Question

Suppose that you walk 15 meters at 30 degrees as measured from the East. Then you walk another 25 meters at 60 degrees from the East what is your net displacement

Answer 1

The net displacement is

`R= √(A^2+B^2+2AB \cos \theta)`

`=√(15^2+25^2+2AB \cos 30^\circ) \\\\=√(225+625+ \cos30^0) \\\\=38.7m`

Step-by-step explanation:

Suppose that you walk 15 meters at 30 degrees as measured from the East. Then you walk another 25 meters at 60 degrees from the East what is your net displacement

Given data

A = 15 m

B = 25 m

Angle between the vectors A and B is θ = 30°

The net displacement is

`R= √(A^2+B^2+2AB \cos \theta)`

`=√(15^2+25^2+2AB \cos 30^\circ) \\\\=√(225+625+ \cos30^0) \\\\=38.7m`