Answer to Question #246760 in Mechanics | Relativity for Mekirin

Question

A fisherman sailed farther out to sea to find a good catch. He sailed directly easy for 55.3km. Then he sailed northeast for 25.0km. From that point, he countinued sailing toward a straight path and found himself 200km, East from his original position.

A. Find the vector representing the third leg of the journey.

B. What was the total displacement of the fisherman at the end of the second leg of the journey?

Answer 1

Explanations & Calculations

• Refer to the figure attached.

• The three displacements can be represented by $\small \vec{AD},\,\vec{DB}\,\,\&\,\,\vec{BC}.$
• Then the vector representing the third leg can be represented by $\small |\vec{BC}| = x.$
• The displacement at the end of the second leg can be represented then by $\small |\vec{AB}| = y.$

• You can find the angle ADB to be $\scriptsize (180-45) =135^0.$

• Use the cosine theorem to the triangle BCD to calculate the unknown x value.

$\qquad\qquad \begin{aligned} \small \cos {BDC}&=\small \frac{BD^2+DC^2 -BC^2 }{2\times BD\times DC }\\ \small \cos{45}&=\small \frac{25^2+144.7^2-x^2 }{2\times 25\times 144.7}\\ \small x &=\small 128.25\,km \end{aligned}$

• Use sine theorem to calculate the angle c.

$\qquad\qquad \begin{aligned} \small \frac{\sin (135-\theta )}{144.7}&=\small \frac{\sin \theta}{25}\\ \small \theta &=\small 7.92^0 \end{aligned}$

• Then the vector representing the third leg is 128.25 km [S 7.92 E].

• Use the same theorem accordingly to the triangle ABD, with the known angle: 135 to calculate y.