While following a treasure map, you start at an old oak tree. You first walk $825m$directly south, then turn and walk $\mathbf{1}\mathbf{.}\mathbf{25}\mathbf{}\mathit{k}\mathit{m}$at $\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$west of north, and finally walk $\mathbf{1}\mathbf{.}\mathbf{00}\mathbf{}\mathit{k}\mathit{m}$at $\mathbf{32}\mathbf{.}\mathbf{0}\mathbf{°}$north of east, where you find the treasure: a biography of Isaac Newton!

(a) To return to the old oak tree, in what direction should you head and how far will you walk? Use components to solve this problem.

(b) To see whether your calculation in part (a) is reasonable, compare it with a graphical solution drawn roughly to scale.

The given data can be expressed below as:

• The distance covered in the south is$825m$ .
• The distance covered in the west of north is $1.25km$.
• The distance covered in the west of north at an angle of $30.0°$.
• The distance covered in the north of east is $1.00km$.
• The distance covered in the north of east at an angle of $32.0°$.

This law states that when the two vectors represent the triangle’s two sides according to the order of the direction and magnitude, then the third side represents the resultant vector.

Drawing a triangle and representing the values of the distance gives the resultant side of the triangle, which shows the direction I have to work and the distance I will cover.

a)

The free body diagram has been drawn below in which the +x direction is to the right, and the +y direction is to the left.

The displacement vectors in the component form are illustrated below,

Let x is the displacement vector to the oak tree from the treasure. However, as the net displacement of the displacement vectors is zero, then the sum of the four vectors is zero.

$x+x_1+x_2+x_3=0$

Hence, substituting the values in the above equation, we get,

Hence, the resultant vector has been illustrated below,

Analyzing the above figure, the distance which I have to cover is:

$$\begin{gathered} \left[X\right]\approx \sqrt{{\left(.223m\right)}^2+{\left(787m\right)}^2} \\ \approx 818m \end{gathered}$$

Hence, to cover this distance, I have to walk south of west at an angle of,

$$\begin{gathered} tan\theta \approx \frac{787}{223} \\ \theta \approx 74.2° \end{gathered}$$

The equation of the angle in the east of south is expressed as:

$\alpha =90°-\theta$

Here $\alpha$is the angle in the east of south.

Substituting the values in the above equation, we get,

$$\begin{gathered} \alpha =90°-74.2° \\ =15.8° \end{gathered}$$

Thus, I have to head in the east of south direction, and I have to walk .

b)

The graphical solution for this problem can be expressed as:

The resultant of the above vector representation is,

$$\begin{gathered} d=\sqrt{{\left(-223m\right)}^2+{\left(-787m\right)}^2} \\ =818m \end{gathered}$$

Thus, it agrees with the calculation of part a.