Chapter 1, Problem 69P

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 P1.69. 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 1.69

To determine

The co-ordinates point of the treasurer buried.

Answer to Problem 69P

The co-ordinates point of the treasurer buried is $\left(10.0\text{m, 16}\text{.0}\text{m}\right)$.

Explanation of Solution

The initial points of the pirate buried his treasure an island is,

  ${\overrightarrow{r}}_A=\left(30.0\widehat{i}-20.0\widehat{j}\right)\text{m}$        (I)

Here, ${\overrightarrow{r}}_A$ is the initial displacement of the pirate buried his treasure at point A.

The displacement of the pirate moved to point B from A is,

  $\begin{array}{c}{\overrightarrow{r}}_B-{\overrightarrow{r}}_A=\left(60.0\widehat{i}+80.0\widehat{j}\right)-\left(30.0\widehat{i}-20.0\widehat{j}\right)\\ =60.0\widehat{i}+80.0\widehat{j}-30.0\widehat{i}+20.0\widehat{j}\\ =30.0\widehat{i}+100\widehat{j}\end{array}$        (II)

Here, ${\overrightarrow{r}}_B$ is the displacement of the pirate buried his treasure at point B.

The displacement covered half of the point from B to A is,

  $\begin{array}{c}{\overrightarrow{r}}_2=\left(30.0\widehat{i}-20.0\widehat{j}\right)+\left(15.0\widehat{i}+50.0\widehat{j}\right)\\ =45.0\widehat{i}+30.0\widehat{j}\end{array}$        (III)

Here, ${\overrightarrow{r}}_2$ is the displacement of the pirate moved half of the above point to the point A.

The displacement from the current position to the point of the tree at C is,

  $\begin{array}{c}{\overrightarrow{r}}_C-{\overrightarrow{r}}_2=\left(-10.0\widehat{i}-10.0\widehat{j}\right)-\left(45.0\widehat{i}+30.0\widehat{j}\right)\\ =-10.0\widehat{i}-10.0\widehat{j}-45.0\widehat{i}-30.0\widehat{j}\\ =-55.0\widehat{i}-40.0\widehat{j}\end{array}$        (IV)

Conclusion:

The distance covered by one third of the distance between current location and point C is,

${\overrightarrow{r}}_3={\overrightarrow{r}}_2+\frac{1}{3}\Delta {\overrightarrow{r}}_{2C}$

Here, ${\overrightarrow{r}}_3$ is the distance covered by one third of the distance between current location and to the point C and ${\overrightarrow{r}}_{2C}$ is the current location.

Substitute the equation (III) and (IV) in above relation.

$\begin{array}{c}{\overrightarrow{r}}_3=45.0\widehat{i}+30.0\widehat{j}+\frac{1}{3}\left(-55.0\widehat{i}-40.0\widehat{j}\right)\\ =45.0\widehat{i}+30.0\widehat{j}-18.3\widehat{i}-13.3\widehat{j}\\ =26.7\widehat{i}+16.7\widehat{j}\end{array}$        (V)

The displacement from the current position to the tree moves towards the point at C is,

  $\begin{array}{c}{\overrightarrow{r}}_D-{\overrightarrow{r}}_3=\left(40.0\widehat{i}-30.0\widehat{j}\right)-\left(26.7\widehat{i}+16.7\widehat{j}\right)\\ =40.0\widehat{i}-30.0\widehat{j}-26.7\widehat{i}-16.7\widehat{j}\\ =13.3\widehat{i}-46.7\widehat{j}\end{array}$        (VI)

The distance traverse by one quarter of the distance between current locations is,

    ${\overrightarrow{r}}_4={\overrightarrow{r}}_3+\frac{1}{4}\left({\overrightarrow{r}}_D-{\overrightarrow{r}}_3\right)$

Here, ${\overrightarrow{r}}_3$ is the distance covered by one third of the distance between current location and to the point C and ${\overrightarrow{r}}_{2C}$ is the current location.

Substitute the equation (V) and (VI) in above relation.

$\begin{array}{c}{\overrightarrow{r}}_4=26.7\widehat{i}+16.7\widehat{j}+\frac{1}{4}\left(13.3\widehat{i}-46.7\widehat{j}\right)\\ =26.7\widehat{i}+16.7\widehat{j}+3.33\widehat{i}-11.7\widehat{j}\\ =30.0\widehat{i}+5.00\widehat{j}\end{array}$        (VII)

The displacement from the current position to the new location at the point E is,

  $\begin{array}{c}{\overrightarrow{r}}_E-{\overrightarrow{r}}_4=\left(-70.0\widehat{i}+60.0\widehat{j}\right)-\left(30.0\widehat{i}+5.00\widehat{j}\right)\\ =-70.0\widehat{i}+60.0\widehat{j}-30.0\widehat{i}-5.00\widehat{j}\\ =-100\widehat{i}+55.0\widehat{j}\end{array}$        (VIII)

The distance covered by one fifth of the distance between current locations to moving distance is,

    $\overrightarrow{r}={\overrightarrow{r}}_4+\frac{1}{5}\left({\overrightarrow{r}}_E-{\overrightarrow{r}}_4\right)$

Here, $\overrightarrow{r}$ is the distance covered by the treasurer and ${\overrightarrow{r}}_E$ is the new location.

Substitute the equation (VII) and (VIII) in above relation.

$\begin{array}{c}\overrightarrow{r}=30.0\widehat{i}+5.00\widehat{j}+\frac{1}{5}\left(-100\widehat{i}+55.0\widehat{j}\right)\\ =30.0\widehat{i}+5.00\widehat{j}-20.0\widehat{i}+11.0\widehat{j}\\ =10.0\widehat{i}+16.00\widehat{j}\end{array}$

Therefore, the co-ordinates point of the treasurer buried is $\left(10.0\text{m, 16}\text{.0}\text{m}\right)$.

To determine

If the pirate is not labeled the trees means what will be happened.

Answer to Problem 69P

The center of mass of the tree distribution is located in same path, in whatever order is taken by the trees.

Explanation of Solution

The certain directions bring the pirates to the average position of the trees.

From part (a) the displacement covered half of the point from B to A is,

  $\begin{array}{c}{\overrightarrow{r}}_2={\overrightarrow{r}}_A+\frac{1}{2}\left({\overrightarrow{r}}_B-{\overrightarrow{r}}_A\right)\\ =\frac{\left({\overrightarrow{r}}_A+{\overrightarrow{r}}_B\right)}{2}\end{array}$        (IX)

From part (a) the distance covered by one third of the distance between current location and point C is,

${\overrightarrow{r}}_3={\overrightarrow{r}}_2+\frac{1}{3}\left({\overrightarrow{r}}_C-{\overrightarrow{r}}_2\right)$

Substitute equation (IX) in above relation.

  $\begin{array}{c}{\overrightarrow{r}}_3=\frac{\left({\overrightarrow{r}}_A+{\overrightarrow{r}}_B\right)}{2}+\frac{1}{3}\left({\overrightarrow{r}}_C-\frac{\left({\overrightarrow{r}}_A+{\overrightarrow{r}}_B\right)}{2}\right)\\ =\frac{{\overrightarrow{r}}_A+{\overrightarrow{r}}_B+{\overrightarrow{r}}_C}{3}\end{array}$        (X)

Conclusion:

From part (a) the distance traverse by one quarter of the distance between current locations is,

    ${\overrightarrow{r}}_4={\overrightarrow{r}}_3+\frac{1}{4}\left({\overrightarrow{r}}_D-{\overrightarrow{r}}_3\right)$

Substitute equation (X) in above relation.

  $\begin{array}{c}{\overrightarrow{r}}_4=\frac{{\overrightarrow{r}}_A+{\overrightarrow{r}}_B+{\overrightarrow{r}}_C}{3}+\frac{1}{4}\left({\overrightarrow{r}}_D-\frac{{\overrightarrow{r}}_A+{\overrightarrow{r}}_B+{\overrightarrow{r}}_C}{3}\right)\\ =\frac{{\overrightarrow{r}}_A+{\overrightarrow{r}}_B+{\overrightarrow{r}}_C+{\overrightarrow{r}}_D}{4}\end{array}$        (IX)

From part (a) the distance covered by one fifth of the distance between current locations to moving distance is,

  $\overrightarrow{r}={\overrightarrow{r}}_4+\frac{1}{5}\left({\overrightarrow{r}}_E-{\overrightarrow{r}}_4\right)$

Substitute equation (IX) in above relation.

  $\begin{array}{c}\overrightarrow{r}=\frac{{\overrightarrow{r}}_A+{\overrightarrow{r}}_B+{\overrightarrow{r}}_C+{\overrightarrow{r}}_D}{4}+\frac{1}{5}\left({\overrightarrow{r}}_E-\frac{{\overrightarrow{r}}_A+{\overrightarrow{r}}_B+{\overrightarrow{r}}_C+{\overrightarrow{r}}_D}{4}\right)\\ =\frac{{\overrightarrow{r}}_A+{\overrightarrow{r}}_B+{\overrightarrow{r}}_C+{\overrightarrow{r}}_D+{\overrightarrow{r}}_E}{5}\end{array}$

Therefore, the center of mass of the tree location is distributed in same path.