Question

A ship sails 50 km due north and then sails 85 km, 45 degrees north east. How far is the ship from its starting point, and what is its exact location? Use the graphical method and analytical method

Answer 1

Answer on Question #64422-Physics-Molecular Physics-Thermodynamics

A ship sails $50\mathrm{km}$ due north and then sails $85\mathrm{km}$ , 45 degrees north east. How far the ship from its starting point and what is its exact location? Use the graphical method and analytical method

Solution

1. Analytical method.

$$\vec {A} = (0, 5 0) m$$ $$\vec {B} = (8 5 \cos 4 5, 8 5 \sin 4 5) = \left(\frac {8 5}{\sqrt {2}}, \frac {8 5}{\sqrt {2}}\right) m.$$ $$\vec {A} + \vec {B} = \left(\frac {8 5}{\sqrt {2}}, 5 0 + \frac {8 5}{\sqrt {2}}\right) m.$$

The distance is

$$d = \sqrt {\left(\frac {8 5}{\sqrt {2}}\right) ^ {2} + \left(5 0 + \frac {8 5}{\sqrt {2}}\right) ^ {2}} = 1 2 5 m.$$

The direction is

$$\theta = \tan^ {- 1} \frac {5 0 + \frac {8 5}{\sqrt {2}}}{\frac {8 5}{\sqrt {2}}} = 6 1 {}^ {\circ} \mathrm {n o r t h o f e a s t}.$$

2. Graphical method.

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