Chapter 5.1, Problem 3PP

To determine

The magnitude of net displacement of a hiker from his initial point.

Answer to Problem 3PP

The magnitude of displacement of the hiker from his initial point is $10\text{ km}$.

Explanation of Solution

Given:

The hiker walks 4.5 km in one direction. Then turns to right at an angle $45°$ and walks 6.4 km.

Formula Used:

For a set of two vectors A and B with included angle $\theta$ and its resultant R, the law of cosines is given by

  $\text{R}=\sqrt{{\text{A}}^2+{\text{B}}^2-2\text{ABcos}\theta }$

Calculation:

The movement of hiker can be graphically represented as in Figure 1 (Observe that, the 4.5 km is shown in negative x-direction and hence the displacement is -4.5 km. )

  

Here, the two individual vectors show the movement of hiker and the resultant of the vectors represented by S shows the displacement of hiker with respect to the initial point. The vector S is the resultant of the two given vectors. Since the given vectors are not perpendicular to each other, the law of cosines is used to find the resultant vector S. Applying the law of cosines for the given vectors, the resultant displacement is obtained as ,

  $\begin{array}{c}\text{S}=\sqrt{{\left(-4.5\right)}^2+{\left(6.4\right)}^2-2\left(-4.5\right)\left(6.4\right)\text{cos45}°}\\ =10.09\text{ km}\\ \simeq \text{10 km}\end{array}$

Conclusion:

The net displacement of the hiker with respect to his initial position is 10 km.