Chapter 5.1, Problem 4PP

To determine

To Calculate: The magnitude and direction of the ant’s displacement.

Explanation of Solution

Given:

Displacement towards south, $d_1=5.0\text{cm}$

Displacement towards southwest, $d_2=4.0\text{cm}$

Formula Used:

Resultant displacement can be obtained by:

  $d=\sqrt{d_1{}^2+d_2{}^2+2d_1{d}_2\cos \theta }$

Where, $\theta$ is the angle between two displacement vectors.

Direction of the resultant vector can be obtained by:

  $\alpha ={\tan }^{-1}\left(\frac{d_2\sin \theta }{d_1+d_2\cos \theta }\right)$

Calculation:

  $\theta =45°$

  $\begin{array}{c}d=\sqrt{d_1{}^2+d_2{}^2+2d_1{d}_2\cos \theta }\\ =\sqrt{{5.0}^2+{4.0}^2+2\left(5.0\right)\left(4.0\right)\cos 45°}\\ =\sqrt{69.28}\\ =8.32\text{cm}\end{array}$

  $\begin{array}{c}\alpha ={\tan }^{-1}\left(\frac{4.0\sin 45°}{5.0+4.0\cos 45°}\right)\\ ={\tan }^{-1}\left(0.36\right)\\ =19.85°\end{array}$

Conclusion:

Thus, the magnitude and direction of the ant’s displacement is $8.32\text{cm}$ and $19.85°$ respectively.