his displacements are given as
[tex]d_1 = 15 km [/tex] East
[tex]d_2 = 8 km[/tex] North
now we can find the direction of displacement by using the concept
[tex]\theta = tan^{-1}\frac{\deta y}{\delta x}[/tex]
[tex]\theta = tan^{-1}\frac{8}{15}[/tex]
[tex]\theta = 28.1 degree[/tex]
magnitude of displacement is given as
[tex]d = \sqrt{x^2 + y^2}[/tex]
[tex]d = \sqrt{15^2 + 8^2}[/tex]
[tex]d= 17 km[/tex]
so the displacement is 17 km at angle 28.1 degree North of East
