Question #169728

Albert is going for a walk follows the path 300m East, then he going to 100m South, after a rest he proceed to 200m,30⁰South of West and lastly, he continues to 150m, 60⁰ North of West and stop. The total trip consists of four straight-line paths. At the end of the walk, what is the person’s resultant displacement measured from the starting point?

1
Expert's answer
2021-03-08T08:16:55-0500

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

dx=300 mcos0+100 mcos270+200 mcos(180+30)+150 mcos(18060)=51.8 m,d_x=300\ m\cdot cos0^{\circ}+100\ m\cdot cos270^{\circ}+200\ m\cdot cos(180^{\circ}+30^{\circ})+150\ m\cdot cos(180^{\circ}-60^{\circ})=51.8\ m,dy=dx=300 msin0+100 msin270+200 msin(180+30)+150 msin(18060)=70.1 m.d_y=d_x=300\ m\cdot sin0^{\circ}+100\ m\cdot sin270^{\circ}+200\ m\cdot sin(180^{\circ}+30^{\circ})+150\ m\cdot sin(180^{\circ}-60^{\circ})=-70.1\ m.Then, the resultant displacement can be found from the Pythagorean theorem:


d=dx2+dy2=(51.8 m)2+(70.1 m)2=87.16 m.d=\sqrt{d_x^2+d_y^2}=\sqrt{(51.8\ m)^2+(-70.1\ m)^2}=87.16\ m.

We can find the angle as follows:


θ=sin1(dyd)=sin1(70.1 m87.16 m)=53.5.\theta=sin^{-1}(\dfrac{d_y}{d})=sin^{-1}(\dfrac{-70.1\ m}{87.16\ m})=-53.5^{\circ}.


The sign minus means that resultant displacement has direction 53.5S of E53.5^{\circ} S\ of\ E.

Therefore, the resultant displacement dd has magnitude 87.16 m and direction of 53.5S of E53.5^{\circ} S\ of\ E.


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