magnitude of the total displacement?

Question

A person walks 25 m west and then 45 m at an angle of
60degrees north of east.

What is the magnitude of the total displacement?

Answer 1

You can do this using vectors:

first walk = 25m to the west, position vector is p1 = -25i

second walk = 45m north of east, position vector is

p2 = [45*cos(60)] i + [45*sin(60)] j = 22.5i + 22.5*sqrt(3)j

Total displacement = p1 + p2 = -25i + 22.5i + 22.5*sqrt(3)j

= -2.5i + 22.5*sqrt(3)j

Magnitude of displacement = magnitude of the vector

= sqrt [ (-2.5)^2 + { 22.5*sqrt(3) }^2 ]

= sqrt { 6.25 + 1518.75 }

= sqrt { 1525 }

= 39.05 m

Answer 2

Let unknown distance be x m
x² = 45² + 25² - 2 x 45 x 25 x cos 60° (cosine rule)
x² = 2650 - 1125
x² = 1528
x = 39.1

Magnitude of displacement = 39.1 m

Answer 3

total east = (-25m) + (45m)cos60 = -2.5 m
total north = (45m)sin60 = 39 m

magnitude = sqrt(2.5^2 + 39^2) = 39 m