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You are camping with two friends, Joe and Karl. Since all three of you like your privacy, you don't pitch your tents close together. Joe's tent is 19.0 m from yours, in the direction 19.0° north of east. Karl's tent is 45.0 m from yours, in the direction 39.0° south of east. What is the distance between Karl's tent and Joe's tent?

2 Answers

7 votes

Final answer:

The distance between Joe's tent and Karl's tent is approximately 36.84 m.

Step-by-step explanation:

To find the distance between Joe's tent and Karl's tent, we can use the concept of vector addition. We first need to break down the given distances and angles into their respective components:

  • Joe's tent: 19.0 m at 19.0° north of east
  • Karl's tent: 45.0 m at 39.0° south of east

Next, we can use the components to find the displacement from Joe's tent to Karl's tent:

  • For Joe's tent: North component = 19.0 m * sin(19.0°) = 6.36 m, East component = 19.0 m * cos(19.0°) = 17.88 m
  • For Karl's tent: North component = -45.0 m * sin(39.0°) = -27.10 m, East component = 45.0 m * cos(39.0°) = 34.37 m

Using the components, we can calculate the displacement from Joe's tent to Karl's tent:

  • North displacement = -27.10 m - 6.36 m = -33.46 m
  • East displacement = 34.37 m - 17.88 m = 16.49 m

Finally, we can use the Pythagorean theorem to find the magnitude of the displacement:

Magnitude = sqrt((-33.46 m)^2 + (16.49 m)^2) = 36.84 m

Therefore, the distance between Joe's tent and Karl's tent is approximately 36.84 m.

User Nimantha
by
7.3k points
4 votes

Answer:

Distance between Karl and Joe is 38.467 m

Solution:

Let us assume that you are at origin

Now, as per the question:

Joe's tent is 19 m away from yours in the direction
19.0^(\circ) north of east.

Now,

Using vector notation for Joe's location, we get:


\vec{r_(J)} = 19cos(19.0^(\circ))\hat{i} + 19sin(19.0^(\circ))\hat{j}


\vec{r_(J)} = 17.96\hat{i} + 6.185\hat{j} m

Now,

Karl's tent is 45 m away from yours and is in the direction
39.0^(\circ)south of east, i.e.,
- 39.0^(\circ) from the positive x-axis:

Again, using vector notation for Karl's location, we get:


\vec{r_(K)} = 45cos(-319.0^(\circ))\hat{i} + 45sin(- 39.0^(\circ))\hat{j}


\vec{r_(K)} = 34.97\hat{i} - 28.32\hat{j} m

Now, obtain the vector difference between
\vec{r_(J)} and
\vec{r_(K)}:


\vec{r_(K)} - \vec{r_(J)} = 34.97\hat{i} - 28.32\hat{j} - (17.96\hat{i} + 6.185\hat{j}) m


\vec{d} = \vec{r_(K)} - \vec{r_(J)} = 17.01\hat{i} - 34.51\hat{j} m

Now, the distance between Karl and Joe, d:

|\vec{d}| = |17.01\hat{i} - 34.51\hat{j}|


d = \sqrt{(17.01)^(2) + (34.51)^(2)} m

d = 38.469 m

The distance between Karl's and Joe's tent is:

User Phrogg
by
7.3k points