Question

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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° south of due east. (a) how far and (b) in what direction (south of due west) must he now travel to reach base camp?
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(b) number
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Explanation:

Step1: Define displacement vectors

Let the desired displacement to base camp be $\vec{D} = 5.8\hat{j}$ km (due north).
The actual displacement is $\vec{A} = 8.3\cos(32^\circ)\hat{i} + 8.3\sin(32^\circ)\hat{j}$ km.
Let the required correction displacement be $\vec{C}$, so $\vec{D} = \vec{A} + \vec{C}$, meaning $\vec{C} = \vec{D} - \vec{A}$.

Step2: Calculate x-component of $\vec{C}$

$C_x = 0 - 8.3\cos(32^\circ)$
$C_x = -8.3\times0.8480 \approx -7.038$ km

Step3: Calculate y-component of $\vec{C}$

$C_y = 5.8 - 8.3\sin(32^\circ)$
$C_y = 5.8 - 8.3\times0.5299 \approx 5.8 - 4.398 \approx 1.402$ km

Step4: Find magnitude of $\vec{C}$

$|\vec{C}| = \sqrt{C_x^2 + C_y^2}$
$|\vec{C}| = \sqrt{(-7.038)^2 + (1.402)^2} \approx \sqrt{49.53 + 1.966} \approx \sqrt{51.496} \approx 7.18$ km

Step5: Find direction of $\vec{C}$

Let $\theta$ be the angle south of west: $\tan\theta = \frac{|C_y|}{|C_x|}$
$\tan\theta = \frac{1.402}{7.038} \approx 0.1992$
$\theta = \arctan(0.1992) \approx 11.3^\circ$

Answer:

(a) 7.2 km
(b) $11.3^\circ$ south of due west