1.78 A ship leaves the island of Guam and sails 285 km at 40.0 degrees north of west. In which direction must it now head and how far must it sail so that its resultant displacement will be 115 km directly east of Guam?

1.78 A ship leaves the island of Guam and sails 285 km at $40.0^o$ north of west. In which direction must it now head and how far must it sail so that its resultant displacement will be 115 km directly east of Guam?

SOLUTION

Let $\vec{A}=285 \ km$, $40.0^o$ north of west

$\vec{B}=$?

$\vec{d}=115 \ km$, east

The resultant displacement $\vec{d}=\vec{A}+\vec{B}$ and Figure 1 shows a vector-addition diagram drawn not to scale, but we don't know what is $\vec{B}$ so we re-arrange the equation and have, $\vec{B}=\vec{d}-\vec{A}$.

Solving for the components of the given vectors, where the angles are measured from the +x-axis towards the +y-axis, we have

$A_x=(285 \ km) \ cos \ (180^o-40^o)=-218.3 \ km$

$A_y=(285 \ km) \ sin \ (180^o-40^o)=183.2 \ km$

$d_x=(115 \ km) \ cos \ (0^o)=115 \ km$

$d_y=(115 \ km) \ sin \ (0^o)=0$

then the components of $\vec{B}$ are,

$B_x=d_x-A_x=115 \ km-(-218.3 \ km)=333.3 \ km$

$B_y=d_y-A_y=0-(183.2 \ km)=-183.2 \ km$

Solving for the magnitude and direction,

$B=\sqrt{(B_x)^2+(B_y)^2}=380 \ km$

$\theta=arctan \ (\frac{B_y}{B_x})=-29^o$ measured clockwise from the +x-axis, or $29^o$ south of east.

Figure 1