Tag Archives: College Physics

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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Grantham PHY220 Week 2 Assignment Problem 8

If a car is traveling at 50 m/s and then stops over 300 meters (while sliding), what is the coefficient of kinetic friction between the tires of the car and the road?

SOLUTION:

Draw the free-body diagram of the car

week 2 problem 8

Consider the vertical direction

\sum F_y=ma_y

F_N-mg=0

F_N=mg

Consider the motion in the horizontal direction

Solve for the acceleration of the car.

v^2=\left(v_0\right)^2+2a_x\Delta x

a_x=\frac{v^2-\left(v_0\right)^2}{2\Delta x}=\frac{0-50^2}{2\left(300\right)}=-4.17\:m/s^2

Solve for the coefficient of kinetic friction

\sum F_x=ma_x

-F_{fr}=ma_x

-\mu _kF_N=ma_x

\mu _k=\frac{ma_x}{-F_N}=\frac{m\:\left(-4.17\right)}{-m\left(9.80\right)}=\frac{4.17}{9.80}

\mu _k=0.43

College Physics by Openstax Chapter 2 Problem 49

You throw a ball straight up with an initial velocity of 15.0 m/s. It passes a tree branch on the way up at a height of 7.00 m. How much additional time will pass before the ball passes the tree branch on the way back down?

Solution:

The known values are a=-9.80\:\text{m/s}^2; v_o=15.0\:\text{m/s}; y=7.00\:\text{m}

The applicable formula is.

y=v_ot+\frac{1}{2}at^2

Using this formula, we can solve it in terms of time, t.

t=\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a}

Substituting the known values, we have

\begin{align*}
t & =\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a} \\
t & =\frac{-15.0\:\text{m/s}\pm \sqrt{\left(15.0\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(7.00\:\text{m}\right)}}{-9.80\:\text{m/s}^2} \\
t&=\frac{-15.0\:\text{m/s}\pm 9.37\:\text{m/s}}{-9.80\:\text{m/s}^2}
\end{align*}

We have two values for time, t. These two values represent the times when the ball passes the tree branch.

 t_1=\frac{-15.0\:m/s+9.37\:m/s}{-9.80\:m/s^2}=0.57\:sec \\
t_2=\frac{-15.0\:m/s-9.37\:m/s}{-9.80\:m/s^2}=2.49\:sec

Therefore, the total time between passing the branch is the difference between 2.49 seconds and 0.57 seconds.

t_2-t_1=2.49  \ \text{s} - 0.57 \ \text{s}=1.92 \ \text{s}  \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Solution Guides to College Physics Chapter 2 Banner

Chapter 2: Kinematics

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Displacement

Problem 1

Problem 2

Problem 3

Problem 4

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Time, Velocity, and Speed

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

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Acceleration

Problem 16

Problem 17

Problem 18

Problem 19

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Motion Equations for Constant Acceleration in One Dimension

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

Problem 37

Problem 38

Problem 39

Problem 40

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Falling Objects

Problem 41

Problem 42

Problem 43

Problem 44

Problem 45

Problem 46

Problem 47

Problem 48

Problem 49

Problem 50

Problem 51

Problem 52

Problem 53

Problem 54

Problem 55

Problem 56

Problem 57

Problem 58

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Graphical Analysis of One-Dimensional Motion

Problem 59

Problem 60

Problem 61

Problem 62

Problem 63

Problem 64

Problem 65

Problem 66

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Solution Guide to College Physics by Openstax Chapter 1 Banner

Chapter 1: Introduction: The Nature of Science and Physics

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Physical Quantities and Units

Problem 1

Problem 2

Problem 3

Problem 4

Problem 5

Problem 6

Problem 7

Problem 8

Problem 9

Problem 10

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Accuracy, Precision, and Significant Figures

Problem 11

Problem 12

Problem 13

Problem 14

Problem 15

Problem 16

Problem 17

Problem 18

Problem 19

Problem 20

Problem 21

Problem 22

Problem 23

Problem 24

Problem 25

Problem 26

Problem 27

Problem 28

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Approximation

Problem 29

Problem 30

Problem 31

Problem 32

Problem 33

Problem 34

Problem 35

Problem 36

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