## Solution:

### Part A

We are required to solve for $\vec{R'}=\vec{A'}+\vec{B}'$

The x-component of the resultant is: $R_x'=A_x'+B_x'$ $R_x'=12\:cos\left(110^{\circ} \right)+\left(20.0\right)cos\left(40^{\circ} \right)=11.217\:m$

The y-component of the resultant is: $R_x'=A_y'+B_y'$ $R_x'=\left(12.0\right)sin\left(110^{\circ} \right)+\left(20.0\right)sin\left(40^{\circ} \right)=24.132\:m$

The resultant is $R'=\sqrt{R_x'+R_y'}$ $R'=\sqrt{\left(11.217\:m\right)^2+\left(24.132\:m\right)^2}$ $R'=26.6\:m$

The compass direction is $\alpha =tan^{-1}\left(\frac{R_y'}{R_x'}\right)$ $\alpha =tan^{-1}\left(\frac{24.132}{11.217}\right)=65.1^{\circ}$

The compass direction is $65.1^{\circ} \:North\:of\:East$

### Part B

The result is the same as that in Part A since the components did not change. Only the order has changed. This is consistent with the fact that $\vec{A}-\vec{B}=-\left(\vec{B}-\vec{A}\right)$

## Solution:

Reversing the order of the vectors, we come up with the triangle as shown below.

We can actually see that the resultant can be solved by considering the triangle formed by the three vectors. The same with the procedures used in the previous problem, we have $\text{R}^2=\text{A}^2+\text{B}^2-2\text{AB}\:\cos 70^{\circ}$ $\text{R}=\sqrt{\text{A}^2+\text{B}^2-2\text{AB}\:\text{cos}\:70^{\circ} }\:$ $\text{R}=\sqrt{\text{(12 m)}^2+\text{(20 m)}^2-2\text{(12 m)(20 m)}\:\text{cos}\:70^{\circ} }\:$ $\text{R}=19.4892\:\text{m}$

To solve for the direction, we can solve for the angle included between B and R first. $\displaystyle \frac{12\:\text{m}}{\sin \text{C}}=\frac{19.4892\:\text{m}}{\sin 70^{\circ} }$ $\displaystyle \text{C}=\sin ^{-1}\left(\frac{12\:\text{m}\:\left(\sin 70^{\circ} \right)}{19.4892\:\text{m}}\right)$ $\text{C}=35.35^{\circ}$

Finally to solve for the direction of the resultant, we have $\theta =40^{\circ} -35.35^{\circ} =4.65^{\circ}$

The compass direction is $4.65^{\circ} \:\text{South of West}$.

This result is actually the same as in the previous problem when the order of the vectors are not reversed. No matter what is the order of the vectors we are adding, the result is still the same. This proves that $\vec{A}+\vec{B}=\vec{B}+\vec{A}$.

## College Physics 3.5 – Resultant of two vectors

#### Suppose you first walk 12.0 m in a direction 20º west of north and then 20.0 m in a direction 40.0º south of west. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B , as in Figure 3.54, then this problem finds their sum R=A+B.) Figure 3.54. The two displacements A and B are added to give a total displacement of R with magnitude R and direction θ measured from the x-axis

## Solution:

To solve for the magnitude and direction of the resultant R, we can refer to the triangle formed by the tree vectors as shown in the figure below.

From the figure above, we are given the two sides A and B, and the included angle measured to be 70°. To solve for the magnitude of R, we can use the cosine law. $\text{R}^2=\text{A}^2+\text{B}^2-2\text{AB}\:\cos 70^{\circ}$ $\text{R}=\sqrt{\text{A}^2+\text{B}^2-2\text{AB}\:\text{cos}\:70^{\circ} }\:$ $\text{R}=\sqrt{\text{(12 m)}^2+\text{(20 m)}^2-2\text{(12 m)(20 m)}\:\text{cos}\:70^{\circ} }\:$ $\text{R}=19.4892\:\text{m}$

To solve for the direction of the resultant, we just need to solve for angle C, then subtract 70° to get the value of θ. We can solve for angle C using the sine law. $\displaystyle \frac{20\:\text{m}}{\sin \text{C}}=\frac{19.4892\:\text{m}}{\sin 70^{\circ} }$ $\displaystyle \sin \:\text{C}=\frac{20\:\text{m}\:\left(\sin \:\:70^{\circ \:}\right)}{19.4892\:\text{m}}$ $\displaystyle \text{C}=\text{sin}^{-1}\left(\frac{20\:\text{m}\:\left(\sin \:\:70^{\circ \:}\right)}{19.4892\:\text{m}}\right)$ $\text{C}=74.65^{\circ}$

Finally, to solve for θ, we subtract 70° from 74.65°. $\theta =74.65^{\circ} -70^{\circ} =4.65^{\circ}$

Therefore, the compass direction of the resultant displacement is 4.65° South of West.

## College Physics 3.4 – Resultant of west and north displacements

#### Suppose you walk 18.0 m straight west and then 25.0 m straight north. How far are you from your starting point, and what is the compass direction of a line connecting your starting point to your final position? (If you represent the two legs of the walk as vector displacements A and B, as in Figure 3.53, then this problem asks you to find their sum R = A + B .) Figure 3.53. The two displacements A and B add to give a total displacement R having magnitude R and direction θ.

## Solution:

Refer to the diagram

The magnitude of the displacement (distance from the starting point to the final position) is given by $R=\sqrt{A^2+B^2}$ $R=\sqrt{\left(18.0\:m\right)^2+\left(25.0\:m\right)^2}$ $R=30.8\:m$

The compass direction of a line connecting your starting point to your final position is given by $\displaystyle \theta =tan^{-1}\left(\frac{B}{A}\right)$ $\displaystyle \theta =tan^{-1}\left(\frac{25.0\:m}{18.0\:m}\right)$ $\displaystyle \theta =54.25^{\circ}$

The compass reading is $54.25^{\circ} ,\:\text{North of West}$.

## Solution:

To solve for the north and east components of the given displacement vector, we need to draw them first by virtue of the parallelogram law. The vector, together with its components, is shown in the figure below.

We can solve the north and east components by considering the right triangle ABC. The east component is the side AC and the north component is the side BC.

The North-component is given by $\displaystyle \text{s}_{\text{north}}=\left(5.0\:\text{km}\right)\text{sin}\:\left(40.0\right)^{\circ}$ $\text{s}_{\text{north}}=3.21\:\text{km}$

The East-component is given by $\displaystyle \text{s}_{\text{east}}=\left(5.0\:\text{km}\right)\text{cos}\left(40.0\right)^{\circ}$ $\text{s}_{\text{east}}=3.83\:\text{km}$

## College Physics 3.2 – Distance vs displacement

#### (b) the magnitude and direction of the displacement from start to finish. Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

## Solution:

### Part A

So, the total distance traveled on path B as shown by the red line is just the sum of the distances. $d=\left(4\times 120\:m\right)+\left(3\times 120\:m\right)+\left(3\times 120\:m\right)$ $d=1200\:m$

### Part B

The displacement is just the shortest distance from the beginning to the end of the path. The displacement has a magnitude given by $s=\sqrt{\left(s_x\right)^2+\left(s_y\right)^2}$ $s=\sqrt{\left(1\times 120\:m\right)^2+\left(3\times 120\:m\right)^2}$ $s=379\:m$

The direction of the displacement is given by $\theta =tan^{-1}\left(\frac{1\times 120\:m}{3\times 120\:m}\right)=18.4^{\circ} ,\:East\:of\:North$

## College Physics 3.1 – Distance and displacement computation

#### (b) the magnitude and direction of the displacement from start to finish. Figure 3.52 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

## Solution:

### Part A

So, the total distance traveled on path A as shown by the green line is just the sum of the distances. $d=\left(120\:m\times 3\right)+\left(120\:m\times 1\right)$ $d=480\:m$

### Part B

The displacement is just the shortest distance from the beginning to the end of the path. The displacement has a magnitude given by $s=\sqrt{\left(1\times 120\:m\right)^2+\left(3\times 120\:m\right)^2}=379\:m$

The direction of the displacement is given by $\theta =tan^{-1}\left(\frac{1\times 120\:m}{3\times 120\:m}\right)=18.4^{\circ} ,\:East\:of\:North$

## Solution:

### Part A

The total number of radians made by the object is the area under the graph. Based on the given graph, the total number of radians is $\displaystyle \theta _{rad}=60\:radians$

Convert this to revolutions, knowing that 2π radians is equal to 1 revolution. $\displaystyle revolutions=60\:rad\times \frac{1\:rev}{2\pi }=\frac{60}{2\pi }rev=9.5\:rev$