Tag Archives: Vector Displacement

College Physics by Openstax Chapter 3 Problem 15


Find the north and east components of the displacement from San Francisco to Sacramento shown in Figure 3.57.

Figure 3.57

Solution:

Consider the following figure.

Using the right triangle formed, we can solve for the east component and the north component. Let SE be the east component and SN be the north component of S.

\begin{align*}
S_E & = S \cos 45^\circ \\
& = \left( 123 \ \text{km} \right) \cos45^\circ \\
& = 86.974 \ \text{km} \\
& \approx 87.0 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}
\begin{align*}
S_N & = S \sin 45^\circ \\
& = \left( 123 \ \text{km} \right) \sin 45^\circ \\
& = 86.974 \ \text{km} \\
& \approx 87.0 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Therefore, the east and north components are equal at about 87.0 km.


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College Physics by Openstax Chapter 3 Problem 14


Find the following for path D in Figure 3.56: (a) the total distance traveled and (b) the magnitude and direction of the displacement from start to finish. In this part of the problem, explicitly show how you follow the steps of the analytical method of vector addition.

Figure 3.56 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side.

Solution:

Part A

Looking at path D, we can see that it moves 2 blocks downward, 6 blocks to the right, 4 blocks upward, and 1 block to the left. Thus, the total distance of path D is

\begin{align*}
\text{distance} & = \left( 2\times 120\ \text{m} \right)+\left( 6\times 120\ \text{m} \right)+\left( 4\times 120\ \text{m} \right)+\left( 1\times 120\ \text{m} \right) \\
& = 1\ 560 \ \text{m} \\
& = 1.56 \times 10^{3} \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Part B

Looking at the initial and final position of path D, the final position is 5 blocks to the right or 600 meters to the right of the initial position, and 2 blocks or 240 meters upward from the initial position. Refer to the figure below.

Using the right triangle, we can solve for the displacement using the Pythagorean Theorem.

\begin{align*}
\text{displacement} & = \sqrt{\left( 600\ \text{m} \right)^2+\left( 240\ \text{m} \right)^2} \\
& = 646.2198 \ \text{m} \\
& \approx 646 \  \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

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College Physics by Openstax Chapter 3 Problem 5


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

Solution:

Consider Figure 3.5A shown below.

Figure 3.5A

Before we can use cosine law to solve for the magnitude of R, we need to solve for the interior angle 𝛽 first. The value of 𝛽 can be calculated by inspecting the figure and use simple knowledge on geometry. It is equal to the sum of 20° and the complement of 40°. That is

\beta = 20^\circ +\left( 90^\circ -40^\circ  \right) = 70^\circ 

We can use cosine law to solve for R.

\begin{align*}
R^2 & =A^2+B^2 -2AB \cos \beta \\
R^2 & = \left( 12.0\ \text{m} \right) ^2+\left( 20.0\ \text{m} \right)^2-2 \left( 12.0\ \text{m} \right) \left( 20.0\ \text{m} \right)
 \cos 70^\circ \\
R & = \sqrt{ \left( 12.0\ \text{m} \right) ^2+\left( 20.0\ \text{m} \right)^2-2 \left( 12.0\ \text{m} \right) \left( 20.0\ \text{m} \right)
 \cos 70^\circ} \\
R  & =19.4892 \ \text{m} \\
R & =19.5 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

We can solve for α using sine law.

\begin{align*}
\frac{\sin \alpha}{B} & = \frac{\sin \beta}{R} \\
\frac{\sin \alpha}{20.0\ \text{m}} & = \frac{\sin 70^\circ }{19.4892 \ \text{m}} \\
\sin \alpha & = \frac{20.0 \ \sin 70^\circ }{19.4892} \\
\alpha & = \sin ^{-1}  \left(  \frac{20.0 \ \sin 70^\circ }{19.4892}  \right) \\
\alpha & = 74.6488 ^\circ 
\end{align*}

Then we solve for the value of θ by subtracting 70° from α.

\theta=74.6488 ^\circ -70 ^\circ = 4.65^\circ

Therefore, the compass reading is

4.65^\circ, \text{South of West} \ \qquad \ {\color{Orange} \left( \text{Answer} \right)}

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College Physics by Openstax Chapter 3 Problem 24


Suppose a pilot flies 40.0 km in a direction 60º north of east and then flies 30.0 km in a direction 15º north of east as shown in Figure 3.61. Find her total distance R from the starting point and the direction θ of the straight-line path to the final position. Discuss qualitatively how this flight would be altered by a wind from the north and how the effect of the wind would depend on both wind speed and the speed of the plane relative to the air mass.

Figure 3.61

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:

Vectorx-componenty-component
A40\:\cos 60^{\circ} =20\:\text{km} 40\:\sin 60^{\circ} =34.6410\:\text{km}
B 30\:\cos 15^{\circ} =28.9778\:\text{km} 30\:\sin 15^{\circ} =7.7646\:\text{km}
Sum 48.9778\: \text{km} 42.4056 \:\text{km}

The table above indicates east and north as positive components, while west and south indicate negative components. The last row is the sum of the components. These are 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:

\begin{align*}
\text{R} & = \sqrt{\left(48.9778\:\text{km}\right)^2+\left(42.4056\:\text{km}\right)^2} \\
\text{R} & = 64.8\:\text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\
\end{align*}

The direction of the resultant is calculated as follows:

\begin{align*}
\theta & =\tan ^{-1}\left(\frac{42.4056}{48.9778}\right) \\
\theta & =40.9^{\circ} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Therefore, the pilot’s resultant displacement is about 64.8 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 traveling southeast.


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College Physics by Openstax Chapter 3 Problem 23


In an attempt to escape his island, Gilligan builds a raft and sets to sea. The wind shifts a great deal during the day, and he is blown along the following straight lines: 2.50 km 45.0º north of west; then 4.70 km 60.0º south of east; then 1.30 km 25.0º south of west; then 5.10 km straight east; then 1.70 km 5.00º east of north; then 7.20 km 55.0º south of west; and finally 2.80 km 10.0º north of east. What is his final position relative to the island?


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:

VectorX-ComponentY-Component
(1) -2.5\:\cos 45^{\circ} =-1.7678\:\text{km} +2.5\:\sin 45^{\circ} =+1.7678\:\text{km}
(2) +4.70\:\cos 60^{\circ} =+2.3500\:\text{km} -4.70\:\sin 60^{\circ} =-4.0703\:\text{km}
(3) -1.30\:\cos 25^{\circ} =-1.1782\:\text{km} -1.30\:\sin 25^{\circ} =-0.5494\:\text{km}
(4) +5.1000\:\text{km} 0
(5) +1.70\:\sin 5^{\circ} =+0.1482\:\text{km} +1.70\:\cos 5^{\circ} =+1.6935\:\text{km}
(6) -7.20\:\cos 55^{\circ} =-4.1298\:\text{km} -7.20\:\sin 55^{\circ} =-5.8979\:\text{km}
(7) +2.80\:\cos 10^{\circ} =+2.7575\:\text{km} +2.80\:\sin 10^{\circ} =+0.4862\:\text{km}
Sum 3.2799\:\text{km} -6.5701\:\text{km}

The table above indicates east and north as positive components, while west and south indicate 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:

\begin{align*}
\text{R} & =\sqrt{\left(3.2799\:\text{km}\right)^2+\left(-6.5701\:\text{km}\right)^2} \\
\text{R} & =7.34\:\text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\
\end{align*}

The direction of the resultant is calculated as follows:

\begin{align*}
\theta & =\tan ^{-1}\left(\frac{6.5701}{3.2799}\right) \\
\theta & =63.47^{\circ} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\
\end{align*}

Therefore, Gilligan is about 7.34 km directed 63.47° South of East from the starting island.


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