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Here you will learn about the average speed formula, including how to calculate the average speed and how to calculate the time or distance given the average speed.

Students will first learn about average speed formula as part of algebra in high school.

The **average speed formula** is the formula used to calculate the average speed of a journey.

The average speed of an object is the total distance the object travels divided by the total amount of time taken.

Average speed is a **compound measure** which is usually given in the units meters per second (m/s), miles per hour (mph) or kilometers per hour (km/h).

You can calculate average speed using the formula,

\text{average speed = }\cfrac{\text{total distance}}{\text{total time}}If you are calculating average speed in mph or km/h, you will need to ensure the time value is in hours before dividing.

For example,

Nadine travels a distance of 84 \, km; in total it takes her 1 \, hour and 30 \, minutes.

To find her average speed, convert the time to hours and use the formula.

\begin{aligned} 1\text{ hour 30 minutes } &= 1 + \cfrac{30}{60}\text{ hours} \\\\ & = {1.5 \mathrm{~hours}} \end{aligned} \begin{aligned}\text { Average speed } & =\cfrac{\text { total distance }}{\text { total time }} \\\\ & =\cfrac{84 \mathrm{~km}}{1.5 \mathrm{~hours}} \\\\ & =56 \mathrm{~km} / \mathrm{h} \end{aligned}You may need to calculate the average speed of a journey which has been broken into multiple parts.

For example,

Jerome is on a journey which totals 250 \, miles.

He travels at 60 \, mph for 2 \, hours before taking a break for 30 \, minutes.

He traveled the remaining distance at a speed of 52 \, mph.

Calculate Jerome’s average speed for the whole journey.

First you need to find how far Jerome traveled in the first part of his journey.

You can rearrange the formula to calculate the distance.

If you multiply both sides of the formula by ‘total time’,

\left(\text { Average speed }=\cfrac{\text { total distance }}{\text { total time }}\right) \times \text { total time }You get the formula for distance.

\begin{aligned}\text { distance } & =\text { speed } \times \text { time. } \\\\ & =60 \mathrm{~mph} \times 2 \text { hours } \\\\ & =120 \text { miles } \end{aligned}Jerome therefore has 250-120=130 \, miles remaining.

Next, find the time taken for the last part of his journey.

If you divide both sides of the formula by ‘speed’,

\cfrac{\text { distance }}{\text { speed }} = \cfrac{\text { speed } \times \text { time }}{\text { speed }} \begin{aligned}\text { Using the formula time } & =\cfrac{\text { distance }}{\text { speed }} \\\\ & =\cfrac{130 \text { miles }}{52 \mathrm{~mph}} \\\\ & =2.5 \text { hours } \end{aligned}You can now apply the average speed formula, ensuring you also include the time Jerome took for a break.

\text{Average speed } =\cfrac{\text { total distance }}{\text { total time }} \begin{aligned} \quad \quad \quad \quad & =\cfrac{250 \text { miles }}{2 \text { hours }+0.5 \text { hours }+2.5 \text { hours }} \\\\ & =50 \mathrm{~mph} \end{aligned}Jerome’s average speed for the whole journey was 50 \, mph.

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DOWNLOAD FREEHow does this relate to high school math?

**Algebra – Creating Equations (HS-A.CED.A.4)**Rearrange formulas to highlight a quantity of interest, using the same reasoning as in solving equations. For example, rearrange Ohm’s law V=IR to highlight resistance R.

In order to use the average speed formula:

**Check what information about the journey has been provided.****Find any missing values required for the total distance and total time, converting the time if required.****Substitute values into the formula, \textbf{ average speed }=\cfrac{\textbf{total distance}}{\textbf{total time}} and calculate.**

Samiya drove from New York City to Roanoke, VA. The journey took 8\text{ hours 30 minutes } and the total distance traveled was 415 \, miles. Find the average speed for her journey.

**Check what information about the journey has been provided.**

There is a total distance of 415 \, miles and a total time of 8\text{ hours 30 minutes. }

2**Find any missing values required for the total distance and total time, converting the time if required.**

You need to convert the time of 8\text{ hours 30 minutes } to just hours.

\begin{aligned} 8\text{ hours 30 minutes } & = 8 + \cfrac{30}{60}\text{ hours} \\\\ & = 8.5 \text{ hours} \end{aligned}3**Substitute values into the formula, \textbf{ average speed }=\cfrac{\textbf{total distance}}{\textbf{total time}} and calculate.**

The average speed of a car traveling round trip from Mumbai, India to Thane, India is 25 \, km/h. If the journey took 2 \, hours and 10 \, minutes, what distance was traveled?

**Check what information about the journey has been provided.**

There is an average speed of 25 \, km/h and an entire trip time of 2 \, hours and 10 \, minutes.

You need to convert 2 \, hours and 10 \, minutes to just hours.

Rearranging the formula you get,

Theo walked from home to school at an average speed of 4.8 \, km/h. It took him 45 \, minutes to walk to school. How far does Theo have to walk to school?

**Check what information about the journey has been provided.**

There is an average speed of 4.8 \, km/h and a total time of 45 \, minutes.

You need to convert the time 45 \, minutes to hours.

Rearranging the formula you get,

Dean biked home from work at an average speed of 12 \, mph. He biked a total of 20 \, miles. Find the total time it took Dean to bike home in hours and minutes.

**Check what information about the journey has been provided.**

There is a total distance of 20 \, miles and an average speed of 12 \, mph.

You have the values you need for the formula.

Rearranging the formula you get,

1 . \overline{6} \, hours is the same as 1 \cfrac{2}{3} \, hours which is 1 \, hour \, 40 \, minutes.

It took Dean 1 \, hour \, 40 \, minutes to cycle home from work.

Desirée drove to meet a friend for lunch. Her journey consisted of driving for 15 \, minutes at 30 \, mph, then 10 \, miles traveling at 50 \, mph.

Find her average speed for the whole journey.

**Check what information about the journey has been provided.**

There are two parts to her journey. There is a time of 15 \, minutes and a speed of 30 \, mph for the first part and a distance of 10 \, miles and a speed of 50 \, mph for the second part.

You need to find the total distance and the total time.

For the first part of the journey, you can find the distance traveled.

For the second part of the journey you can find the time taken.

The total distance is 7.5+10=17.5 \, miles.

The total time is 0.25+ 0.2=0.45 \, hours

\text {Average speed } =\cfrac{\text { total distance }}{\text { total time }}

\begin{aligned} \;\; & =\cfrac{17.5 \text { miles }}{0.45 \text { hours }} \\\\ & =38.9 \mathrm{~mph} \text { (rounded to the nearest tenth)} \end{aligned}

Tanner wants to meet a friend at the library. From his home to the library is a 15 \, minutes bike ride at 10 \, mph.

Returning, he will walk with his friend at 6 \, mph. If Tanner and his friend stay at the library for 20 \, minutes, what is the average speed for Tanner’s roundtrip from the entire trip?

**Check what information about the journey has been provided.**

There are three parts to his journey. There is a time of 15 \, minutes and speed of 10 \, mph for the first part, 20 \, minutes of 0 \, mph at the library and the return trip home at 6 \, mph.

You need to find the total distance and the total time.

For the first part of the journey you can find the distance traveled.

The second part of the journey was spent at the library, so 0 \, miles were moved.

Converting the 20 \, minutes to hours, \, \cfrac{20}{60}, shows that they spent 0 . \overline{3} \, hours at the library.

For the third part of the journey you can find the time taken. You know from the first part of the journey, that the distance from the library to Tanner’s home is 2.5 \, miles.

The total distance is 2.5+0+2.5=5 \, miles.

The total time is 0.25+0 . \overline{3}+0.41 \overline{6}=1 \text { hour }

\begin{aligned}\text { Average speed } & =\cfrac{\text { total distance }}{\text { total time }} \\\\
& =\cfrac{5 \text { miles }}{1 \text { hour }} \\\\
& =5 \mathrm{~mph} \end{aligned}

- Start with situations that have one speed, before moving on to ones that involve multiple time intervals with different speeds.

- Provide activities that have students recording time and distance. This could be as simple as using a stopwatch to time each student walking 100 \, m. Or as advanced as using robotics equipment or GPS technology.

- If using worksheets, make sure to choose ones that have a variety of units of time, units of distance and units of speed. They should also vary in what information is being solved for, such as the speed of the car, travel time, starting point, etc.

**Creating the wrong units**

Since speed is a compound unit, it is important to pay attention to the units in every calculation. Sometimes an equation creates a compound unit and sometimes it cancels out common units.

For example,

- \frac{20 \text { miles }}{2 \text { hours }}=10 \frac{\text { miles }}{\text { hour }} \leftarrow The equation sets up the units as \frac{\text { miles }}{\text { hour, }} which is often represented as “miles per hour” or “mph”
- \frac{2.5 \mathrm{~miles}}{6 \mathrm{~mph}}=0.41 \overline{6} \text { hours } \leftarrow The equation sets up the units as

\frac{\text { miles }}{\text { mph }}=\frac{\text { miles }}{\frac{\text { miles }}{\text { hour }}}=\text { miles } \times \frac{\text { hour }}{\text { miles }}=\text { hour. } The units of miles cancel out, leaving ‘hours’.

- \frac{20 \text { miles }}{2 \text { hours }}=10 \frac{\text { miles }}{\text { hour }} \leftarrow The equation sets up the units as \frac{\text { miles }}{\text { hour, }} which is often represented as “miles per hour” or “mph”

**Assuming instantaneous speed or constant speed for entire trip**

An average speed of 30 \, mph does not automatically mean that the object was instantaneously going 30 \, mph at the start of the period of time.

It does not also mean that the object was traveling at exactly 30 \, mph for the total duration of the movement. Let’s look at two situations where 30 \, mph is the average speed.

\quad 1) A car travels for 25 \, minutes at 30 \, mph the entire trip.

\quad 2) A car travels for 25 \, minutes at 10 \, mph and then 25 \, minutes at 50 \, mph.

Example #2 changes speeds, but still has an average of 30 \, mph.

**Finding the average speed from a journey with varying speeds by finding the mean value of the different speeds**

A journey was broken into three parts and the speed for the three parts were 30 \, mph, 40 \, mph and 70 \, mph. A common error would be to find the average speed by finding the mean of 30, 40 and 70.

This would not necessarily give the correct answer because you do not know the duration or distance of each part of the journey.

If the parts of the journey each had different durations, finding the mean of the values would not give the correct average speed. You need to find the total time and the total distance.

**The total time is not converted to hours correctly**

A common error is to incorrectly convert time to hours.

For example,

1 \, hour \, 30 \, minutes is**not**1.3 \, hours .

To convert, write 1 \, hour \, 30 \, minutes as the mixed number 1 \cfrac{30}{60}, where the numerator is the minutes passed and the denominator is the whole, in this case 60 \, minutes.

Then write the mixed number as a decimal: 1 \cfrac{30}{60}=1.5

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1. A journey takes a total of 1 \, hour and 30 \, minutes and covers a distance of 90 \, miles. What was the average speed for the journey?

117 \, mph

60 \, mph

135 \, mph

69.23 \, mph

First, convert 1 \, hour and 30 \, minutes to just hours.

\begin{aligned} 1 \text { hour and 30 minutes } &=1 \cfrac{30}{60} \text { hours } \\\\ 1 \cfrac{30}{60} \, hours &=1.5 \, hours \end{aligned}

\begin{aligned} \text { Average speed }& =\cfrac{\text { total distance }}{\text { total time }} \\\\ & =\cfrac{90 \text { miles }}{1.5 \text { hours }} \\\\ & =60 \mathrm{~mph} \end{aligned}

2. A journey takes a total of 3 \, hours and 45 \, minutes and covers a distance of 330 \, miles. What was the average speed for the journey?

95.65 \, mph

101.54 \, mph

88 \, mph

1,237.5 \, mph

First, convert 3 \, hours and 45 \, minutes to just hours.

\begin{aligned} 3 \text { hours and 45 minutes } &=3 \cfrac{45}{60} \text { hours } \\\\ 3 \cfrac{45}{60} \, hours &=3.75 \, hours \end{aligned}

\begin{aligned} \text { Average speed }& =\cfrac{\text { total distance }}{\text { total time }} \\\\ & =\cfrac{330 \text { miles }}{3.75 \text { hours }} \\\\ & =88 \mathrm{~mph} \end{aligned}

3. A journey has an average speed of 55 \, km/h. The total time taken was 4 \, hours and 24 \, minutes. Find the distance traveled.

242 \, km

233.3 \, km

220 \, km

12.5 \, km

First, convert 4 \, hours and 24 \, minutes to just hours.

\begin{aligned} 4 \text { hours and 24 minutes } &=4 \cfrac{24}{60} \text { hours } \\\\ 4 \cfrac{24}{60} \, hours &=4.4 \, hours \end{aligned}

\text {Total distance } =\text { average speed } \times \text { total time }

\begin{aligned} \quad \quad \quad \quad & =55 \mathrm{~km} / \mathrm{h} \times 4.4 \mathrm{~h} \\\\ & =242 \mathrm{~km}\end{aligned}

4. A journey of 120 \, km was completed with an average speed of 36 \, km/h. Find the time taken in hours and minutes.

3 \, hours \, 33 \, minutes

3.33 \, hours

3 \, hours \, 30 \, minutes

3 \, hours \, 20 \, minutes

\begin{aligned} \text { Total time }&=\cfrac{\text { total distance }}{\text { average speed }} \\\\
\text { time } & =\cfrac{120 \mathrm{~km}}{36 \mathrm{~km} / \mathrm{h}}=3 . \overline{3} \text { hours } \end{aligned}

3 . \overline{3} \text { hours } is 1 \cfrac{1}{3} \text { hours }.

The total time taken is 1 \, hour \, 20 \, minutes.

5. A car travels at 30 \, mph for 20 \, minutes and then 54 \, mph for 40 \, minutes. Find the average speed for the whole journey.

46 \, mph

42 \, mph

84 \, mph

1.4 \, mph

This journey is in two parts. You need the total distance and the total time to calculate the average speed for the whole journey. For both parts, you are given the speed and time. You use these to find the distance traveled for each part.

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

Distance for part 1 is 30 \mathrm{~mph} \times \cfrac{20}{60} \text { hours }=10 \text { miles. }

Distance for part 2 is 54 \mathrm{~mph} \times \cfrac{40}{60} \text { hours }=36 \text { miles }

The total distance is 10+36=46 \, miles.

The total time is

20 minutes + \, 40 minutes = 60 minutes = 1 hour.

The average speed for the whole journey is

\text { average speed }=\cfrac{\text { distance }}{\text { time }}=\cfrac{46}{1}=46 \mathrm{~mph}.

6. A car travels 40 \, miles in 45 \, minutes before stopping for a 30 \, minute rest. It then travels at 60 \, mph for a further 51 \, minutes. Find the average speed for the whole journey.

56.875 \, mph

43 . \overline{3} \, mph

56 . \overline{6} \, mph

50 \, mph

This journey is in three parts. For part one, you are given the distance and time. For the second part, you are given the time at rest. For the third part, you are given the speed and time.

You need the total distance and the total time to calculate the average speed for the whole journey.

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

Distance for part 1 is given in the question, 40 \, miles.

Distance for part 2 is given in the question, 0 \, miles.

Distance for part 3 is 60 \mathrm{~mph} \times \cfrac{51}{60} \text { hours }=51 \text { miles }

The total distance is 40+0+51=91 \, miles.

The total time is

45 minutes + \, 30 minutes + \, 51 minutes = 126 minutes = 2.1 hours.

The average speed for the whole journey is

\text { average speed }=\cfrac{\text { distance }}{\text { time }}=\cfrac{91 \text { miles }}{2.1 \text { hours }}=43 . \overline{3} \mathrm{~mph}

Average velocity is similar to average speed, but velocity accounts for direction.

The average speed equation is the same as the average speed formula, which is \text { average speed }=\cfrac{\text { distance }}{\text { time. }}

The abbreviation ‘avg’ is for the word average and it is sometimes used for average speed formula activities.

The base SI units for time and distance are seconds and meters, but hours and kilometers are often used for longer time and distance calculations.

This is one way to represent the time passed and distance traveled. It can also be used to calculate average speed.

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