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*}

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.

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.

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.

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*}