Chapter 3, Problem 67CP

A pirate has buried his treasure on an island with five trees located at the points (30.0 m, −20.0 m), (60.0 m, 80.0 m). (−10.0 m, −10.0 m), (40.0 m, −30.0 m), and (−70.0 m, 60.0 m), all measured relative to some origin, as shown in Figure P3.46. His ship’s log instructs you to start at tree A and move toward tree B, but to cover only one-half the distance between A and B. Then move toward tree C, covering one-third the distance between your current location and C. Next move toward tree D, covering one-fourth the distance between where you are and D. Finally move toward tree E, covering one-fifth the distance between you and E, stop, and dig. (a) Assume you have correctly determined the order in which the pirate labeled the trees as A, B, C, D, and E as shown in the figure. What are the coordinates of the point where his treasure is buried? (b) What If? What if you do not really know the way the pirate labeled the trees? What would happen to the answer if you rearranged the order of the trees, for instance, to B (30 m, −20 m), A (60 m, 80 m), E (−10 m, −10 m), C (40 m, −30 m), and D (−70 m, 60 m)? State reasoning to show that the answer does not depend on the order in which the trees are labeled.

Figure P3.46

To determine

The coordinates of the treasure points.

Answer to Problem 67CP

The coordinates of the treasure points are $\left(10\widehat{\text{i}}+16\widehat{\text{j}}\right)\text{m}$.

Explanation of Solution

Section 1:

To determine: The first position vector moving from tree A to tree B.

Answer: The first position vector moving from tree A to tree B is $\left(45.0\widehat{\text{i}}\text{+30}\text{.0}\widehat{\text{j}}\right)\text{m}$.

Given information:

The coordinates of tree A are $\left(30.0\text{m,}-\text{20}\text{.0}\text{m}\right)$, the coordinates of tree B are $\left(60.0\text{m,80}\text{.0}\text{m}\right)$, the coordinates of tree C are $\left(-10.0\text{m,}-\text{10}\text{.0}\text{m}\right)$, the coordinates of tree D are $\left(40.0\text{m,}-3\text{0}\text{.0}\text{m}\right)$ and the coordinates of tree E are $\left(-70.0\text{m,60}\text{.0}\text{m}\right)$.

Formula to calculate the first position vector moving from tree A to tree B as per condition is,

$\overrightarrow{{\text{d}}_1}=\overrightarrow{\text{A}}+\frac{\overline{\text{B}}-\overrightarrow{\text{A}}}{2}$

• $\overrightarrow{{\text{d}}_{\text{1}}}$ is the position vector of first position.

Substitute $\left(60.0\widehat{\text{i}}+\text{80}\text{.0}\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{\text{B}}$ and $\left(30.0\widehat{\text{i}}-\text{20}\text{.0}\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{\text{A}}$ in the above equation.

$\begin{array}{c}\overrightarrow{{\text{d}}_1}=\left(30.0\widehat{\text{i}}-\text{20}\text{.0}\widehat{\text{j}}\right)\text{m}+\frac{\left(60.0\widehat{\text{i}}+\text{80}\text{.0}\widehat{\text{j}}\right)\text{m}-\left(30.0\widehat{\text{i}}-\text{20}\text{.0}\widehat{\text{j}}\right)\text{m}}{2}\\ =\left(45.0\widehat{\text{i}}\text{+30}\text{.0}\widehat{\text{j}}\right)\text{m}\end{array}$

Section 2:

To determine: The second position vector moving from first position to tree C.

Answer: The second position vector moving from first position to tree C is $\left(26.67\widehat{\text{i}}+16.67\widehat{\text{j}}\right)\text{m}$.

Given information:

The coordinates of tree A are $\left(30.0\text{m,}-\text{20}\text{.0}\text{m}\right)$, the coordinates of tree B are $\left(60.0\text{m,80}\text{.0}\text{m}\right)$, the coordinates of tree C are $\left(-10.0\text{m,}-\text{10}\text{.0}\text{m}\right)$, the coordinates of tree D are $\left(40.0\text{m,}-3\text{0}\text{.0}\text{m}\right)$ and the coordinates of tree E are $\left(-70.0\text{m,60}\text{.0}\text{m}\right)$.

Formula to calculate the second position vector moving from first location to tree C as per condition is,

$\overrightarrow{{\text{d}}_2}=\overrightarrow{{\text{d}}_1}+\frac{\overrightarrow{\text{C}}-\overrightarrow{{\text{d}}_1}}{3}$

• $\overrightarrow{{\text{d}}_2}$ is the position vector of second position.

Substitute $\left(-10.0\widehat{\text{i}}-1\text{0}\text{.0}\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{\text{C}}$ and $\left(45.0\widehat{\text{i}}\text{+30}\text{.0}\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{{\text{d}}_{\text{1}}}$ in the above equation.

$\begin{array}{c}\overrightarrow{{\text{d}}_2}=\left(45.0\widehat{\text{i}}+5\text{0}\text{.0}\widehat{\text{j}}\right)\text{m}\frac{\left(-10.0\widehat{\text{i}}-1\text{0}\text{.0}\widehat{\text{j}}\right)\text{m}-\left(45.0\widehat{\text{i}}+3\text{0}\text{.0}\widehat{\text{j}}\right)\text{m}}{3}\\ =\left(26.67\widehat{\text{i}}+16.67\widehat{\text{j}}\right)\text{m}\end{array}$

Section 3:

To determine: The third position vector moving from second position to tree D.

Answer: The third position vector moving from second position to tree D is $\left(30\widehat{\text{i}}+\text{5}\widehat{\text{j}}\right)\text{m}$.

Given information:

The coordinates of tree A are $\left(30.0\text{m,}-\text{20}\text{.0}\text{m}\right)$, the coordinates of tree B are $\left(60.0\text{m,80}\text{.0}\text{m}\right)$, the coordinates of tree C are $\left(-10.0\text{m,}-\text{10}\text{.0}\text{m}\right)$, the coordinates of tree D are $\left(40.0\text{m,}-3\text{0}\text{.0}\text{m}\right)$ and the coordinates of tree E are $\left(-70.0\text{m,60}\text{.0}\text{m}\right)$.

Formula to calculate the third position vector moving from second location to tree D as per condition is,

$\overrightarrow{{\text{d}}_{\text{3}}}=\overrightarrow{{\text{d}}_{\text{2}}}+\frac{\overrightarrow{\text{D}}-\overrightarrow{{\text{d}}_{\text{2}}}}{\text{4}}$

• $\overrightarrow{{\text{d}}_{\text{3}}}$ is the position vector of third position.

Substitute $\left(40.0\widehat{\text{i}}-3\text{0}\text{.0}\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{\text{D}}$ and $\left(26.67\widehat{\text{i}}+16.67\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{{\text{d}}_{\text{2}}}$ in the above equation.

$\begin{array}{c}\overrightarrow{{\text{d}}_3}=\left(26.67\widehat{\text{i}}+16.67\widehat{\text{j}}\right)\text{m}+\frac{\left(40.0\widehat{\text{i}}-3\text{0}\text{.0}\widehat{\text{j}}\right)\text{m}-\left(26.67\widehat{\text{i}}+16.67\widehat{\text{j}}\right)\text{m}}{4}\\ =\left(30\widehat{\text{i}}+\text{5}\widehat{\text{j}}\right)\text{m}\end{array}$

Section 4:

To determine: The fourth position vector moving from third position to tree E.

Answer: The fourth position vector moving from third position to tree E is $\left(10\widehat{\text{i}}+16\widehat{\text{j}}\right)\text{m}$.

Given information:

The coordinates of tree A are $\left(30.0\text{m,}-\text{20}\text{.0}\text{m}\right)$, the coordinates of tree B are $\left(60.0\text{m,80}\text{.0}\text{m}\right)$, the coordinates of tree C are $\left(-10.0\text{m,}-\text{10}\text{.0}\text{m}\right)$, the coordinates of tree D are $\left(40.0\text{m,}-3\text{0}\text{.0}\text{m}\right)$ and the coordinates of tree E are $\left(-70.0\text{m,60}\text{.0}\text{m}\right)$.

Formula to calculate the fourth position vector moving from third location to tree E as per condition is,

$\overrightarrow{{\text{d}}_{\text{4}}}=\overrightarrow{{\text{d}}_{\text{3}}}+\frac{\overrightarrow{\text{E}}-\overrightarrow{{\text{d}}_{\text{3}}}}{\text{5}}$

• $\overrightarrow{{\text{d}}_{\text{4}}}$ is the position vector of fourth position.

Substitute $\left(-70.0\widehat{\text{i}}+\text{60}\text{.0}\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{\text{E}}$ and $\left(30\widehat{\text{i}}+\text{5}\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{{\text{d}}_{\text{3}}}$ in the above equation.

$\begin{array}{c}\overrightarrow{{\text{d}}_4}=\left(30\widehat{\text{i}}+\text{7}\text{.33}\widehat{\text{j}}\right)\text{m}\frac{\left(-70.0\widehat{\text{i}}+\text{60}\text{.0}\widehat{\text{j}}\right)\text{m}-\left(30\widehat{\text{i}}+\text{7}\text{.33}\widehat{\text{j}}\right)\text{m}}{5}\\ =\left(10\widehat{\text{i}}+16\widehat{\text{j}}\right)\text{m}\end{array}$

Conclusion:

Therefore, the coordinates of the treasure points are $\left(10\widehat{\text{i}}+16\widehat{\text{j}}\right)\text{m}$.

To determine

The new coordinates of the treasure points.

Answer to Problem 67CP

The new coordinates of the treasure points are $\left(10\widehat{\text{i}}+16\widehat{\text{j}}\right)\text{m}$.

Explanation of Solution

Section 1:

To determine: The first position vector moving from tree A to tree B.

Answer: The first position vector moving from tree A to tree B is $\left(45.0\widehat{\text{i}}\text{+30}\text{.0}\widehat{\text{j}}\right)\text{m}$.

Given information:

The coordinates of tree A are $\left(60.0\text{m,80}\text{.0}\text{m}\right)$, the coordinates of tree B are $\left(30.0\text{m,}-\text{20}\text{.0}\text{m}\right)$, the coordinates of tree C are $\left(40.0\text{m,}-3\text{0}\text{.0}\text{m}\right)$, the coordinates of tree D are $\left(-70.0\text{m,60}\text{.0}\text{m}\right)$ and the coordinates of tree E are $\left(-10.0\text{m,}-\text{10}\text{.0}\text{m}\right)$.

Formula to calculate the first position vector moving from tree A to tree B as per condition is,

$\overrightarrow{{\text{D}}_1}=\overrightarrow{\text{A}}+\frac{\overrightarrow{\text{B}}-\overrightarrow{\text{A}}}{2}$

• $\overrightarrow{{\text{D}}_{\text{1}}}$ is the position vector of first position.

Substitute $\left(30.0\widehat{\text{i}}-\text{20}\text{.0}\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{\text{B}}$ and $\left(60.0\widehat{\text{i}}+\text{80}\text{.0}\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{\text{A}}$ in the above equation.

$\begin{array}{c}\overrightarrow{{\text{D}}_1}=\left(60.0\widehat{\text{i}}+\text{80}\text{.0}\widehat{\text{j}}\right)\text{m}\frac{\left(30.0\widehat{\text{i}}-\text{20}\text{.0}\widehat{\text{j}}\right)\text{m}-\left(60.0\widehat{\text{i}}+\text{80}\text{.0}\widehat{\text{j}}\right)\text{m}}{2}\\ =\left(45.0\widehat{\text{i}}+\text{30}\text{.0}\widehat{\text{j}}\right)\text{m}\end{array}$

Section 2:

To determine: The second position vector moving from first position to tree C.

Answer: The second position vector moving from first position to tree C is $\left(43.33\widehat{\text{i}}+10\widehat{\text{j}}\right)\text{m}$.

Given information:

The coordinates of tree A are $\left(60.0\text{m,80}\text{.0}\text{m}\right)$, the coordinates of tree B are $\left(30.0\text{m,}-\text{20}\text{.0}\text{m}\right)$, the coordinates of tree C are $\left(40.0\text{m,}-3\text{0}\text{.0}\text{m}\right)$, the coordinates of tree D are $\left(-70.0\text{m,60}\text{.0}\text{m}\right)$ and the coordinates of tree E are $\left(-10.0\text{m,}-\text{10}\text{.0}\text{m}\right)$.

Formula to calculate the second position vector moving from first location to tree C as per condition is,

$\overrightarrow{{\text{D}}_2}=\overrightarrow{{\text{D}}_1}+\frac{\overrightarrow{\text{C}}-\overrightarrow{{\text{D}}_1}}{3}$

• $\overrightarrow{D_2}$ is the position vector of second position.

Substitute $\left(40.0\widehat{\text{i}}-3\text{0}\text{.0}\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{\text{C}}$ and $\left(45.0\widehat{\text{i}}+\text{30}\text{.0}\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{{\text{D}}_{\text{1}}}$ in the above equation.

$\begin{array}{c}\overrightarrow{{\text{D}}_{\text{2}}}=\left(45.0\widehat{\text{i}}+\text{30}\text{.0}\widehat{\text{j}}\right)\text{m}+\frac{\left(40.0\widehat{\text{i}}-3\text{0}\text{.0}\widehat{\text{j}}\right)\text{m}-\left(45.0\widehat{\text{i}}+\text{30}\text{.0}\widehat{\text{j}}\right)\text{m}}{3}\\ =\left(43.33\widehat{\text{i}}+10\widehat{\text{j}}\right)\text{m}\end{array}$

Section 3:

To determine: The third position vector moving from second position to tree D.

Answer: The third position vector moving from second position to tree D is $\left(15\widehat{\text{i}}+\text{22}\text{.5}\widehat{\text{j}}\right)\text{m}$.

Given information:

The coordinates of tree A are $\left(60.0\text{m,80}\text{.0}\text{m}\right)$, the coordinates of tree B are $\left(30.0\text{m,}-\text{20}\text{.0}\text{m}\right)$, the coordinates of tree C are $\left(40.0\text{m,}-3\text{0}\text{.0}\text{m}\right)$, the coordinates of tree D are $\left(-70.0\text{m,60}\text{.0}\text{m}\right)$ and the coordinates of tree E are $\left(-10.0\text{m,}-\text{10}\text{.0}\text{m}\right)$.

Formula to calculate the third position vector moving from second location to tree D as per condition is,

${\overrightarrow{\text{D}}}_3={\overrightarrow{\text{D}}}_2+\frac{\overrightarrow{\text{D}}-{\overrightarrow{\text{D}}}_2}{4}$

• $\overrightarrow{{\text{D}}_{\text{3}}}$ is the position vector of third position.

Substitute $\left(-70.0\widehat{\text{i}}+60.\text{0}\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{\text{D}}$ and $\left(43.33\widehat{\text{i}}+10\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{{\text{D}}_{\text{2}}}$ in the above equation.

$\begin{array}{c}\overrightarrow{{\text{D}}_3}=\left(43.33\widehat{\text{i}}+10\widehat{\text{j}}\right)\text{m}+\frac{\left(-70.0\widehat{\text{i}}+6\text{0}\text{.0}\widehat{\text{j}}\right)\text{m}-\left(43.33\widehat{\text{i}}+10\widehat{\text{j}}\right)\text{m}}{4}\\ =\left(15\widehat{\text{i}}+\text{22}\text{.5}\widehat{\text{j}}\right)\text{m}\end{array}$

Section 4:

To determine: The fourth position vector moving from third position to tree E.

Answer: The fourth position vector moving from third position to tree E is $\left(10\widehat{\text{i}}+16\widehat{\text{j}}\right)\text{m}$.

Given information:

The coordinates of tree A are $\left(60.0\text{m,80}\text{.0}\text{m}\right)$, the coordinates of tree B are $\left(30.0\text{m,}-\text{20}\text{.0}\text{m}\right)$, the coordinates of tree C are $\left(40.0\text{m,}-3\text{0}\text{.0}\text{m}\right)$, the coordinates of tree D are $\left(-70.0\text{m,60}\text{.0}\text{m}\right)$ and the coordinates of tree E are $\left(-10.0\text{m,}-\text{10}\text{.0}\text{m}\right)$.

Formula to calculate the fourth position vector moving from third location to tree E as per condition is,

$\overrightarrow{{\text{D}}_4}=\overrightarrow{{\text{D}}_3}+\frac{\overrightarrow{\text{E}}-\overrightarrow{{\text{D}}_3}}{5}$

• $\overrightarrow{D_4}$ is the position vector of fourth position.

Substitute $\left(-10.0\widehat{\text{i}}-1\text{0}\text{.0}\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{\text{E}}$ and $\left(15\widehat{\text{i}}+\text{22}\text{.5}\widehat{\text{j}}\right)\text{m}$ for $\overrightarrow{{\text{D}}_{\text{3}}}$ in the above equation.

$\begin{array}{c}\overrightarrow{{\text{D}}_4}=\left(15\widehat{\text{i}}+\text{22}\text{.5}\widehat{\text{j}}\right)\text{m}+\frac{\left(-10.0\widehat{\text{i}}-1\text{0}\text{.0}\widehat{\text{j}}\right)\text{m}-\left(15\widehat{\text{i}}+\text{22}\text{.5}\widehat{\text{j}}\right)\text{m}}{5}\\ =\left(10\widehat{\text{i}}+16\widehat{\text{j}}\right)\text{m}\end{array}$

Conclusion:

Therefore, the coordinates of the treasure points are $\left(10\widehat{\text{i}}+16\widehat{\text{j}}\right)\text{m}$. The coordinates of the treasure point is same in both the part so, the answer does not depend on the order in which the trees are labeled.