Tag Archives: One-Dimensional Kinematics with Constant Acceleration

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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