\(\displaystyle \frac {9\text{km}}{1\text{h}} \times \frac {0.6214\text{mi}}{1\text{km}} = 5.6 \text{mph}\)

\(\displaystyle \frac {65\text{mi}}{1\text{h}} \times \frac {1.609\text{km}}{1\text{mi}} = 104.59\text{km/h}\)

\(\displaystyle \frac {80\text{mi}}{1\text{h}} \times \frac{1\text{h}}{60\text{min}} \times\frac {1609.344\text{m}}{1\text{mi}} = 2145.8 \frac {\text{m}}{\text{min}}\)

Written Solution:

Step 1: Start with what you know on top and bottom of the fraction (numerator and denominator, respectively).

The train is traveling at a speed of 80 miles per hour, which can be written as \( \dfrac {80\:\text {mi}}{1\:\text {hour}}\)

Step 2: Determine what you want to get in the end (on top and bottom).

We want to know how fast the train is going in \( \dfrac {\text{m}} {\text{min}}\)

Step 3: Determine what conversion factor(s) to use. You will need more than one (at least one for the top and one for the bottom).

We know the following:

\(1\; mi = 1609.344\; m\), so we will either use \(\dfrac{1\text{mi}}{1609.344\text{m}} \) or \(\dfrac{1609.344\text{m}}{1\text{mi}}\)

\(1\; hour = 60\; min\), so we will either use \(\dfrac{1\text{hour}}{60\text{min}}\) \(\dfrac{60\text{min}}{1\text{hour}}\)

Step 4: Multiply by 1 in the form of the conversion factor that cancels out the unwanted units.

We know that we need to change hours to minutes and miles to meters. We need to decide which conversion factors will allow us to do that.

\(\displaystyle \frac{80\text{mi}}{1\text{hour}}\times\:?\:\times\:?\:=\frac{?\:\text{min}}{?\:\text{m}}\)

Note that we chose the conversion factors that will enable us to cancel out the existing units and change them to the desired units. (For example, hour is in the bottom of our original fraction, so we chose the conversion factor that has hour in the top so they will cancel out, etc.)

\(\displaystyle \frac{80\text{mi}}{1\text{hour}}\times\frac{1\text{hour}} {60\text{min}}\:\times\frac{1609.344\text{m}}{1\text{mi}}\)

Next, we cancel out miles:



Then cancel out hours:



Using the zig-zag method we make the calculations in a zig-zag pattern. Remember, any time we move to the denominator we divide. And any time we move to the numerator, we multiply the following:



\(80 \div 1 \times 1 \div 60\; min \times 1609.344\; m \div 1 = 2145.792\)

So the train is traveling at a rate of \(\displaystyle \frac{2145.8\:\text{m}}{1\:\text{min}} = 2145.8\frac{\text{meters}}{\text{minute}}\) when rounded to the nearest tenth.

\(\displaystyle \frac{20\text{ft}}{1\text {min}}\times\frac{1\text{min}}{60\text{sec}}\times\frac{1\text{m}}{3.2808\text{ft}} = 0.102 \frac{\text{m}}{\text{sec}}\)

\(\displaystyle \frac{100\text{m}}{9.58\text{sec}}\times\frac{3.2808\text{ft}}{1\text{m}}\times\frac{60\text{sec}}{1\text{min}} = 2055 \frac{\text{ft}}{\text{min}}\)

\( \displaystyle \frac{11\text{m}}{1\text{sec}}\times\frac{1\text{km}}{1000\text{m}}\times\frac{60\text{sec}}{1\text{min}}\times\frac{60\text{min}}{1\text{h}} = 39.6 \frac{\text{km}}{\text{h}}\)

Written Solution:

Step 1: Start with what you know (on top and bottom).

The horse is galloping at a speed of 11 meters per second, which can be written as \( \dfrac {11\:\text{m}}{1\:\text{sec}}\)

Step 2: Determine what you want to get in the end (on top and bottom).

We want to know how fast the horse is going in \( \dfrac {\text{km}}{\text{h}}\)

Step 3: Determine what conversion factor(s) to use. You will need more than one: at least one for the top (numerator) and one for the bottom (denominator).

We know the following:

1 km = 1000 m, so we will either use \(\dfrac{1\text{km}}{1000\text{m}}\) or \(\dfrac{1000\text{m}}{1\text{km}}\)

1 min = 60 sec, so we will either use \(\dfrac{1\text{min}}{60\text{sec}}\) or \(\dfrac{60\text{sec}}{1\text{min}}\)

1 hour = 60 min, so we will either use \(\dfrac{1\text{hour}}{60\text{min}}\) or \(\dfrac{60\text{min}}{1\text{hour}}\)

Step 4: Multiply by 1 in the form of the conversion factor that cancels out the unwanted units.

We know that we need to change seconds to minutes to hours and meters to kilometers. We need to decide which conversion factors will allow us to do that.

\(\displaystyle \frac{11\:\text{m}}{1\:\text{sec}}\:\times\:?\:\times\:?\:\times\:?\:=\frac{?\:\text{km}}{?\:\text{hour}}\)

Note that we chose the conversion factors that will enable us to cancel out the existing units and change them to the desired units. (For example, seconds is in the bottom of our original fraction, so we chose the conversion factor with seconds in the top so they will cancel out, etc.)

\(\displaystyle \frac{11\:\text{m}}{1\:\text{sec}}\times\frac{60\:\text{sec}}{1\:\text{min}}\times\frac{60\:\text{min}}{1\:\text{hour}}\times\frac{1\:\text{km}}{1000\:\text{m}}\)

Next, we cancel out seconds:

\(\displaystyle \frac{11\:\text{m}}{1\:\cancel{{\color{Red} \text{sec}}}}\times\frac{60\:\cancel{{\color{Red} \text{sec}}}}{1\:\text{min}}\times\frac{60\:\text{min}}{1\:\text{hour}}\times\frac{1\:\text{km}}{1000\:\text{m}}\)

Then cancel out meters:

\(\displaystyle \frac{11\:\cancel{{\color{Purple} \text{m}}}}{1\:\cancel{{\color{Red} \text{sec}}}}\times\frac{60\:\cancel{{\color{Red} \text{sec}}}}{1\:\text{min}}\times\frac{60\:\text{min}}{1\:\text{hour}}\times\frac{1\:\text{km}}{1000\:\cancel{{\color{Purple} \text{m}}}}\)

Then cancel minutes:

\(\displaystyle \frac{11\:\cancel{{\color{Blue} \text{m}}}}{1\:\cancel{{\color{Red} \text{sec}}}}\times\frac{60\:\cancel{{\color{Red} \text{sec}}}}{1\:\cancel{{\color{Teal} \text{min}}}}\times\frac{60\:\cancel{{\color{Teal} \text{min}}}}{1\:\text{hour}}\times\frac{1\:\text{km}}{1000\:\cancel{{\color{Blue} \text{m}}}}\)

Using the zig-zag method we make the calculations in a zig-zag pattern. Remember, any time we move to the denominator we divide. Any time we move to the numerator, we multiply the following:



\(11 \div 1 \times 60 /div 1 \times 60 \div 1 hour \times 1 km \div 1000 = 39.6 kilometers\; per\; hour\)

So the horse is traveling at a rate of \(\dfrac{39.6\:\text{km}}{\text{hr}}\) or 39.6 km/h.