Answer to Question #125695 in Classical Mechanics for Quanda

Question

A ship travels 20km in a direction 30 degrees south of east, then turns and goes due south for another 40km, then turns again 30 degrees of South of East and goes another 20km. What is the ship's displacement from its original position?

Answer 1

Suppose journey started from origin.

Now,

$x=40\cos(30^\circ)\:km\\ y=40\sin(30^\circ)+40\:km$

Thus, Coordinate of the final position is $(-x,-y)$ ,hence displacement is

$\sqrt{(-x-0)^2+(-y-0)^2}=69.28 \:\text{km}$