College Physics by Openstax Chapter 3 Problem 16

Solve the following problem using analytical techniques: Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B, as in Figure 3.58, then this problem asks you to find their sum R=A+B.)

Figure 3.58 The two displacements A and B add to give a total displacement R having magnitude R and direction θ.

Solution:

Considering the right triangle formed by the vectors A, B, and R. We can solve for the magnitude of R using the Pythagorean Theorem. That is

R=A2+B2=(18.0 m)2+(25.0 m)2=30.806 m≈30.8 m  (Answer)\begin{align*} R & = \sqrt{A^2+B^2} \\ & = \sqrt{\left( 18.0\ \text{m} \right)^2+\left( 25.0\ \text{m} \right)^2} \\ & =30.806 \ \text{m} \\ & \approx 30.8 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Then we solve for the compass direction by solving the value θ using the same right triangle.

θ=arctan⁡(BA)=arctan⁡(25.0 m18.0 m)=54.246∘≈54.2∘\begin{align*} \theta & = \arctan\left( \frac{B}{A} \right) \\ & = \arctan\left( \frac{25.0\ \text{m}}{18.0\ \text{m}} \right) \\ & = 54.246 ^\circ \\ & \approx 54.2 ^ \circ \end{align*}

Therefore, the compass direction of the resultant is 54.2° North of West.

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