Category Archives: Engineering Mathematics Blog

College Physics by Openstax Chapter 3 Problem 4

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 A⃗ \vec{A} and B⃗ \vec{B} , as in Figure 3.53, then this problem asks you to find their sum R⃗=A⃗+B⃗\vec{R}=\vec{A}+\vec{B} .)

Figure 3.53

Solution:

Figure 3.4A

Consider Figure 3.54A.

The resultant of the two vectors A⃗\vec{A} and B⃗ \vec{B} is labeled R⃗\vec{R}. This R⃗\vec{R} is directed θ∘\theta ^{\circ} from the x-axis.

We shall use the right triangle formed to solve for the unknowns.

Solve for the magnitude of the resultant.

R=A2+B2R=(18.0 m)2+(25.0 m)2R=30.8 m  (Answer)\begin{align*} R & = \sqrt{A^2 +B^2} \\ R & = \sqrt{\left(18.0 \ \text{m} \right)^2+\left( 25.0 \ \text{m} \right)^2} \\ R & = 30.8 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

Solve for the value of θ \theta .

θ=arctan⁡(BA)θ=arctan⁡(25.0 m18.0 m)θ=54.2∘\begin{align*} \theta & = \arctan \left( \frac{B}{A} \right) \\ \theta & = \arctan \left( \frac{25.0 \ \text{m}}{18.0 \ \text{m}} \right) \\ \theta & = 54.2^\circ \end{align*}

We need the complementary angle for the compass angle.

90∘−54.2∘=35.8∘\begin{align*} 90^\circ -54.2^\circ =35.8^\circ \end{align*}

Therefore, the compass angle reading is

35.8∘,W of N  (Answer)\begin{align*} 35.8^\circ , \text{W of N} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}
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College Physics by Openstax Chapter 3 Problem 3

Find the north and east components of the displacement for the hikers shown in Figure 3.50.

Figure 3.50

Solution:

Refer to Figure 3-3-A for the north and east components of the displacement s of the hikers.

Figure 3-3-A

Considering the right triangle formed. The north component is computed as

snorth=(5.00 km)sin⁡40∘snorth=3.21 km  (Answer)\begin{align*} \text{s}_{\text{north}} & = \left( 5.00 \ \text{km} \right)\sin 40^\circ \\ \text{s}_{\text{north}} & = 3.21 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}

Using the same right triangle, the east component is computed as follows.

seast=(5.00 km)cos⁡40∘seast=3.83 km  (Answer)\begin{align*} \text{s}_{\text{east}} & = \left( 5.00 \ \text{km} \right)\cos 40^\circ \\ \text{s}_{\text{east}} & = 3.83 \ \text{km} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}
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College Physics by Openstax Chapter 3 Problem 2

Find the following for path B in Figure 3.52:
(a) The total distance traveled, and
(b) The magnitude and direction of the displacement from start to finish.

Figure 3.54 The various lines represent paths taken by different people walking in a city. All blocks are 120 m on a side
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

The total distance traveled is 

d=(4×120m)+(3×120 m)+(3×120 m)d=1200 m  (Answer)\begin{align*} \text{d} & = \left(4 \times 120 \text{m} \right) + \left(3 \times 120\ \text{m} \right) + \left(3 \times 120\ \text{m} \right) \\ \text{d} & = 1 200\ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}

Part B

The magnitude of the displacement is 

s=(sx)2+(sy)2s=(1×120 m)2+(3×120 m)2s=(120 m)2+(360 m)2s=379 m  (Answer)\begin{align*} \text{s} & = \sqrt{\left( s_x \right)^2+\left( s_y \right)^2} \\ \text{s} & = \sqrt{\left( 1 \times 120\ \text{m} \right)^2+ \left( 3 \times 120 \ \text{m} \right)^2} \\ \text{s} & = \sqrt{\left( 120\ \text{m} \right)^2+ \left( 360 \ \text{m} \right)^2} \\ \text{s} & = 379 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)} \end{align*}

The direction is

θ=arctan⁡(sysx)θ=arctan⁡(360 m120 m)θ=71.6∘, N of E (Answer)\begin{align*} \theta & = \arctan \left( \frac{s_y}{s_x} \right) \\ \theta & = \arctan \left( \frac{360\ \text{m}}{120\ \text{m}} \right) \\ \theta & = 71.6^\circ , \ \text{N of E} \ \qquad {\color{DarkOrange} \left( \text{Answer} \right)}\\ \end{align*}
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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 st = 0 \ \text{s}, x=0 mx = 0 \ \text{m}, and when t=5 st = 5 \ \text{s}, x=17.5 mx = 17.5 \ \text{m}. The velocity at t=2.5 s is

velocity=slopev=ΔxΔtv=x2−x1t2−t1v=17.5 m−0 m5 s−0 sv=3.5 m/s  (Answer)\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 st = 10 \ \text{s}, x=2.5 mx=2.5 \ \text{m}. Considering the points at t=5 s and t=10 s, the slope at 7.5 s is

velocity=slopev=ΔxΔtv=x2−x1t2−t1v=2.5 m−17.5 m10 s−5 sv=−3.0 m/s  (Answer)\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 st=0 \ \text{s}, v=166 m/sv=166 \ \text{m/s} and when t=20 st=20 \ \text{s}, v=230 m/sv=230 \ \text{m/s}. Therefore, the acceleration is

acceleration=slopeacceleration=ΔvΔta=v2−v1t2−t1a=230 m/s−166 m/s20 s−0 sa=3.2 m/s2\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 st=20\ \text{s}, x=7 mx=7\ \text{m} and when t=40 st=40\ \text{s}, x=12 mx=12\ \text{m}.

slope=ΔxΔtslope=x2−x1t2−t1slope=12 m−7 m40 s−20 sslope=0.25 m/s\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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Header Photos for Elementary Differential Equations Chapter 9 Special Second Ordered Equations

Chapter 9: Special Second-Ordered Equations

Supplementary Problems

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

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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 st=20\ \text{s}, x=7 mx=7\ \text{m} and when t=40 st=40\ \text{s}, x=12 mx=12\ \text{m}.

slope=ΔxΔtslope=x2−x1t2−t1slope=12 m−7 m40 s−20 sslope=0.25 m/s\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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