return to the old oak tree, in what direction should you head and how far will you walk? Use components to solve this problem. (b) To see whether your calculation in part (a) is reasonable, check it with a graphical solution drawn roughly to scale.
SOLUTION
Let $\vec{A} =825 \ m$, south
$\vec{B} =1.25 \ km$, $30.0^o$ west of north $= 1.25 \ x \ 10^3 \ m$, $30.0^o$ west of north
$\vec{C} =1.00 \ km$, $40.0^o$ north of east $= 1.00 \ x \ 10^3 \ m$, $40.0^o$ north of east
Since I will be returning to where I started then my total displacement $\vec{d}=0$, and let my last displacement be $\vec{D}$,
$\vec{d}=0=\vec{A} +\vec{B}+ \vec{C} + \vec{D} $
Re-arranging the equation since we don't know what $\vec{D}$ is, we have
$\vec{D}=-\vec{A} -\vec{B}-\vec{C}$
Now, solve for the components of the given angles where the angles are measured from the +x-axis rotating towards +y-axis, we have
$A_x= (825 \ m) \ cos \ 270^o=0$
$A_y= (825 \ m) \ sin \ 270^o=-825 \ m$
$B_x= (1.25 \ x \ 10^3 \ m) \ cos \ (90^o+30^o)=-625.0 \ m$
$B_y= (1.25 \ x \ 10^3 \ m) \ sin \ (90^o+30^o)=1.083 \ x \ 10^3 \ m$
$C_x= (1.00 \ x \ 10^3 \ m) \ cos \ (40^o)=766.0 \ m$
$C_y= (1.00 \ x \ 10^3 \ m) \ sin \ (40^o)=642.8 \ m$
Then we can solve for the components of the last displacement, $\vec{D}$$D_x=-A_x-B_x-C_x=0-(-625.0 \ m)-766.0 \ m=-144.0 \ m$
$D_y=-A_y-B_y-C_y=-(-825 \ m)-1.083 \ x \ 10^3 \ m-642.8 \ m=-900.8$
Solving for the magnitude and direction,
$D=\sqrt{(D_x)^2+(D_y)^2}=912 \ m$
$tan \ \theta=\frac{D_y}{D_x}$
$\theta=81^o$ measured counterclockwise from the -x-axis, or $8.9^o$ west of south.
A graphical solution is shown below,
