Chapter 5, Problem 61A

To determine

To Find: The hiker’s net displacement in the order A, B, C graphically.

Answer to Problem 61A

14.43 km

Explanation of Solution

The graphically representation of the hiker’s displacements below-

Given:

According to the question

  $\begin{array}{c}A=\left(8\cos {60}^0,8\sin {60}^0\right)\\ =\left(4,6.93\right)\\ B=\left(7\cos 0^0,7\sin 0^0\right)\\ =\left(7,0\right)\\ C=\left(4\cos {315}^0,4\sin {315}^0\right)\\ =\left(2.83,-2.83\right)\end{array}$

Formula used:

So, the sum of all positions

  $\begin{array}{l}\text{Sum}=(4+7+2.83,6.93+0-2.83)\\ \text{Sum}=(13.83,4.1)\end{array}$

Calculation:

The magnitude of sum

  $\begin{array}{c}\text{Magnitude of Sum}=\sqrt{{(13.83)}^2+{(4.1)}^2}\\ =\sqrt{208.1}\\ =14.43\text{ km}\end{array}$

Conclusion:

Hence, the displacement of hiker’s is 14.43 km.