Answer to Question #125187 in Physics for qwerty

Question #125187
A sailor in a small boat encounters shifting winds. She sails 8 km south, then 15 km 30 E of N, and then 12 km 25 N of W. Use any method to determine the magnitude of her resultant displacement in kilometers
1
Expert's answer
2020-07-06T15:25:36-0400

The resultant displacement is shown on the figure (OC).





Let's denote the origin as O. Then, using the law of cosines, from the triangle AOB find:


OB2=OA2+AB22OAABcos(30°)=82+152281522119.2OB10.9kmOB^2 = OA^2 + AB^2 - 2OA\cdot AB\cdot \cos(30\degree) = 8^2 + 15^2 - 2\cdot 8\cdot 15\cdot \dfrac{\sqrt{2}}{2} \approx119.2\\ OB \approx 10.9 km

Using the law of sines, from the triangle AOB find:


OBA=arcsin(8kmsin30°10.9km)21.5°\angle OBA = \arcsin{(\dfrac{8km\cdot \sin 30\degree}{10.9 km})} \approx 21.5\degree

From the right triangle ADB, the angle DBA is equal to 180-90-30 = 60 degrees. Thus, the angle CBO will be:


CBO=DBA+25°OBA=60°+25°21.5°=63.5°\angle CBO = \angle DBA+25\degree-\angle OBA = 60\degree+25\degree-21.5\degree = 63.5\degree

Then, using the law of cosines, from the triangle OCB find:


OC2=OB2+CB22OBCBcos(63.5°)==10.92+122210.912cos(63.5°)116.9OC10.8kmOC^2 = OB^2 + CB^2 - 2OB\cdot CB\cdot \cos(63.5\degree) =\\ = 10.9^2 + 12^2 - 2\cdot 10.9\cdot 12\cdot \cos(63.5\degree)\approx116.9\\ OC \approx 10.8 km

Answer. The resultant displacement is 18.8 km.


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