Answer to Question #160669 in Mechanics | Relativity for ferly mae

Question

Given the following velocities: 380 m/s North, 400 m/s East, 520 m/s 35o North of West, 440 m/s 40o South of West, and 200 m/s 60o South of East. Solve for the resultant velocity.

Answer 1

Answer

Fitstly we divided into components

So then

X component of velocity

$v_x=400+100-410-337$

=-247m/s

Y component

$v_y=-173+380+298-283$

=222m/s

So resultant velocity

$v=\sqrt{v_x^2+v_y^2}=332.1m/s$