him the displacement, calculated using the method of components, that will return him to his hut.
SOLUTION
Let $\vec{A}=40 \ steps$, northeast
$\vec{B}=80 \ steps$, $60^o$ north of west
$\vec{C}=50 \ steps$, south
a) Figure 1.76a below shows a vector-addition diagram of $\vec{R}=\vec{A}+\vec{B}+\vec{C}$ |
| Figure 1.76a |
b) Since the explorer will return to where he started then her total displacement is zero and let his last displacement be $\vec{D}$. Figure 1.76b showed a vector-addition diagram of $\vec{R}=0=\vec{A}+\vec{B}+\vec{C}+\vec{D}$. Rearranging for us to obtain the magnitude and direction $\vec{D}$, and solving for the components of the given vectors where the angles used are measured from the +x-axis and rotates toward the +y-axis,
$\vec{D}=-\vec{A}-\vec{B}-\vec{C}$, then
$A_x=(40 \ steps) \ cos \ (360^o-45^o)=28.28 \ steps$$A_y=(40 \ steps) \ sin \ (360^o-45^o)=-28.28 \ steps$
$B_x=(80 \ steps) \ cos \ (180^o-60^o)=-40.0 \ steps$
$B_y=(80 \ steps) \ sin \ (180^o-60^o)=69.28 \ steps$
$C_x=(50 \ steps) \ cos \ (270^o)=0$
$C_y=(50 \ steps) \ sin \ (270^o)=-50 \ steps$
The components of the last displacement $\vec{D}$
$D_x=-A_x-B_x-C_x=-28.28 \ steps-(-40.0 \ steps)-0=11.72 \ steps$
$D_y=-A_y-B_y-C_y=-(-28.28 \ steps)-69.28 \ steps-(-50 \ steps)=9 \ steps$
Using the Pythagorean theorem and trigonometry to solve for the magnitude and direction of $\vec{D}$, we have
$D=\sqrt{(D_x)^2+(D_y)^2}=15 \ steps$
$tan \ \theta= \frac{D_y}{D_x}$
$\theta=38^o$ measured counterclockwise from the +x-axis of our imaginary xy-plane in the figure below. Or $38^o$ north of east.
 |
| Figure 1.76b |
