Question

A hiker walks 15km due east then heads due north for 8km . What is the direction of the resultant vector?

Answer 1

his displacements are given as

`d_1 = 15 km` East

`d_2 = 8 km` North

now we can find the direction of displacement by using the concept

`\theta = tan^(-1)(\deta y)/(\delta x)`

`\theta = tan^(-1)(8)/(15)`

`\theta = 28.1 degree`

magnitude of displacement is given as

`d = √(x^2 + y^2)`

`d = √(15^2 + 8^2)`

`d= 17 km`

so the displacement is 17 km at angle 28.1 degree North of East