Question

A hiker walks 4.5 km at an angle of 45° north of west. Then the hiker walks 4.5 km south. What is the magnitude,angle, and direction of the hiker's total
displacement?

Answer 1

Answer:

Magnitude of the hiker's total  displacement = 3.5 km

Angle and direction of the hiker's total  displacement = 22 degree south of west

Explanation:

The first Displacement in vector form is

$s_1$ = -4.5 cos 45i +4.5 sin 45j

= -3.182i + 3.182j

The second Displacement in vector form is  

$s_2$= -4.5j

The total displacement is ,

$s = s_1+ s_2$

= -3.182i + 3.182j - 4.5j

=-3.182i-1.318j

The magnitude of the displacement is,

$s= \sqrt{(-3.182)^2 + (-1.318)^2$

$s=\sqrt{10.125+1.7371}$

$s=\sqrt{11.862}$

s= 3.444

The direction is,

$\theta = tan^{-1}(\frac{1.318}{3.182})$

$\theta = tan^{-1}(0.4142)$

$\theta = 22.49^{\circ}$