Tag Archives: College Physics Complete Solution Manual

College Physics by Openstax Chapter 2 Problem 18


A commuter backs her car out of her garage with an acceleration of 1.40 m/s2 .

(a) How long does it take her to reach a speed of 2.00 m/s?

(b) If she then brakes to a stop in 0.800 s, what is her deceleration?


Solution:

Part A

The formula for acceleration is

\overline{a}=\frac{\Delta v}{\Delta t}

If we rearrange the formula by solving for \Delta t, in terms of velocity and acceleration, we come up with

\Delta t=\frac{\Delta v}{\overline{a}}

Substituting the given values, we have

\begin{align*}
\Delta t & =\frac{\Delta v}{\overline{a}} \\
\Delta t & = \frac{v_f-v_0}{\overline{a}} \\
\Delta t & =\frac{2.00 \ \text{m/s}-0 \ \text{m/s}}{1.40 \ \text{m/s}^2} \\
\Delta t & =1.43 \ \text{seconds} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The formula for acceleration (deceleration) is

\overline{a}=\frac{\Delta v}{\Delta t}

Then substituting all the given values, we have

\begin{align*}
\overline{a} & = \frac{v_f-v_0}{\Delta t} \\
\overline{a} & = \frac{0 \ \text{m/s}-2\ \text{m/s}}{0.8 \ \text{m/s}^2} \\
\overline{a} & = -2.50 \ \text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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College Physics by Openstax Chapter 2 Problem 17


Dr. John Paul Stapp was U.S. Air Force officer who studied the effects of extreme deceleration on the human body. On December 10, 1954, Stapp rode a rocket sled, accelerating from rest to a top speed of 282 m/s (1015 km/h) in 5.00 s, and was brought jarringly back to rest in only 1.40 s! Calculate his

(a) acceleration and

(b) deceleration.

Express each in multiples of g (9.80 m/s2) by taking its ratio to the acceleration of gravity.


Solution:

Part A

The formula for acceleration is

\begin{align*}
\overline{a} & =\frac{\Delta v}{\Delta t} \\
\overline{a} & = \frac{v_f-v_0}{t_f-t_0} \\
\end{align*}

Substituting the given values

\begin{align*}
\overline{a} & =\frac{282\:\text{m/s}-0\:\text{m/s}}{5.00\:\sec } \\
\overline{a} & =56.4\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The deceleration is 

\begin{align*}
\overline{a} & =\frac{0\:\text{m/s}-282\:\text{m/s}}{1.40\:\text{s}} \\
\overline{a} & =-201.43\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

In expressing the computed values in terms of g, we just divide them by 9.80.

The acceleration is

\overline{a}=\frac{56.4}{9.80}=5.76\text{g} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

The deceleration is

\overline{a}=\frac{201.43}{9.80}=20.55\text{g} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

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College Physics by Openstax Chapter 2 Problem 16


A cheetah can accelerate from rest to a speed of 30.0 m/s in 7.00 s. What is its acceleration?


Solution:

The formula for acceleration is 

\begin{align*}
\overline{a} & =\frac{\text{change in velocity}}{\text{change in time}}\\
\overline{a}  & =\frac{\Delta \text{v}}{\Delta t} \\
\overline{a}  & =\frac{v_f-v_0}{t_f-t_0} \\
\end{align*}

Substituting the given values

\begin{align*}
\overline{a} & =\frac{30.0\:\text{m/s}-0\:\text{m/s}}{7.00\:\text{s}} \\
& =4.29\:\text{m/s}^2 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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College Physics by Openstax Chapter 2 Problem 15


The planetary model of the atom pictures electrons orbiting the atomic nucleus much as planets orbit the Sun. In this model, you can view hydrogen, the simplest atom, as having a single electron in a circular orbit 1.06×10-10 m in diameter.

(a) If the average speed of the electron in this orbit is known to be 2.20×106 m/s, calculate the number of revolutions per second it makes about the nucleus.

(b) What is the electron’s average velocity?


Solution:

Part A

The formula to be used is

\begin{align*}
\text{average speed} & =\frac{\text{distance}}{\text{time}} \\
r & = \frac{d}{t}
\end{align*}

Rearranging the formula–solving for the distance

\begin{align*}
d=r\times t
\end{align*}

Substituting the given values for 1 second period

\begin{align*}
d & = \left(2.20\times 10^6\:\text{m/s}\right)\left(1\:\text{s}\right) \\
& =2.20\times 10^6\:\text{meters}
\end{align*}

This is the total distance traveled in 1 sec.

With the given radius, the total distance traveled in 1 revolution is

\begin{align*}
1\:\text{revolution} & =2\pi \text{r} \\
& =\pi \text{d} \\
&=\pi \left(1.06\times 10^{-10}\text{m}\right)
\end{align*}

Therefore, the total number of revolutions traveled in 1 second is

\begin{align*}
\text{no. of revolutions} & = \frac{\text{total distance}}{\text{distance in 1 revolution}} \\
& = \frac{2.20\times 10^6}{\pi \left(1.06\times 10^{-10}\right)} \\
& =6.61\times 10^{15} \ \text{revolutions} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

In one complete revolution, the electron will go back to its original position. Thus, there is no net displacement. Therefore,

\begin{align*}
\overline{v} & =\frac{\Delta x}{\Delta t} \\
\overline{v} & =0 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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College Physics by Openstax Chapter 2 Problem 14


A football quarterback runs 15.0 m straight down the playing field in 2.50 s. He is then hit and pushed 3.00 m straight backward in 1.75 s. He breaks the tackle and runs straight forward another 21.0 m in 5.20 s. Calculate his average velocity

(a) for each of the three intervals and

(b) for the entire motion.


Solution:

Part A

The average velocity for each interval is computed using the formula

\overline{v}=\frac{\Delta x}{\Delta t}

For the first interval

 \overline{v_1}=\frac{15.0\:\text{meters}}{2.50\:\sec }=6.00\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

For the second interval

\overline{v_2}=\frac{-3.00\:\text{meters}}{1.75\:\sec }=-1.71\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

For the third interval

\overline{v_3}=\frac{21.0\:\text{m}}{5.20\:\text{sec}}=4.04\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)

Part B

For the entire motion, we need displacement from the beginning to the end.

\begin{align*}
\overline{v}& =\frac{\Delta x}{\Delta t} \\
& =\frac{15\:\text{m}-3\:\text{m}+21\:\text{m}}{2.50\:\text{s}+1.75\:\text{s}+5.20\:\text{s}} \\
& =\frac{33\:\text{m}}{9.45\:\text{s}} \\
& =3.49\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

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College Physics by Openstax Chapter 2 Problem 13


Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice was so loud in the astronaut’s space helmet that it was picked up by the astronaut’s microphone and transmitted back to Earth. It is reasonable to assume that the echo time equals the time necessary for the radio wave to travel from the Earth to the Moon and back (that is, neglecting any time delays in the electronic equipment). Calculate the distance from Earth to the Moon given that the echo time was 2.56 s and that radio waves travel at the speed of light 3.00×108 m/s.


Solution:

The total distance traveled is computed using the formula

\text{distance}=\text{speed}\times \text{time}

Therefore, the total distance traveled is

\begin{align*}
\text{distance} & =\left(3.00\times 10^8\:\text{m/s}\right)\left(2.56\:\sec \right) \\
& =768\:000\:000\:\text{meters}
\end{align*}

Hence, the distance between the Earth and Moon is equal to the total distance traveled divided by 2.

\begin{align*}
\text{Earth-Moon Distance} & =\frac{768\:000\:000\:\text{m}}{2} \\
& =384\:000\:000\:\text{meters} \\
& = 384 \ 000 \  \text{km} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The distance between the Earth and Moon is approximately 384 thousand kilometers.


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College Physics by Openstax Chapter 2 Problem 10


Tidal friction is slowing the rotation of the Earth. As a result, the orbit of the Moon is increasing in radius at a rate of approximately 4 cm/year. Assuming this to be a constant rate, how many years will pass before the radius of the Moon’s orbit increases by 3.84×106 m(1%)?


Solution:

From the formula \overline{v}=\frac{\Delta x}{\Delta t}, we can solve for \Delta t as follows

\begin{align*}
\Delta \text{t} & = \frac{\Delta x}{\overline{v}} \\
& = \frac{3.84\times 10^6\:\text{m}}{4\:\text{cm/year}}\times \frac{100\:\text{cm}}{1\:\text{m}} \\
& =96\:000\:000\:\text{years} \\
& =96.0\times 10^6\:\text{years} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

It will take about 96 million years.


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College Physics by Openstax Chapter 2 Problem 11


A student drove to the university from her home and noted that the odometer reading of her car increased by 12.0 km. The trip took 18.0 min.

(a) What was her average speed?

(b) If the straight-line distance from her home to the university is 10.3 km in a direction 25° S of E, what was her average velocity?

(c) If she returned home by the same path 7 h 30 min after she left, what were her average speed and velocity for the entire trip?


Solution:

Part A

The average speed is 

\begin{align*}
\text{speed} & = \frac{\text{distance}}{\text{time}}\\
&= \frac{12\:\text{km}}{18\:\text{mins}}\times \frac{60\:\text{mins}}{1\:\text{hr}} \\
& =40\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

The average velocity is

\begin{align*}
\overline{v} & =\frac{\Delta \text{x}\:}{\Delta \text{t}} \\
& =\frac{10.3\:\text{km}}{18.0\:\min \:}\times \frac{60\:\text{mins}}{1\:\text{hr}} \\
&=34.33\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The direction of the velocity is 25° S of E \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right).

Part C

The average speed is 

\begin{align*}
\text{speed} & = \frac{\text{distance}}{\text{time}}\\
& =\frac{12.0\:\text{km}\times 2}{7.5\:\text{hr}} \\
& =3.2\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

And the average velocity is

\begin{align*}
\overline{v}=0 \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

The average velocity is zero since the total displacement is zero.


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College Physics by Openstax Chapter 2 Problem 9


On May 26, 1934, a streamlined, stainless steel diesel train called the Zephyr set the world’s nonstop long-distance speed record for trains. Its run from Denver to Chicago took 13 hours, 4 minutes, 58 seconds, and was witnessed by more than a million people along the route. The total distance traveled was 1633.8 km. What was its average speed in km/h and m/s?


Solution:

The total time of travel is converted to seconds.

\begin{align*}
\text{t} & =\left(13\:\text{h}\:\times \frac{3600\:\text{s}}{1\:\text{hr}}\right)+\left(4\:\text{mins}\:\times \frac{60\:\text{s}}{1\:\min }\right)+58\:\sec \\
\text{t} & =47\:098\:\text{seconds}
\end{align*}

The total time of travel in hours

 \text{t}=\left(47\:098\:\text{seconds}\right)\left(\frac{1\:\text{h}}{3600\:\sec }\right)=13.0828\:\text{hours}

Therefore, the average speed in km/hr is

\begin{align*}
\text{speed in km/hr} & =\frac{\text{distance traveled}}{\text{time}} \\
& =\frac{1633.8\:\text{km}}{13.0828\:\text{hr}} \\
& =124.88\:\text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

And the average speed in m/s is

\begin{align*}
\text{speed in m/s} & =\frac{1\:633\:800\:\text{m}}{47\:098\:\text{s}} \\
& =34.689\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

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Problem 2-8: Motion of land mass around the San Andreas fault

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PROBLEM:

Land west of the San Andreas fault in southern California is moving at an average velocity of about 6 cm/y northwest relative to land east of the fault. Los Angeles is west of the fault and may thus someday be at the same latitude as San Francisco, which is east of the fault. How far in the future will this occur if the displacement to be made is 590 km northwest, assuming the motion remains constant?


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SOLUTION:

From the formula \overline{v}=\frac{\Delta x}{\Delta t}, we can solve for \Delta t as follows

\begin{align*}
\Delta t & =\frac{\Delta x}{\overline{v}} \\ \\
& =\frac{5.90 \times 10^{5}\ \text{m}}{6\ \text{cm/year}}\times \frac{100\ \text{cm}}{1\ \text{m}} \\ \\
& = 9.83 \times 10^{6}\ \text{years} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)\\
\end{align*}

Therefore, it will take about 9.83 million years.


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