What you need to know

Speed is a measure of quickly something moves, so it is calculated by dividing distance (how far that thing travelled) by time (how long it took to travel that far).

\text{speed } = \dfrac{\text{distance}}{\text{time}}

Typically, we measure distance in metres (m), kilometres (km), or miles (sometimes also ‘m’, but it should always be clear in the question whether it’s metres or miles), and we measure time in seconds (s) or hours (h).

As a result, the units we used to measure speed are compound units (for more information, see here (https://mathsmadeeasy.co.uk/gcse-maths-revision/conversions-gcse-revision-and-worksheets/)). The unit of choice will depend on what units are used to measure the values you’re given. If, in a particular question, time is measured in seconds and distance is measured in metres, then the speed we calculate would be measured “metres per second”, which is denoted either by \text{ms}^{-1} or \text{m/s}. If the starting units were different, then the resulting unit of speed would be different. Let’s see an example.

Example: A truck travels 110 miles in 2 hours. What is the average speed of the bus?

Speed is equal to the distance divided by the time, so this will be our calculation, but what will the unit of our answer be? In the question, the distance is given in miles and the time in hours, so the speed should be in miles per hour, which is denoted here like \text{m/h}.

\text{Speed } = \dfrac{110}{2} = 55\text{ m/h}

This is one of the more common units of speed, but it varies, so watch out for the units given in the question. In general, the units of speed will be in the form “[units of distance] per [units of volume]”.

Note: in reality, the truck’s speed will fluctuate during the 2 hours and won’t always be going at 55 m/h, but we don’t know all that information, hence why we calculate average speed instead.

On top of calculating speed, you will also be expected to be able to rearrange this formula and use it to find distance (when given the speed and time) and time (when given the speed and distance). From now on, let s be speed, t be time, and d be distance. Then, our original equation looks like:

s=\dfrac{d}{t}

If we multiply both sides of the equation by t, then swap the left and right-hand side, we get

d=s\times t

So, we see that we can calculate distance by multiplying the speed by the time. Furthermore, if we then divide both sides by s and swap the left and right-hand side, we get

t=\dfrac{d}{s}

Therefore, we can calculate time by dividing the distance by the speed. It’s good practice rearranging this formula to get it in the form that you want, but a quick way to remember how to calculate one of these values using the other two is to refer to the triangle below.

 

The way to use this triangle is as follows. Take your finger and cover up the letter which represents the thing that you’re trying to calculate. Then, the triangle will tell you what to do with the other two quantities to get the value you want.

 

For example, if we want to calculate the speed, then we construct this triangle and cover up the ss (since that’s what we want). Then, we see that what’s left over is “dd over tt”, or in other words, dd divided by tt will give us the speed. Learning this triangle and how to use it makes your life a lot easier.

Example: Jesse throws a ball that moves at an average speed of 35 metres per second and travels for a total of 4.5 seconds. Work out the distance travelled by the ball.

 

We’re looking for distance, so, constructing the triangle and covering up the d, we get

 

Therefore, to calculate the distance we must multiply the speed by the time. So

 

\text{distance }=35\times 4.5=157.5\text{ s}

Note: this question works fine because all the units match up. If, for example, speed was given in ‘metres per second’ but time was given in ‘hours’, we couldn’t work out the distance straight away. We would first have to either convert the time to ‘seconds’ so it matched the speed, or convert the speed into ‘metres per hour’ so it would match the time.

 

A good knowledge of how speed, distance, and time are related is important not only for questions like these, but also when working with distance-time graphs (https://mathsmadeeasy.co.uk/gcse-maths-revision/distance-time-graphs-gcse-revision-and-worksheets/) and velocity-time graphs (https://mathsmadeeasy.co.uk/gcse-maths-revision/velocity-time-graphs-gcse-revision-and-worksheets/).

Example Questions

1) Skyler runs a new personal best in the 100-metre sprint. Her average speed over the course of the race is 8.5 metres per second. How long, to 2dp, did it take her to run the race?

 

Answer

We are looking for time, so covering up the t on the triangle above, we can see that we need to divide distance, d, by speed, s.

 

Doing so, we get

 

\text{time }=\dfrac{d}{s}=\dfrac{100}{8.5}=11.76\text{ second (2dp)}

2) Gustavo is driving a bus along a motorway with a speed limit of 70 miles per hour. In 30 minutes, he travels 36 miles. Is his average speed during this period exceeding the speed limit?

 

Answer

We are looking for speed, so covering up the s on the triangle above, we can see that we need to divide distance, d, by time, t.

 

However, before we can do that, we need to sort out the units. The speed limit is in ‘miles per hour’, but the time we want to calculate with is given in ‘minutes’. Rather than converting the speeds to be the same unit afterwards, we can immediately fix this converting the minutes into hours. This is easy, as 30 minutes is half an hour, or 0.5 hours. So, we get that Gustavo’s average speed is

 

\text{speed }=\dfrac{d}{t}=\dfrac{36}{0.5}=72\text{ miles per hour}

 

So, yes, Gustavo is exceeding the speed limit.

Speed-distance-time questions are a staple of the old and new GCSE Maths. It is one of the questions types which hasn’t changed as drastically as the others. The most common mistake made by students is always to do with units. The speed, distance, time questions on this page is suitable revision for the new GCSE Maths 9-1 specifications.