# This note has been used to help create the Speed, Distance, and Time wiki

## Definition

The relationship between **speed**, **distance**, and **time** is expressed in this equation

\[\mbox{Speed} = \frac{\mbox{Distance}}{\mbox{Time}},\]

which can also be rearranged as

\[\mbox{Time} = \frac{\mbox{Distance}}{\mbox{Speed}}\]

and

\[\mbox{Distance} = \mbox{Speed} \times \mbox{Time}.\]

## Technique

Speed, distance, and time problems ask us to solve for one of the three variables given certain information. In these problems, we deal with objects moving at either constant speeds or average speeds.

Most problems will give values for two variables and ask us to solve for the third. For example:

## Bernie boards a train at 1:00 PM and gets off at 5:00 PM. During this trip, the train traveled 360 kilometers. What was the train's average speed in kilometers per hour?

In this problem, the total time is 4 hours and the total distance is 360 km, which we can plug into the equation: \[\begin{align}

\mbox{Speed} &= \frac{\mbox{Distance}}{\mbox{Time}}\\

&= \frac{360~\mbox{km}}{4~\mbox{h}}\\

&= 90~\mbox{km/h}. \quad_\square

\end{align}\]

When working with these problems, always pay attention to the units for speed, distance, and time. Here is a problem where we need to convert units to get the correct answer:

## A horse is trotting along at a constant speed of 8 miles per hour. How many miles will it travel in 45 minutes?

The equation for calculating distance is \[\mbox{Distance} = \mbox{Speed} \times \mbox{Time},\]

but we won't arrive at the correct answer if we just multiply 8 and 45 together, as the answer would be in units of \(\mbox{miles} \times \mbox{minute} / \mbox{hour}\). To fix this, we incorporate a unit conversion. \[\begin{align}

\mbox{Distance} &= \frac{8~\mbox{miles}}{~\mbox{hour}} \times 45~\mbox{minutes} \times \frac{1~\mbox{hour}}{60~\mbox{minutes}} \\

&= 6~\mbox{miles} \quad_\square

\end{align}\]

Alternatively, we can convert the speed to units of miles per minute and calculate for distance: \[\mbox{Distance} = \frac{2}{15}~\frac{\mbox{miles}}{\mbox{minute}} \times 45~\mbox{minutes} = 6~\mbox{miles}\]

or we can convert time to units of hours before calculating: \[\mbox{Distance} = 8~\frac{\mbox{miles}}{\mbox{hour}} \times \frac{3}{4}~\mbox{hours} = 6~\mbox{miles}.\]

Any of these methods will give the correct units and answer.

In more involved problems, we may find it convenient to use variables such as \(v\), \(d\), and \(t\) for speed, distance, and velocity respectively.

## Application and Extensions

## Albert and Danny are running in a long-distance race. Albert runs at 6 miles per hour while Danny runs at 5 miles per hour. You may assume they run at a constant speed throughout the race. When Danny reaches the 25 mile mark, Albert is exactly 40 minutes away from finishing. What is the race's distance in miles?

Let's begin by calculating how long it takes for Danny to run 25 miles. Let \[\begin{align}

\mbox{Time} &= \frac{\mbox{Distance}}{\mbox{Speed}}\\

&= \frac{25~\mbox{miles}}{5~\mbox{miles/hour}}\\

&= 5~\mbox{hours}. \end{align}\] So, it will take Albert \(5~\mbox{hours} + 40~\mbox{minutes}\), or \(\frac{17}{3}~\mbox{hours}\), to finish the race. Now we can calculate the race's distance: \[\begin{align}

\mbox{Distance} &= \mbox{Speed} \times \mbox{Time} \\

&= (6~\mbox{miles/hour}) \times (\frac{17}{3}~\mbox{hours}) \\

&= 34~\mbox{miles}\quad_\square

\end{align}\]

## A cheetah spots a gazelle 300 m away and sprints towards it at 100 km/h. At the same time, the gazelle runs away from the cheetah at 80 km/h. How many seconds does it take for the cheetah to catch the gazelle?

Let's set up equations representing the distance the cheetah travels and the distance the gazelle travels. If we set distance \(d\) equal to \(0\) as the cheetah's starting point, we have: \[\begin{align}

d_\text{cheetah} &= 100t \\

d_\text{gazelle} &= 0.3 + 80t.

\end{align}\] Note that time \(t\) here is in units of hours, and 300 m was coverted to 0.3 km.The cheetah catches the gazelle when \[\begin{align}

d_\text{cheetah} &=d_\text{gazelle} \\

100t &= 0.3 + 80t \\

20t &= 0.3 \\

t &= 0.015~\mbox{hours}.

\end{align}\] Converting that answer to seconds, we find that the cheetah catches the gazelle in \(54~\mbox{seconds}\).\(_\square\)

## Comments

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TopNewestWrite the answer in a separate section so parents like me can access them. This is great for my children and he has learnt lots. Thanks – Maths Perera · 3 years, 1 month ago

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