Tag Archives: College Physics Solutions

College Physics by Openstax Chapter 3 Problem 25


A projectile is launched at ground level with an initial speed of 50.0 m/s at an angle of 30.0º above the horizontal. It strikes a target above the ground 3.00 seconds later. What are the x and y distances from where the projectile was launched to where it lands?


Solution:

Since we do not know the exact location of the projectile after 3 seconds, consider the following arbitrary figure:

The path of the projectile from the ground to a point 3 seconds later.

From the figure, we can solve for the components of the initial velocity.

\begin{align*}
\text{v}_{\text{ox}} &=\left(50\:\text{m/s}\right)\cos 30^{\circ} \\
& =43.3013\:\text{m/s}
\\
\\
\text{v}_{\text{oy}} & =\left(50\:\text{m/s}\right)\sin 30^{\circ} \\
&=25\:\text{m/s}
\\
\end{align*}

So, we are asked to solve for the values of x and y. To solve for the value of the horizontal displacement, x, we shall use the formula x=voxt. That is,

\begin{align*}
\text{x} & =\text{v}_{\text{ox}}\text{t} \\
\text{x} & =\left(43.3013\:\text{m/s}\right)\left(3\:\text{s}\right) \\
\text{x} & =130\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

To solve for the vertical displacement, y, we shall use the formula y=voyt+1/2at2. That is

\begin{align*}
\text{y} & =\text{v}_{\text{oy}}\text{t}+\frac{1}{2}\text{a}\text{t}^2 \\
\text{y} & =\left(25\:\text{m/s}\right)\left(3\:\text{s}\right)+\frac{1}{2}\left(-9.81\:\text{m/s}^2\right)\left(3\:\text{s}\right)^2 \\
\text{y} & =30.9\:\text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Therefore, the projectile strikes a target at a distance 129.9 meters horizontally and 30.9 meters vertically from the launching point.


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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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College Physics by Openstax Chapter 3 Problem 12


Find the components of vtot along a set of perpendicular axes rotated 30º counterclockwise relative to those in Figure 3.55.

The figure shows v_A directed 22.5° from the positive x-axis, and v_B started from the head of v_A and is directed 23.0° from the resultant. The resultant is given to be 6.72 m/s and is directed 26.5° from v_A. In total, the resultant is measured 49° from the positive x-axis.
Figure 3.55

Solution:

By isolating the vtot from the rest of the other vectors, we come up with the following figure.

The isolated resultant velocity

The resultant velocity has a magnitude of 6.72 m/s and is directed 49° from the positive x-axis. Now, we shall create another set of axes rotated at 30° counterclockwise. We call the axes x’ and y’ axes. The figure is shown below.

The resultant velocity with the rotated axes.

From the figure, we can see that the resultant velocity is 19° from the x’ axis. Therefore, the x’ and y’ components are:

\text{x'-component}=\left(6.72\:\text{m/s}\right)\cos 19^{\circ} =6.35\:\text{m/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
 \text{y'-component}=\left(6.72\:\text{m/s}\right)\sin 19^{\circ} =2.19\:\text{m/s}\ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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College Physics by Openstax Chapter 3 Problem 11


Find the components of vtot along the x- and y-axes in Figure 3.55.

The figure shows v_A directed 22.5° from the positive x-axis, and v_B started from the head of v_A and is directed 23.0° from the resultant. The resultant is given to be 6.72 m/s and is directed 26.5° from v_A. In total, the resultant is measured 49° from the positive x-axis.
Figure 3.55

Solution:

By isolating the vtot from the rest of the other vectors, we come up with the following figure. Also, the x and y-components are shown.

The resultant velocity and its x and y components

The resultant velocity has a magnitude of 6.72 m/s and is directed 49° from the positive x-axis. To solve for the x and y components, we just need to solve the legs of the right triangle formed by the three vectors. That is,

 \text{x-component}=\left(6.72\:\text{m/s}\right)\cos 49^{\circ} =4.41\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\text{y-component}=\left(6.72\:\text{m/s}\right)\sin 49^{\circ} =5.07\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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College Physics by Openstax Chapter 3 Problem 10


Find the magnitudes of velocities vA and vB in Figure 3.55

The figure shows v_A directed 22.5° from the positive x-axis, and v_B started from the head of v_A and is directed 23.0° from the resultant. The resultant is given to be 6.72 m/s and is directed 26.5° from v_A. In total, the resultant is measured 49° from the positive x-axis.
Figure 3.55

Solution:

Basically, we are given an oblique triangle. First, we shall determine the value of the interior angle at the intersection of vA and vB. We can solve this knowing that the sum of the interior angles of a triangle is 180°.

To solve for vA and vB, we will use the sine law.

\begin{align*}
 \frac{\text{v}_{\text{A}}}{\sin 23^{\circ} } & =\frac{6.72\:\text{m/s}}{\sin 130.5^{\circ} } \\
\text{v}_{\text{A}} & =\frac{6.72\:\text{m/s}\:\sin \:23^{\circ }\:}{\sin \:130.5^{\circ }\:} \\
\text{v}_{\text{A}} & =3.45\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

Using the same law to solve for the value of vB, we have

\begin{align*}
\frac{\text{v}_{\text{B}}}{\sin 26.5^{\circ} } & =\frac{6.72\:\text{m/s}}{\sin 130.5^{\circ} } \\
\text{v}_{\text{B}} & =\frac{6.72\:\text{m/s}\:\sin \:26.5^{\circ }\:\:}{\sin \:130.5^{\circ }\:} \\
\text{v}_{\text{B}} & =3.94\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

\end{align*}

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College Physics by Openstax Chapter 3 Problem 9


Show that the sum of the vectors discussed in Example 3.2 gives the result shown in Figure 3.24.

Figure 3.24

Solution:

So, we are given the two vectors shown below.

Vectors A and B

If we use the graphical method of adding vectors, we can join the two vectors using head-tail addition and come up with the following:

Figure 3.9B: Vectors A and B added graphically

The resultant is drawn from the tail of the first vectors (the origin) to the head of the last vector. The resultant is shown in red in the figure below.

Solve for the value of the angle 𝛼 by geometry.

\alpha = 66^\circ +\left( 180^\circ-112^\circ \right) = 134^\circ

Solve for the magnitude of the resultant using 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( 27.5 \ \text{m} \right)^2+\left( 30.0 \ \text{m} \right)^2-2\left( 27.5\ \text{m} \right)\left( 30.0\ \text{m} \right) \cos 134^\circ} \\
R & =52.9380 \ \text{m} \\
R & = 52.9 \ \text{m} \ \qquad \ {\color{DarkOrange} \left( \text{Answer} \right)}
\end{align*}

Solve for 𝛽 using sine law.

\begin{align*}
\frac{\sin \beta}{B} & = \frac{\sin \alpha}{R} \\
\beta & = \sin ^{-1} \left( \frac{B \sin \alpha }{R} \right) \\
\beta & = \sin ^{-1} \left( \frac{30.0\ \text{m} \sin 134^\circ}{52.9380 \ \text{m}} \right) \\
\beta & = 24.0573^\circ
\end{align*}

Finally, solve for 𝜃.

\theta = 66^\circ+24.0573^\circ = 90.1^\circ \ \qquad \ {\color{Orange} \left( \text{Answer} \right)}

The result is in conformity with that in figure 3.24 shown on the question shown above.


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College Physics by Openstax Chapter 3 Problem 8


Show that the order of addition of three vectors does not affect their sum. Show this property by choosing any three vectors A, B, and C, all having different lengths and directions. Find the sum A + B + C then find their sum when added in a different order and show the result is the same. (There are five other orders in which A, B, and C can be added; choose only one.)


Solution:

Consider the three vectors shown in the figures below:

Vector A

Vector B

Vector C

First, we shall add them A+B+C. Using the head-tail or graphical method of vector addition, we have the figure shown below.

Figure 3.8B: The resultant force of A+B+C

Now, let us try to find the sum of the three vectors by reordering vectors A, B, and C. Let us try to find the sum of C+B+A in that order. The result is shown below.

Figure 3.8C: The resultant of 3 vectors added in different order.

We can see that the resultant is the same directed from the origin upward. This proves that the resultant must be the same even if the vectors are added in different order.


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College Physics by Openstax Chapter 2 Problem 66


Figure 2.68 shows the position graph for a particle for 6 s. (a) Draw the corresponding Velocity vs. Time graph. (b) What is the acceleration between 0 s and 2 s? (c) What happens to the acceleration at exactly 2 s?

position graph for a particle for 6 s.
Figure 2.68

Solution:

Part A

The velocity of the particle is the slope of the position vs time graph. Since the position graph is composed of straight lines, we can say that the velocity is constant for several time ranges.

Time RangeSlope of the Position vs Time Graph
0 to 2 seconds=\frac{2-0}{2-0}=1\:\text{m/s}
2 to 3 seconds=\frac{-3-2}{3-2}=\frac{-5}{1}=-5\:\text{m/s}
3 to 5 seconds=0 \ \text{m/s}
5 to 6 seconds=\frac{-2-\left(-3\right)}{6-5}=\frac{1}{1}=1\:\text{m/s}

Based on the data in the table, we can draw the velocity diagram

velocity vs time graph
velocity vs time graph

Part B

Since the velocity is constant between 0 seconds and 2 seconds, we say that the acceleration is 0.

Part C

Since there is a sudden change in velocity at exactly 2 seconds in a very short amount of time, we say that the acceleration is undefined in this case.


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College Physics by Openstax Chapter 2 Problem 65


A graph of v(t) is shown for a world-class track sprinter in a 100-m race. (See Figure 2.67). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at t=5 s? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?

A graph of  v(t)  is shown for a world-class track sprinter in a 100-m race.
Figure 2.67

Solution:

Part A

To find the average velocity over the straight line graph of the velocity vs time shown, we just need to locate the midpoint of the line. In this case, the average speed for the first 4 seconds is

v_{\text{ave}}=6\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

Looking at the graph, the velocity at exactly 5 seconds is 12 m/s.

Part C

If we are given the velocity-time graph, we can solve for the acceleration by solving for the slope of the line.

Consider the line from time zero to time, t=4 seconds. The slope, or acceleration, is

a=\text{slope}=\frac{12\:\text{m/s}-0\:\text{m/s}}{4\:\text{s}}=3\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part D

For the first 4 seconds, the distance traveled is equal to the area under the curve.

\text{distance}=\frac{1}{2}\left(4\:\sec \right)\left(12\:\text{m/s}\right)=24\:\text{m}

So, the sprinter traveled a total of 24 meters in the first 4 seconds. He still needs to travel a distance of 76 meters to cover the total racing distance. At the constant rate of 12 m/s, he can run the remaining distance by

\text{t}=\frac{\text{distance}}{\text{velocity}}=\frac{76\:\text{m}}{12\:\text{m/s}}=6.3\:\sec

Therefore, the total time of the sprint is

\text{t}_{\text{total}}=4\:\sec +6.3\:\sec =10.3\:\sec \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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