1.72 A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction.

1.72 A sailor in a small sailboat encounters shifting winds. She sails 2.00 km east, then 3.50 km southeast, and then an additional distance in an unknown direction. Her final position is 5.80 km directly east of the starting point (Fig. P1.72). Find the magnitude and direction of the third leg of the journey. Draw the vector-addition diagram and show that it is in qualitative agreement with

your numerical solution.

SOLUTION

Let $\vec{A} = 2.00 \ km$, east

$\vec{B} = 3.50 \ km, 45^o$ south of east

$\vec{C} = $?

$\vec{R} = 5.80 \ km, east$.

Vector $\vec{R}=\vec{A} +\vec{B} +\vec{C}$, but we don't know what is $\vec{C}$ so we have to rearrange this equation and have, $\vec{C}=\vec{R} -\vec{A} -\vec{B}$. Solving for the components of the given vectors where the angles are measured from the +$x$-axis rotating towards +$y$-axis we have,

$R_x=(5.80 \ km) \ cos \ 0^o=5.80 \ km$

$R_y=(5.80 \ km) \ sin \ 0^o=0$

$A_x=(2.00 \ km) \ cos \ 0^o=2.00 \ km$

$A_y=(2.00 \ km) \ sin\ 0^o=0$

$B_x=(3.50 \ km) \ cos \ (360^o-45^o)=2.47 \ km$

$B_y=(3.50 \ km) \ sin \ (360^o-45^o)=-2.47 \ km$

Then the components of $\vec{C}$ are,

C_x=R_x-A_x-B_x=5.80 \ km-2.00 \ km)-2.47 \ km=1.33 \ km

C_y=R_y-A_y-B_y=0-0-(-2.47 \ km)=2.47 \ km

Then use the Pythagorean theorem and trigonometry to find the magnitude and direction,

$C=\sqrt{(C_x)^2+(C_y)^2}=2.81 \ km$

$tan \ \theta=\frac{C_y}{C_x}$

$\theta=61.7^o$ measured counterclockwise from the +$x$-axis. Or it is $90^o-61.7^o=28.0^o$ east of north.

The vector-addition diagram of $\vec{C}=\vec{R} -\vec{A} -\vec{B}$,