Question

A hiker walks 3.3km at an angle of 40 degrees north of west. Then, the hiker walks 3.4km at an angle of 60 degrees north of west. What is the hiker's total displacement

Answer 1

Answer

6.6 km

The description of the problem is shown in the attached figure, where the line "d" represents the final displacement vector.

First, the trekker walked 3.3km in a 40 ° direction, as shown in the figure. We can write this vector in its Cartesian coordinates:

$-3.3sin (40)x + 3.3cos (40)y$

Then the hiker walked 3.4 km in a 60 degree northwest direction.

We can write this as a vector in its Cartesian coordinates:

$-3.4sin (60)x + 3.4cos (60)y$.

When adding this two vectors we will obtain the final displacement "d"

$d = [- 3.3sin(40) -3.4sin (60)]x + [3.3cos(40) + 3.3cos (60)]y$

$d = -5.07x +4.23y$

To obtain the magnitude of this vector we calculate its module:

$\sqrt{5.07 ^2 +4.23 ^ 2}$

Then the magnitude of the final displacement was:

6.6 km