\left(\frac{5}{\sqrt{2}}\right)^2 + \left(3 + \frac{5}{\sqrt{2}}\right)^2} = \sqrt{\frac{25}{2} + 9 + \frac{30}{\sqrt{2}} + \frac{25}{2}} = \sqrt{34 + 15\sqrt{2}} \cong 7.43$

Answer:

7.43km

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