Tag Archives: Physics Solutions

College Physics by Openstax Chapter 3 Problem 17

Repeat Exercise 3.16 using analytical techniques, but reverse the order of the two legs of the walk and show that you get the same final result. (This problem shows that adding them in reverse order gives the same result—that is, B + A = A + B .) Discuss how taking another path to reach the same point might help to overcome an obstacle blocking your other path.

Figure 3.58 The two displacements A and B add to give a total displacement R having magnitude R and direction θ.

Solution:

Considering the right triangle formed by the vectors A, B, and R. We can solve for the magnitude of R using the Pythagorean Theorem. That is

\begin{align*}
R & = \sqrt{A^2+B^2} \\
& = \sqrt{\left( 18.0 \text{m} \right)^2+\left( 25.0 \text{m} \right)^2} \\
& =30.806  \text{m} \\
& \approx 30.8  \text{m}  \qquad  {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Then we solve for the compass direction by solving the value θ using the same right triangle. 

\begin{align*}
\theta & = \arctan \left( \frac{B}{A} \right)  \\
& = \arctan \left( \frac{25.0 \text{m}}{18.0 \text{m}} \right) \\
& = 54.246 ^\circ \\
& \approx  54.2 ^ \circ \\
\end{align*}

Therefore, the compass direction of the resultant is 54.2° North of West.

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College Physics by Openstax Chapter 3 Problem 16

Solve the following problem using analytical techniques: 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 3.58, then this problem asks you to find their sum R=A+B.)

Figure 3.58 The two displacements A and B add to give a total displacement R having magnitude R and direction θ.

Solution:

Considering the right triangle formed by the vectors A, B, and R. We can solve for the magnitude of R using the Pythagorean Theorem. That is

\begin{align*}
R & = \sqrt{A^2+B^2} \\
& = \sqrt{\left( 18.0\ \text{m} \right)^2+\left( 25.0\ \text{m} \right)^2} \\
& =30.806 \ \text{m} \\
& \approx 30.8 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Then we solve for the compass direction by solving the value θ using the same right triangle. 

\begin{align*}
\theta & = \arctan\left( \frac{B}{A} \right) \\
& = \arctan\left( \frac{25.0\ \text{m}}{18.0\ \text{m}} \right) \\
& = 54.246 ^\circ \\
& \approx  54.2 ^ \circ 
\end{align*}

Therefore, the compass direction of the resultant is 54.2° North of West.

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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 7

(a) Repeat the problem two problems prior, but for the second leg you walk 20.0 m in a direction 40.0º north of east (which is equivalent to subtracting B from A —that is, to finding R’=A−B ). (b) Repeat the problem two problems prior, but now you first walk 20.0 m in a direction 40.0º south of west and then 12.0 m in a direction 20.0º east of south (which is equivalent to subtracting A from B —that is, to finding R”=B−A=−R’ ). Show that this is the case.

Solution:

Refer to this problem.

Part A

Consider Figure 3-7A

Figure 3-7A

First, we solve for the value of α using a simple geometry. 

\alpha = 40^\circ+\left( 90^\circ-20^\circ \right) = 110^\circ

We can solve for R using the cosine law.

\begin{align*}
R^2 & = A^2 +B^2-2AB \cos \alpha \\
R & = \sqrt{A^2 +B^2-2AB \cos \alpha} \\
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 110^\circ} \\
R & =  26.6115 \ \text{m} \\
R & = 26.6 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Before we can solve for θ, we need to solve for 𝛽 first using the sine law.

\begin{align*}
\frac{\sin \beta}{B} & = \frac{\sin \alpha }{R} \\
\sin \beta & = \frac{B \  \sin \alpha}{R} \\
\beta & = \sin ^{-1} \left( \frac{B \  \sin \alpha}{R} \right) \\
\beta & = \sin ^{-1} \left( \frac{20.0 \ \text{m}\ \sin 110^{\circ}}{26.6115 \ \text{m}} \right) \\
\beta & = 44.9290^\circ
\end{align*}

Now, we can solve for θ. 

\begin{align*}
\theta & = \left( 90^\circ + 20^\circ \right)-\beta \\
\theta & = 110^\circ - 44.9290^\circ \\
\theta & = 65.071 \\
\theta & = 65.1 ^\circ
\end{align*}

Therefore, the compass reading is

65.1^\circ, \text{North of East} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}

Part B

Refer to Figure 3-7B

Figure 3-7B

First, we solve for the value of α using a simple geometry. 

\alpha = 40^\circ+\left( 90^\circ-20^\circ \right) = 110^\circ

We can solve for R using the cosine law. 

\begin{align*}
R^2 & = A^2 +B^2-2AB \cos \alpha \\
R & = \sqrt{A^2 +B^2-2AB \cos \alpha} \\
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 110^\circ} \\
R & =  26.6115 \ \text{m} \\
R & = 26.6 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Before we can solve for θ, we need to solve for 𝛽 first using the sine law. 

\begin{align*}
\frac{\sin \beta}{A} & = \frac{\sin \alpha }{R} \\
\sin \beta & = \frac{A \  \sin \alpha}{R} \\
\beta & = \sin ^{-1} \left( \frac{A \  \sin \alpha}{R} \right) \\
\beta & = \sin ^{-1} \left( \frac{12.0 \ \text{m}\ \sin 110^{\circ}}{26.6115 \ \text{m}} \right) \\
\beta & = 25.0708^\circ
\end{align*}

Now, we can solve for θ. 

\begin{align*}
\theta & = 40^\circ + \beta \\
\theta & = 40^\circ +  25.0708^\circ\\
\theta & = 65.0708^\circ\\
\theta & = 65.1 ^\circ
\end{align*}

Therefore, the compass reading is 

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

This is consistent with Part A because (A-B) = -(B-A).

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College Physics by Openstax Chapter 2 Problem 64

(a) Take the slope of the curve in Figure 2.64 to find the jogger’s velocity at t=2.5 s. (b) Repeat at 7.5 s. These values must be consistent with the graph in Figure 2.65.

Figure 2.64

Solution:

Part A

To find the slope at t=2.5 s, we need the position values at t= 0 s and t=5 s. When t = 0 \ \text{s}, x = 0 \ \text{m}, and when t = 5 \ \text{s}, x = 17.5 \ \text{m}. The velocity at t=2.5 s is

\begin{align*}

\text{velocity} & =\text{slope} \\
\text{v} & =\frac{\Delta x}{\Delta t} \\
\text{v} & = \frac{x_2-x_1}{t_2-t_1} \\
\text{v} & = \frac{17.5\ \text{m}-0\ \text{m}}{5 \ \text{s}-0\ \text{s}} \\
\text{v} & =3.5 \ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\

\end{align*}

Part B

When t = 10 \ \text{s}, x=2.5 \ \text{m}. Considering the points at t=5 s and t=10 s, the slope at 7.5 s is

\begin{align*}

\text{velocity} & =\text{slope} \\
\text{v} & =\frac{\Delta x}{\Delta t} \\
\text{v} & = \frac{x_2-x_1}{t_2-t_1} \\
\text{v} & = \frac{2.5\ \text{m}-17.5\ \text{m}}{10 \ \text{s}-5\ \text{s}} \\
\text{v} & =-3.0 \ \text{m/s} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \\

\end{align*}
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College Physics by Openstax Chapter 2 Problem 63

Construct the position graph for the subway shuttle train as shown in Figure 2.18(a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.

Figure 2.18

Solution:

The figure with the corresponding examples are shown on this page: https://openstax.org/books/college-physics/pages/2-4-acceleration#import-auto-id2590556

The position-vs-time graph of the train’s motion is also graphed in the first figure here: https://openstax.org/apps/archive/20210713.205645/resources/a697c43432cdf2d09a02df47d2b746283b841fcd

(a) Position of the train over time. Notice that the train’s position changes slowly at the beginning of the journey, then more and more quickly as it picks up speed. Its position then changes more slowly as it slows down at the end of the journey. In the middle of the journey, while the velocity remains constant, the position changes at a constant rate. (b) The velocity of the train over time. The train’s velocity increases as it accelerates at the beginning of the journey. It remains the same in the middle of the journey (where there is no acceleration). It decreases as the train decelerates at the end of the journey. (c) The acceleration of the train over time. The train has positive acceleration as it speeds up at the beginning of the journey. It has no acceleration as it travels at constant velocity in the middle of the journey. Its acceleration is negative as it slows down at the end of the journey.

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College Physics by Openstax Chapter 2 Problem 62

By taking the slope of the curve in Figure 2.63, verify that the acceleration is 3.2 m/s2 at t=10 s .

Figure 2.63

Solution:

To solve for the slope of the curve at t=10 s, we need two points — 1 just before and 1 just after.

When t=0 \ \text{s}, v=166 \ \text{m/s} and when t=20 \ \text{s}, v=230 \ \text{m/s}. Therefore, the acceleration is

\begin{align*}
\text{acceleration} & =\text{slope} \\
\text{acceleration} & =\frac{\Delta v}{\Delta t} \\
\text{a} & = \frac{v_2-v_1}{t_2-t_1} \\
\text{a} & = \frac{230\ \text{m/s}-166\ \text{m/s}}{20\ \text{s}-0\ \text{s}} \\
\text{a} & =3.2\ \text{m/s}^2 \\

\end{align*}

Note that the values are approximated to satisfy the given acceleration in the problem statement. The values may differ from one’s answer due to some uncertainties of a graph.

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College Physics by Openstax Chapter 2 Problem 61

Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t=30.0 s is approximately 0.24 m/s.

Figure 2.62

Solution:

We can obviously see from the graph that it is a straight line or approximately a straight line. In this case, the slope is constant.

To get an approximate slope at t=30 s, we can use the values at t=20 s and t=40 s. When t=20\ \text{s}, x=7\ \text{m} and when t=40\ \text{s}, x=12\ \text{m}.

\begin{align*}

\text{slope} & =\frac{\Delta x}{\Delta t} \\
\text{slope} & = \frac{x_2-x_1}{t_2-t_1} \\
\text{slope} & = \frac{12\ \text{m}-7\ \text{m}}{40\ \text{s}-20\ \text{s}} \\
\text{slope} & =0.25\ \text{m/s}

\end{align*}

Although not equal, the computed slope is almost the same with 0.24 m/s. This is due to the fact that values are uncertainties when using graphs. The difference is not really significant for this case.

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College Physics by Openstax Chapter 2 Problem 60

Using approximate values, calculate the slope of the curve in Figure 2.62 to verify that the velocity at t=30.0 s is approximately 0.24 m/s. Assume all values are known to 2 significant figures.

Figure 2.62

Solution:

We can obviously see from the graph that it is a straight line or approximately a straight line. In this case, the slope is constant.

To get an approximate slope at t=30 s, we can use the values at t=20 s and t=40 s. When t=20\ \text{s}, x=7\ \text{m} and when t=40\ \text{s}, x=12\ \text{m}.

\begin{align*}

\text{slope} & =\frac{\Delta x}{\Delta t} \\
\text{slope} & = \frac{x_2-x_1}{t_2-t_1} \\
\text{slope} & = \frac{12\ \text{m}-7\ \text{m}}{40\ \text{s}-20\ \text{s}} \\
\text{slope} & =0.25\ \text{m/s}

\end{align*}

Although not equal, the computed slope is almost the same with 0.24 m/s. This is due to the fact that values are uncertain when using graphs. The difference is not really significant for this case.

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