Answer
The magnitude and direction of the fourth displacement $D$ is 144m ,$40.9^{\circ}$ South of West.
Work Step by Step
We use a coordinate system where North is along the (+)ve y-direction and East is along the (+)ve x-direction.
Let $A,B,C,D$ be the displacement vectors where $D$ is the unknown vector. Their resultant displacement vector is $R=A+B+C+D$. $->(1)$
Now,
$A_{x}=180cos(180^{\circ})=-180m$, $A_{y}=180sin(180^{\circ})=0$
$B_{x}=210cos(315^{\circ})=148.5m ,B_{y}=210sin(315^{\circ})=-148.5m$
$C_{x}=280cos(60^{\circ})=140m,C_{y}=280sin(60^{\circ})=242.5m$
Since ,she ends up in the position where she started ,so from (1)
$R=0$
$=>A+B+C+D=0$
$=>D=-(A+B+C)$
So, $D_{x}=-(A_{x}+B_{x}+C_{x})$ and $D_{y}=-(A_{y}+B_{y}+C_{y})$
$D_{x}=-(-180+148.5+140)m=-108.5m$ and $D_{y}=-(-148.5+242.5)m=-94.0m$
Thus,$D=-(108.5m)i-(94.0m)j$ and it lies in the 3rd quadrant.
The magnitude of $D$ is given by
$|D|=\sqrt ((-108.5)^2+(-94.0)^2)=143.6m\approx144m$
The direction of $D$ is obtained from
$tan(\theta)=\frac{D_{y}}{D_{x}}=\frac{-94.0}{-108.5}$
$\theta=tan^{-1}(\frac{94.0}{108.5})=40.9^{\circ}$
Since $D$ lies in the 3rd quadrant ,$\theta$ is off by $180^{\circ}$.So
$\theta=180^{\circ}+40.9^{\circ}=220.9^{\circ}$
Therefore ,the magnitude and direction of $D$ is 144m ,$40.9^{\circ}$ South of West.
From the vector-addition diagram approximate to scale,we get
$|D|=144m$ and $\theta=40^{\circ} $South of West
Thus,this value is in qualitative agreement with our numerical solution.
