Question

A map in a pirate's log gives directions to the location of a buried treasure. the starting location is an old oak tree. according to the map, the treasure's location is found by proceeding 24 paces north from the oak tree and then 40 paces northwest. at this location, an iron pin is sunk in the ground. from the iron pin, walk 14 paces south and dig. how far (in paces) from the oak tree is the spot at which digging occurs?

Answer 1

The distance from the oak tree to the spot where digging occurs is approximately 36 paces.

Step-by-step explanation:

To find the distance from the oak tree to the spot where digging occurs, we need to determine the displacement from the oak tree to the iron pin and then from the iron pin to the spot where digging occurs.

According to the map, the treasure is located by proceeding 24 paces north from the oak tree and then 40 paces northwest. So, the displacement from the oak tree to the iron pin can be represented as a vector A with magnitude 24 and an angle of 45 degrees.

From the iron pin, we walk 14 paces south, which can be represented as a vector B with magnitude 14 and an angle of 270 degrees. To find the overall displacement, we need to add vectors A and B.

Using trigonometry and vector addition, we can find that the distance from the oak tree to the spot where digging occurs is approximately 36 paces.

Answer 2

the displacements or the instructions as given as follows

1). 24 paces North

2). 40 Paces North West

3). 14 paces South

now all these displacement is written in vector form

we can say north is +y direction and South is - Y direction

while East is + x direction and West is - x direction

Now by the vector form

`\vec d_1 = 24 \hat j`

`\vec d_2 = 40cos45 \hat i + 40 sin45\hat j`

`\vec d_2 = 28.3\hat i + 28.3\hat j`

`\vec d_3 = -14\hat j`

now the net displacement from starting point is given as

`\vec d = \vec d_1 +\vec d_2 + \vec d_3`

`\vec d = 24\hat j + 28.3\hat i + 28.3\hat j - 14 \hat j`

`\vec d = 28.3 \hat i + 38.3\hat j`

magnitude is given as

`d = √(28.3^2 + 38.3^2)`

`d = 47.6 paces`

direction is given as

`\theta = tan^(-1)(38.3)/(28.3)`

`\theta = 53.53 degree`

so it is 47.6 Paces from starting point at 53.53 degree North of East