## Solution:

The pilot’s displacement is characterized by 2 vectors, A and B, as depicted in Figure 3.61. To determine her total displacement R from the starting point, we need to add the two given vectors. To do this, we individually get the x and y components of each vector. This is presented in the table that follows:

The table above indicates east and north as positive components, while west and south indicates negative components. The last row is the sum of the components. This is also the x and y components of the resultant vector.

To calculate the magnitude of the resultant, we simply use the Pythagorean Theorem as follows:

$\displaystyle \text{R}=\sqrt{\left(48.9778\:\text{km}\right)^2+\left(42.4056\:\text{km}\right)^2}$

$\displaystyle \text{R}=64.7847\:\text{km}$     ◀

The direction of the resultant is calculated as follows:

$\displaystyle \theta =\tan ^{-1}\left(\frac{42.4056}{48.9778}\right)$

$\displaystyle \theta =40.9^{\circ}$     ◀

Therefore, the pilot’s resultant displacement is about 64.7847 km directed 40.9° North of East from the starting island.

##### Discussion:

If the wind speed is less than the speed of the plane, it is possible to travel to the northeast, but she will travel more to the east than without the wind. If the wind speed is greater than the speed of the plane, then it is no longer possible for the plane to travel to the northeast, it will end up travelling southeast.

## Solution:

Gilligan’s displacement is characterized by 7 vectors. To determine his final position relative to the starting point, we simply need to add the vectors. To do this, we individually get the x and y components of each vector. This is presented in the table that follows:

The table above indicates east and north as positive components, while west and south indicates negative components. The last row is the sum of the components. This is also the x and y components of the resultant vector.

To calculate the magnitude of the resultant, we simply use the Pythagorean Theorem as follows:

$\displaystyle \text{R}=\sqrt{\left(3.2799\:\text{km}\right)^2+\left(-6.5701\:\text{km}\right)^2}$

$\displaystyle \text{R}=7.3433\:\text{km}$

The direction of the resultant is calculated as follows:

$\displaystyle \theta =\tan ^{-1}\left(\frac{6.5701}{3.2799}\right)$

$\displaystyle \theta =63.47^{\circ}$

Therefore, Gilligan is about 7.3433 km directed 63.47° South of East from the starting island.

## Solution:

### Part A

From the given statement, you first walk 18.0 m straight west and then 25.0 straight south. These vectors are represented by the graph shown below.

To solve for the resultant, we simply need to use the Pythagorean theorem to solve for the hypotenuse of the right triangle formed. That is,

$\displaystyle \text{R}=\sqrt{\text{A}^2+\text{B}^2}$

$\displaystyle \text{R}=\sqrt{\left(18.0\:\text{m}\right)^2+\left(25.0\:\text{m}\right)^2}$

$\displaystyle \text{R}=30.8\:\text{m}$

To solve for the angle, θ, we shall use the tangent function.

$\displaystyle \theta =\tan ^{-1}\left(\frac{\text{B}}{\text{A}}\right)$

$\displaystyle \theta =\tan ^{-1}\left(\frac{25.0\:\text{m}}{18.0\:\text{m}}\right)$

$\displaystyle \theta =54.25^{\circ}$

Thus, the value of 𝜙 can be calculated as

$\displaystyle \phi=90^{\circ} -54.25^{\circ}$

$\displaystyle \phi=35.75^{\circ}$

Therefore, the compass direction of the resultant is 35.75° West of South.

### Part B

From the statement, you walk 25.0 m north first and then 18.0 m east. This is represented by the figure shown below.

So, we have a right triangle with legs 25.0 m and 18.0 m. We are tasked to solve for the value of R, and the angle 𝜙 for the compass direction. The value of R can be solved using the Pythagorean Theorem as in Part A.

$\displaystyle \text{R}=\sqrt{\text{A}^2+\text{B}^2}$

$\displaystyle \text{R}=\sqrt{\left(18.0\:\text{m}\right)^2+\left(25.0\:\text{m}\right)^2}$

$\displaystyle \text{R}=30.8\:\text{m}$

To solve for the angle, θ, we shall use the tangent function.

$\displaystyle \theta =\tan ^{-1}\left(\frac{\text{B}}{\text{A}}\right)$

$\displaystyle \theta =\tan ^{-1}\left(\frac{18.0\:\text{m}}{25.0\:\text{m}}\right)$

$\displaystyle \theta =35.75^{\circ}$

Therefore, the compass direction of the resultant is 35.75° East of North.

## Farmer wants to Fence off his Four-Sided Plot with missing Side| Vector Addition and Subtraction| Analytical Method| College Physics| Problem 3.22

#### A farmer wants to fence off his four-sided plot of flat land. He measures the first three sides, shown as A, B, and C in the figure, and then correctly calculates the length and orientation of the fourth side D. What is his result?

SOLUTION:

We know that the sum of the 4 vectors is zero. So, Vector D is calculated as

$D=-A-B-C$

We solve for the x-component first. The x-component of vector D is

$D_x=-A_x-B_x-C_x$

$D_x=-\left(4.70\:km\right)cos\left(-7.50^{\circ} \right)-\left(2.48\:km\right)cos106^{\circ} -\left(3.02\:km\right)cos161^{\circ}$

$D_x=-1.12\:km$

We solve for the y-component.

$D_y=-A_y-B_y-C_y$

$D_y=-\left(4.70\:km\right)sin\left(-7.50^{\circ} \right)-\left(2.48\:km\right)sin106^{\circ} -\left(3.02\:km\right)sin161^{\circ}$

$D_y=-2.75\:km$

Since we already know the x and y component of vector D, we can finally solve for the distance of vector D.

$D=\sqrt{\left(D_x\right)^2+\left(D_y\right)^2}$

$D=\sqrt{\left(-1.12\:km\right)^2+\left(-2.75\:km\right)^2}$

$D=2.97\:km$

The corresponding direction of vector D is

$\theta =tan^{-1}\left|\frac{D_x}{D_y}\right|$

$\theta =tan^{-1}\left|\frac{1.12\:km}{2.75\:km}\right|$

$\theta =22.2^{\circ} \:W\:of\:S$

## Solution:

### Part A

The component along the south direction is

$D_s=Rsin\theta =\left(32.0\:km\right)sin\:35^{\circ} =18.4\:km$

The component along the west direction is

$D_w=Rcos\theta =\left(32.0\:km\right)cos35^{\circ} =26.2\:km$

### Part B

Consider the following figure with the rotated axes x’-y’.

The component along the southwest direction is

$D_{SW}=Rcos\theta '=\left(32.0\:km\right)cos10^{\circ} =31.5\:km$

The component along the northwest direction is

$D_{NW}=Rsin\theta '=\left(32.0\:km\right)cos10^{\circ} =5.56\:km$

## Solution:

Since the figure is closed, we know that the sum of the three vectors is equal to 0. So, we have

$\vec{A}+\vec{B}+\vec{C}=0$

Solving for C, we have

$\vec{C}=-\vec{A}-\vec{B}$

Next, we solve for the x and y components of Vectors A and B.

For Vector A, the x and y components are

$A_x=\left(80\:m\right)cos\left(21^{\circ} \right)=74.6864\:m$

$A_y=-\left(80\:m\right)sin\left(21^{\circ} \right)=-28.6694\:m$

For Vector B, the x and y components are

$B_x=-\left(105\:m\right)sin\:11^{\circ} =-20.0349\:m$

$B_y=\left(105\:m\right)cos\:11^{\circ} =103.0709\:m$

Solve for the components of Vector C.

$C_x=-A_x-B_x=-74.6864\:m\:-\left(-20.0349\:m\right)=-54.6515\:m$

$C_y=-A_y-B_y=-\left(-28.6694\:m\right)-\left(103.0709\:m\right)=-74.4015\:m$

Therefore, the length of Vector C is

$C=\sqrt{\left(C_x\right)^2+\left(C_y\right)^2}$

$C=\sqrt{\left(-54.6515\:m\right)^2+\left(-74.4015\:m\right)^2}$

$C=92.3167\:m$

The direction of Vector C is

$\theta =tan^{-1}\left(\frac{C_y}{C_x}\right)=tan^{-1}\left(\frac{74.4015\:m}{54.6515\:m}\right)=53.7^{\circ} \:\:South\:of\:West$

## Solution:

### Part A

Consider the following figure:

The east distance is the component in the horizontal direction.

$D_E=7.50\:km\:\cdot sin\:\left(15^{\circ} \right)$

$D_E=1.94\:km$

The north distance is the vertical component

$D_E=7.50\:km\cdot cos\left(15^{\circ} \right)$

$D_E=7.24\:km$

### Part B

Based from the figure, we can easily see that the order is reversible in the addition of vectors. We say that $D_E+D_N=D_N+D_E$

## Solution:

Consider the right triangle formed by the legs A, B, and R. We know that A is 18 m, B is 25 m, and we are solving for the magnitude of R. We can do this by using the Pythagorean Theorem. That is

$R=\sqrt{A^2+B^2}$

$R=\sqrt{\left(18\:m\right)^2+\left(25\:m\right)^2}$

$R=\sqrt{324+625}$

$R=\sqrt{949}$

$R=30.8\:m$

So, the distance is about 30.8 meters from the starting point. To solve for the value of the unknown angle, φ, we can use the tangent function. That is

$tan\:\phi =\frac{A}{B}$

$tan\:\phi =\frac{18\:m}{25\:m}$

$\phi =tan^{-1}\left(\frac{18}{25}\right)$

$\phi =35.75^{\circ}$

Therefore, the compass angle is $35.75^{\circ} \:West\:of\:North$

## Solution:

Consider the right triangle formed by the legs A, B, and R. We know that A is 18 m, B is 25 m, and we are solving for the magnitude of R. We can do this by using the Pythagorean Theorem. That is

$R=\sqrt{A^2+B^2}$

$R=\sqrt{\left(18\:m\right)^2+\left(25\:m\right)^2}$

$R=\sqrt{324+625}$

$R=\sqrt{949}$

$R=30.8\:m$

So, the distance is about 30.8 meters from the starting point. To solve for the value of the unknown angle, θ, we can use the tangent function. That is

$tan\:\theta =\frac{B}{A}$

$tan\:\theta =\frac{25\:m}{18\:m}$

$\theta =tan^{-1}\left(\frac{25}{18}\right)$

$\theta =54.25^{\circ}$

So, the compass reading, can be solved by taking the complimentary angle, $\phi$ as shown in the figure.

$\phi =90^{\circ} -54.25^{\circ}$

$\phi =35.75^{\circ}$

Therefore, the compass angle is $35.75^{\circ} \:West\:of\:North$

## Solution:

### North Component

The North component is given by

$S_N=123\:km\:\cdot sin\left(45^{\circ} \right)=86.97\:km$

### East Component

The east component is

$S_E=123\:km\:\cdot cos\:\left(45^{\circ} \right)=86.97\:km$