Chapter 5.1, Problem 5PP

To determine

Net displacement of S.

Answer to Problem 5PP

Displacement of S is 0.87 km in direction of $77°$ west of north.

Explanation of Solution

Given:

Displacement is 0.40 km in a direction of $60°$ west of north then 0.50 km due to west.

Formula used:

The x-component and y-component of a vector R are:

  $\begin{array}{l}{R}_x=R\cos \theta \\ R_y=R\sin \theta \end{array}$

Calculation:

Let west and north directions are considered as position vectors.

Therefore first displacement in west direction,

  $\begin{array}{c}{d}_{1w}=d_1\sin \theta \\ =0.40\times \sin 60°\\ =0.3464\text{ km}\end{array}$

And first displacement in north direction is;

  $\begin{array}{c}{d}_{1n}=d_1\cos \theta \\ =0.40\times \cos 60°\\ =0.2\text{ km}\end{array}$

After that S walks 0.50 km due to west.

Therefore second displacement in west direction,

  $d_{2w}=0.5\text{ km}$

And second displacement in north direction is,

  $d_{2n}=0\text{ km}$

Now total displacement in west direction is,

  $\begin{array}{c}{d}_w=d_{1w}+d_{2w}\\ =0.3464+0.5\\ =0.85\text{ km}\end{array}$

And total displacement in north direction is,

  $\begin{array}{c}{d}_n=d_{1n}+d_{2n}\\ =0.2+0\\ =0.2\text{ km}\end{array}$

Therefore net displacement is,

  $\begin{array}{c}d=\sqrt{d_n^2+d_w^2}\\ =\sqrt{{0.85}^2+{0.2}^2}\\ =\sqrt{0.7625}\\ =0.87\text{ km}\end{array}$

Direction of displacement is given by,

  $\begin{array}{c}\theta ={\tan }^{-1}\left(\frac{d_w}{d_n}\right)\\ ={\tan }^{-1}\left(\frac{0.85}{0.20}\right)\\ =77°\end{array}$

Conclusion:

Therefore displacement of S is 0.87 km in direction of $77°$ west of north.