Resultant displacement is 29.2 km at [tex]83.1^{\circ}[/tex] north of west
Explanation:
To solve the problem, we have to use the rules of vector addition, resolving first each vector along the x- and y- direction.
Taking east as positive x direction and north as positive y- direction, we have:
- First displacement is 25.5 km east, therefore its components are
[tex]A_x = 25.5 km\\A_y = 0 km[/tex]
- Second displacement is 41.0 km northwest, so its components are
[tex]B_x = (41.0)cos(135^{\circ})=-29.0 km\\B_y =(41.0)sin(135^{\circ})=29.0 km[/tex]
So, the components of the resultant displacement are
[tex]R_x=A_x+B_x=25.5+(-29.0)=-3.5 km\\R_y=A_y+B_y=0+29.0=29.0 km[/tex]
And so, the magnitude is calculated using Pythagorean's theorem:
[tex]R=\sqrt{R_x^2+R_y^2}=\sqrt{(-3.5)^2+(29.0)^2}=29.2 km[/tex]
And the direction is given by
[tex]\theta=tan^{-1}(\frac{R_y}{|R_x|})=tan^{-1}(\frac{29.0}{3.5})=83.1^{\circ}[/tex]
Where the angle is measured from the west direction, since Rx is negative.
Learn more about displacement:
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