Tag Archives: Physics Solutions

College Physics by Openstax Chapter 2 Problem 12

The speed of propagation of the action potential (an electrical signal) in a nerve cell depends (inversely) on the diameter of the axon (nerve fiber). If the nerve cell connecting the spinal cord to your feet is 1.1 m long, and the nerve impulse speed is 18 m/s, how long does it take for the nerve signal to travel this distance?

Solution:

The time of travel is computed based on the formula

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

Therefore, the time of travel is

\begin{align*}
\text{time} & =\frac{1.1\:\text{m}}{18\:\text{m/s}} \\
& =0.0611\:\text{seconds} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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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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Problem 2-6: The average speed and average velocity of a spinning helicopter blade

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PROBLEM:

A helicopter blade spins at exactly 100 revolutions per minute. Its tip is 5.00 m from the center of rotation.

(a) Calculate the average speed of the blade tip in the helicopter’s frame of reference.

(b) What is its average velocity over one revolution?

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SOLUTION:

Part A

The average speed of the blade tip is equal to the distance traveled divided by the time elapsed.

\begin{align*}
\text{speed} & =\frac{ \text{distance traveled}}{ \text{time elapsed}} \\ \\
& =\frac{2\pi \text{r}}{ \text{t}} \\ \\
& =\frac{2\pi \left(5.00\text{m}\right)}{60\:\text{s}}\times 100\:\text{rev} \\ \\
& =52.36\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}

Part B

After one revolution, the tip of the blade is at the same position as it is originally. This means that the displacement is zero. Thus, the velocity is zero.

\text{v}=0\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
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Problem 1-30: The number of generations passed since the year 0 AD

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PROBLEM:

A generation is about one-third of a lifetime. Approximately how many generations have passed since the year 0 AD?

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SOLUTION:

We are looking for generations passed in history. The general assumptions are:

  • There are 1011 seconds in 1 history
  • In 1 generation, there is 1/3 of a lifetime.
  • In half a lifetime, there are 109 seconds

Therefore, the number of generations passed is

\begin{align*}
1\ \text{history} & =1\:\text{history}\times \frac{10^{11}\:\text{sec}}{1\:\text{history}}\times \frac{1\:\text{generation}}{\frac{1}{3}\:\text{lifetime}}\times \frac{0.5\:\text{lifetime}}{10^9\:\text{sec}} \\
& =150\:\text{generations} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{align*}
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Problem 1-24: Calculating uncertainties in distance, time and speed of a marathon runner

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PROBLEM:

A marathon runner completes a 42.188-km course in 2 h, 30 min, and 12 s. There is an uncertainty of 25 m in the distance traveled and an uncertainty of 1 s in the elapsed time.

(a) Calculate the percent uncertainty in the distance.

(b) Calculate the uncertainty in the elapsed time.

(c) What is the average speed in meters per second?

(d) What is the uncertainty in the average speed?

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SOLUTION:

Part A

The percent uncertainty in the distance is

\begin{align*}
\text{\%\:uncertainty}_{\text{distance}} & =\frac{25\:\text{m}}{42.188\:\text{km}}\times \frac{1\:\text{km}}{1000\:\text{m}}\times 100\% \\
 & =0.0593\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Part B

The uncertainty in time is

\begin{align*}
\text{\%\:uncertainty}_{\text{time}} & =\frac{1\:\text{s}}{9012\:\text{s}}\times 100\% \\
& =0.0111\% \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Part C

The average speed is

\begin{align*}
\text{average speed} & =\frac{42.188\:\text{km}}{9012\:\text{s}}\times \frac{1000\:\text{m}}{1\:\text{km}} \\
& = 4.681\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}

Part D

The percent uncertainty in the speed is the combination of uncertainties of distance and time.

\begin{align*}
\text{\%\:uncertainty}_{\text{speed}} & =\text{\%\:uncertainty}_{\text{distance}}+\text{\%\:uncertainty}_{\text{time}} \\
& =0.0593\%+0.0111\% \\
&  =0.0704\% \\
\end{align*}

Therefore, the uncertainty in the speed is

\begin{align*}
\delta _{speed} & =\frac{0.0704\%}{100\%}\times 4.681\:\text{m/s} \\
& = 0.003\:\text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right) \\
\end{align*}
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Problem 1-2: Converting car speed of 33 m/s to kilometers per hour and determining if it exceeds the speed limit

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PROBLEM:

A car is traveling at a speed of 33 m/s.
(a) What is its speed in kilometers per hour?
(b) Is it exceeding the 90 km/h speed limit?

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SOLUTION:

Part A

\begin{aligned}
33 \ \text{m/s} & =33\ \frac{\text{m}}{\text{s}} \times \frac{1\ \text{km}}{1000 \ \text{m}} \times \frac{3600\ \text{s}}{1 \ \text{hr}} \\
\\
& =118.8 \ \text{km/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{aligned}

Part B

At 118.8 km/h, the car is traveling faster than the speed limit of 90 km/h. (Answer)

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Problem 1-1: Converting 100 km/h to meters per second and miles per hour

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PROBLEM:

The speed limit on some interstate highways is roughly 100 km/h.
(a) What is this in meters per second?
(b) How many miles per hour is this?

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SOLUTION:

Part A

\begin{aligned}
100 \  \frac{ \text{km}}{\text{hour}} & =100 \ \frac{ \bcancel{\text{km}}}{\bcancel{\text{hour}}} \times \frac{1000 \ \text{m}}{1 \ \bcancel{\text{km}}} \times \frac{1 \ \bcancel{\text{hour}}}{3600 \ \text{sec}}\\
\\
&=27.7 \ \text{m/s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{aligned}

Part B

\begin{aligned}
100 \  \frac{ \text{km}}{\text{hour}} & =100 \ \frac{ \bcancel{\text{km}}}{\text{hour}} \times\frac{1 \ \text{mile}}{1.609\ \bcancel{\text{km}}} \\
\\
&=62.2 \ \text{mi/hr} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
\end{aligned}
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Video Solution:

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