Question

A sailboat sails $2.0\text{km}$ east, then $4.0\text{km}$ southeast, then an additional distance in an unknown direction. Its final position is $5.0\text{km}$ directly east of the starting point. Find the magnitude and direction of the third leg of the journey. (Given, $2\sqrt{2}=2.83$)

• A. $0.17\text{km},{\tan }^{-1}\left(16.65\right)$ East of North
• B. $4\text{km},{\tan }^{-1}\left(16.65\right)$ East of North
• C. $2.84\text{km},{\tan }^{-1}\left(16.65\right)$ North of East
• D. $1\text{km},{\tan }^{-1}\left(1\right)$ North of East

Answer 1

The correct option is C $2.84\text{km},{\tan }^{-1}\left(16.65\right)$ North of East

Taking East as $+x$-axis and North as $+y$-axis, then the displacement vectors are expressed as:

displacement-$1$ of the sailboat, ${\overrightarrow{d}}_1=2.0\hat{i}\text{km}$

displacement-$2$ of the sailboat, ${\overrightarrow{d}}_2=4.0\cos {45}^{\circ }\hat{i}-4.0\sin {45}^{\circ }\hat{j}\text{km}=2.83\hat{i}-2.83\hat{j}\text{km}$

displacement-$3$ of the sailboat, ${\overrightarrow{d}}_3$ (to be found)

Net displacement of the sailboat, $\overrightarrow{d}=5.0\hat{i}\text{km}$

The net displacement is expressed as:

$\overrightarrow{d}={\overrightarrow{d}}_1+{\overrightarrow{d}}_2+{\overrightarrow{d}}_3$
$5.0\hat{i}\text{km}=\left(2.0\hat{i}\text{km}\right)+(2.83\hat{i}-2.83\hat{j}\text{km})+{\overrightarrow{d}}_3$

${\overrightarrow{d}}_3=(0.17\hat{i}+2.83\hat{j})\text{km}$

The magnitude of the third leg of the journey is calculated as:

$|{\overrightarrow{d}}_3|=\sqrt{{0.17}^2+{2.83}^2}\approx 2.84\text{km}$

The direction of the third leg of the journey is calculated as:

$\theta ={\tan }^{-1}\left(\frac{2.83}{0.17}\right)$

$\theta ={\tan }^{-1}\left(16.65\right)$ North of East