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

$\theta = tan^{-1}\frac{\deta y}{\delta x}$

$\theta = tan^{-1}\frac{8}{15}$

$\theta = 28.1 degree$

magnitude of displacement is given as

$d = \sqrt{x^2 + y^2}$

$d = \sqrt{15^2 + 8^2}$

$d= 17 km$

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