Here student moves as
[tex]d_n = 1 miles[/tex] North
[tex]d_w = 0.4 miles[/tex] West
[tex]d_s = 0.1 miles[/tex] South
Now total distance moved by the student is just the algebraic sum of all
So it is
[tex]d = d_n + d_w + d_s = 1 + 0.4 + 0.1 = 1.5 miles[/tex]
So here we can also find the displacement
displacement is the straight line distance between initial and final position
so net displacement here towards north direction is given as
[tex]d_n = 1 - 0.1 = 0.9 miles[/tex] North
net displacement towards west is
[tex]d_w = 0.4 miles[/tex]
now the net displacement will be given by Pythagoras Theorem
[tex]d = \sqrt{d_n^2 + d_w^2} [/tex]
[tex]d = \sqrt{0.9^2 + 0.4^2} = 0.98 miles[/tex]
and direction is given as
[tex]\theta = tan^{-1}\frac{0.9}{0.4}[/tex]
[tex]\theta = 66 deree[/tex] North of west
