Question #178637

Katherine goes on an adventure to the tower beyond the Aguna forest. She starts her trip 

by travelling 25 km 45 degrees south of east and camps to spend the night. On the next day, 

Katherine continues her journey by following 60 degrees north of east, which is 40 km long, 

and finally reaches the tower. Determine the magnitude and direction of Katherine’s resultant 

displacement. Also, express the resultant displacement in terms of unit vectors.



Expert's answer

Let's first find xx- and yy-components of the resultant displacement:


dx=25 km⋅cos(360∘−45∘)+40 km⋅cos60∘=37.68 km,d_x=25\ km\cdot cos(360^{\circ}-45^{\circ})+40\ km\cdot cos60^{\circ}=37.68\ km,dy=25 km⋅sin(360∘−45∘)+40 km⋅sin60∘=16.96 km.d_y=25\ km\cdot sin(360^{\circ}-45^{\circ})+40\ km\cdot sin60^{\circ}=16.96\ km.

Then, we can find the resultant displacement from the Pythagorean theorem:


d=dx2+dy2=(37.68 km)2+(16.96 km)2=41.32 km.d=\sqrt{d_x^2+d_y^2}=\sqrt{(37.68\ km)^2+(16.96\ km)^2}=41.32\ km.

We can find the direction of the resultant displacement from the geometry:


θ=cos−1(dxd),\theta=cos^{-1}(\dfrac{d_x}{d}),θ=cos−1(37.68 km41.32 km)=24.2∘ N of E.\theta=cos^{-1}(\dfrac{37.68\ km}{41.32\ km})=24.2^{\circ}\ N\ of\ E.

Also, we can express the resultant displacement in terms of unit vectors:


d=(37.68i+16.96j) km.d=(37.68i+16.96j)\ km.
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