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 5.5 km, but when the snow clears, he discovers that he actually traveled 8.3 km at 52° north of due east. (a) how far and (b) in what direction (south of due west) must he now travel to reach base camp? (a) number 1 7.2 units km (b) number 11.3 units ° (degrees) etextbook and media hint save for later attempts: 1 of 10 used submit answer

Explanation:

Step1: Define coordinate system

Let east be +x, north be +y. Target position: $\vec{T} = (0, 5.5)$ km. Actual position: $\vec{A} = (8.3\cos52^\circ, 8.3\sin52^\circ)$ km.

Step2: Calculate required displacement

Displacement needed: $\vec{D} = \vec{T} - \vec{A}$
First compute components:
$x$-component: $0 - 8.3\cos52^\circ \approx 0 - 8.3\times0.6157 \approx -5.110$ km
$y$-component: $5.5 - 8.3\sin52^\circ \approx 5.5 - 8.3\times0.7880 \approx 5.5 - 6.540 \approx -1.040$ km

Step3: Find magnitude of displacement

Magnitude of $\vec{D}$: $|\vec{D}| = \sqrt{(-5.110)^2 + (-1.040)^2}$
$|\vec{D}| \approx \sqrt{26.11 + 1.082} \approx \sqrt{27.19} \approx 5.21$ km

Step4: Find direction of displacement

Direction $\theta$ (south of west): $\tan\theta = \frac{|\text{y-component}|}{|\text{x-component}|} = \frac{1.040}{5.110}$
$\theta \approx \arctan(0.2035) \approx 11.5^\circ$

Answer:

(a) 5.2 km
(b) 11.5 °(degrees)