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 50 o north of due east. (a) How far and (b) in what direction must he now travel to reach base camp

Answer 1

Answer:

The explorer should travel to reach base camp to 5.02 Km at 4.28° south of due west.

Explanation:

Using trigonometric function like Sen(Ф), Cos(Ф) and Tan(Ф) we can get distance and direction that the explorer should travel to reach base camp. When we discompound the vector $X = 7.8*Cos(50) = 5.01 Km$ y $y = 7.8 * Sen (50) - 5.6 = 5.975 - 5.6 (Km) = 0.375 (Km)$ so that $Tan (\alpha ) = \frac{0.375}{5.01} = 4.28$; $\alpha = Arctang (\frac{0.375}{5.01}) = 4.28 (degree)$ to get how far we use Pythagorean theorem so $R^{2} = x^{2}+y^{2}$ so that $R=\sqrt{0.375^{2}+5.01^{2} } =5.02 (Km)$

Answer 2

Answer:

Part a)

$d_2 = 5.02 km$

Part b)

$\theta = 4.33 degree$ South of West

Explanation:

As we know that the displacement of the explorer while return to camp is given as

$d = 7.8 km$ at an angle 50 degree North of East

so it is given as

$d = 7.8 cos50 \hat i + 7.8 sin50 \hat j$

so here we have

$d_1 = 5.01\hat i + 5.98 \hat j$

so now we need the resultant displacement of 5.6 km due North

so we can say

$d = d_1 + d_2$

$5.6 \hat j = 5.01\hat i + 5.98 \hat j + d_2$

$d_2 = -5.01\hat i - 0.38\hat j$

so magnitude of the displacement is

$d_2 = \sqrt{5.01^2 + 0.38^2}$

$d_2 = 5.02 km$

Part b)

Direction of the displacement is given as

$\theta = tan^{-1}\frac{y}{x}$

$\theta = tan^{-1}\frac{-0.38}{-5.01}$

$\theta = 4.33 degree$ South of West