Question

An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) while returning to base camp. He was supposed to travel due north for 5.6 km , but when the snow clears, he discovers that he actually traveled 7.8 km at $\mathbf{50}\mathbf{°}$north of due east. (a) How far and (b) in what direction must he now travel to reach base camp?

Short answer

(a) The magnitude of the displacement to reach base camp is 5.0 km

(b) The angle of the displacement to reach base camp $4.3°$south of west.

Step-by-step solution

To understand the concept of the problem

Consider the distance between the original position and the base camp as vector $\overrightarrow{A}$and the distance between the new position and original position as vector $\overrightarrow{B}$. The distance between the new position and the base camp can be computed by subtracting vector B from vector A. The magnitude of the vector which would be the distance between the new location and the base camp is to be calculated by the general formula of magnitude of the vector. Similarly, angle between components of the vector, can be found as well.

Formulae

$$\begin{gathered} \overrightarrow{C}=\overrightarrow{A}‐\overrightarrow{B} \\ C=\left|\overrightarrow{C}\right|=\sqrt{{\left(\overrightarrow{C_x}\right)}^2+{\left(\overrightarrow{C_y}\right)}^2} \\ \theta ={\tan }^{-1}\left(\frac{\overrightarrow{C_y}}{\overrightarrow{C_x}}\right) \end{gathered}$$

Given

$$\begin{gathered} \overrightarrow{A}=\left(5.6\mathrm{km}\right)\hat{\mathrm{j}},90° \\ \overrightarrow{\mathrm{B}}=\left(7.8\mathrm{km}\right),50° \end{gathered}$$

To find x and y components of C→

$\overrightarrow{A}=\left(5.6\mathrm{km}\right)\hat{\mathrm{j}},90°$or

$\overrightarrow{A}=\left(5.6\mathrm{km}\right)\cos \left(90\right)\hat{\mathrm{i}}+\left(5.6\mathrm{km}\right)\sin \left(90\right)\hat{\mathrm{j}}$

$\overrightarrow{B}=\left(7.8\mathrm{km}\right),50°$ or

$\overrightarrow{B}=\left(7.8\mathrm{km}\right)\cos \left(50\right)\hat{\mathrm{i}}+\left(10\mathrm{m}\right)\sin \left(50\right)\hat{\mathrm{j}}$

The displacement vector $\overrightarrow{C}$is

$$\begin{gathered} \overrightarrow{C}=\overrightarrow{A}-\overrightarrow{B} \\ =\left[\left(5.6\mathrm{km}\right)\cos \left(90\right)\hat{\mathrm{i}}+\left(5.6\mathrm{km}\right)\sin \left(90\right)\hat{\mathrm{j}}\right]-\left[\left(7.8\mathrm{km}\right)\cos \left(50\right)\hat{\mathrm{i}}+\left(10\mathrm{m}\right)\sin \left(50\right)\hat{\mathrm{j}}\right] \\ \overrightarrow{C}=\left(-5.01\mathrm{km}\right)\hat{\mathrm{i}}-\left(0.38\mathrm{km}\right)\hat{\mathrm{j}} \end{gathered}$$

To calculate magnitude of  C→

The magnitude of $\overrightarrow{C}$is

$\sqrt{{\left(-5.01\right)}^2+{\left(-0.38\right)}^2}=5.0\mathrm{km}$

Therefore, magnitude of the displacement to reach base camp is$5.0\mathrm{km}$

To calculate angle between C→ and x axis

The angle is

${\tan }^{-1}\left(\frac{-0.38}{-5.01}\right)=4.3°$ South of due west

Therefore, angle of the displacement to reach base camp $4.3°$south of west.