## What you need to know

A velocity-time graph (or speed-time graph) is a way of visually expressing a journey. With speed on the y-axis and time on the x-axis, a speed-time graph tells us how someone/something’s speed has changed over a period of time.

We are going to be using velocity-time graphs to find two things, primarily: **total distance, **and **acceleration**. There are a few tools you’ll need to comfortable with to do this. Firstly, click here (areas of shapes gcse revision and worksheets for info on how to find the areas of shapes, and secondly click here (gradients straight line graphs revisio) and here (gradients of graphs revision) for info on how to find the gradients of graphs. We’ll see an example, and it should soon make a bit more sense why these things are required.

**Example: **The speed-time graph below describes a 50-second car journey. Work out the total distance travelled by the car and work out the maximum acceleration of the car during the 50 seconds.

Before we get stuck in to answering this, let’s go through the journey described by the graph. Firstly, the car **accelerated** from 0 to 15m/s over the first 10 seconds (because the line is straight, the acceleration is **constant**). Then, the line is flat, meaning the car’s speed was not changing for 10 seconds – it was moving at **constant speed**. Next, the car accelerated up to 25m/s over the next 10 seconds, and finally it spent the last 20 seconds **decelerating **back down to 0m/s.

Now, we want to work out the distance travelled. On a speed-time graph, the **distance travelled is the area under the graph**. To work out the area under this graph, we will break it into 4 shapes: A, B, C, and D -two triangles, a rectangle, and a trapezium are all shapes that we can work out the area of. So, we get

\text{A}=\dfrac{1}{2}\times10 \times15= 75\text{m}

\text{B}=10\times15=150\text{m}

C=\dfrac{1}{2}(15+25)\times10=200 \text{m}

D=\dfrac{1}{2}\times20\times25=250\text{m}

Therefore, the **total distance travelled** is: 75+150+200+250=675\text{m}. Next up, the acceleration. To do this, consider that acceleration is a measure of how quickly something’s speed is increasing. Therefore, given that the gradient is a **rate of change **of y (speed) with respect to x (time), we can work out the acceleration by finding the gradient of the graph.

Now, the question asks for “maximum acceleration”, so we can rule out certain parts of the journey. Specifically, the part where the graph is flat – there is no acceleration here – and the part where the graph slopes downward – it is decelerating here, so can’t be the maximum acceleration.

It might be obvious to you which of the two sections of the graph is steeper, but it isn’t always, so we’ll work out both just to be sure. Firstly, the first 10 seconds: the car’s speed goes from 0 to 15m/s and it takes 10 seconds, so we get

\text{acceleration between 0s and 10s = gradient}=\dfrac{15-0}{10-0}=1.5\text{m/s}^2

Then, the other portion we’re interested in is between 20 and 30 seconds. During this period, the speed increases from 15 to 25, so we get

\text{acceleration between 20s and 30s = gradient}=\dfrac{25-15}{30-20}=1\text{m/s}^2

The first one is larger, so the **maximum acceleration** is 1.5\text{m/s}^2.

Next, we’re going to see an example where the lines aren’t so straight. Make sure you know how to work out gradients of curved graphs before proceeding.

**Example: **Below is a speed-time graph of the first 4 seconds of someone running a race. Work out the average acceleration over the 4 seconds and work out the instantaneous acceleration 2 seconds in.

So, we’ve seen that to work out acceleration, we need to work out gradient.

To work out the **average acceleration **over the 4 seconds, we will draw a line from where the graph is at 0s to where the graph is at 4s and find the gradient of it. To work out the **instantaneous acceleration** 2 seconds in, we will draw a tangent to the line after 2 seconds and work out the gradient of that. The result of drawing these lines is below.

So, we get that average acceleration to be

\text{gradient}=\dfrac{9-0}{4-0}=2.25\text{m/s}^2

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Then, we get the instantaneous acceleration to be

\text{gradient}=\dfrac{8.4-4.8}{3.2-1}=1.64\text{m/s}^2\text{ (3sf)}

On top of this, you may be asked to **draw a velocity-time graph** from a description of a journey/period of time. Have a go at the first question below to see what this is like.

### Example Questions

1) A ball is placed at rest at the top of a hill. It travels with constant acceleration for the first 12 second and reaches a speed of 4m/s. It then **decelerates** at a constant rate of 0.1\text{m/s}^2 for 20 seconds. It then travels at a constant speed for a further 18 seconds.

Draw a speed-time graph for the ball over the course of this 50 seconds.

So, we will firstly draw a straight from the origin to (12, 4), since after 12 seconds, it’s reached 4m/s. Then, for the next part we’re told the deceleration is 0.1\text{m/s}^2 for 20 seconds. So, if the speed decreases by 0.1 every second, after 20 seconds it will be

0.1\times20=2\text{m/s}

Therefore, by 32 seconds in the speed is 2m/s, so we will draw a straight line from (12, 4) to (32, 2). Finally, a constant speed will be represented by a flat line that goes until the 50 second point, still at 2m/s. The result should look like the graph below.

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2) Below is a speed-time graph of a cyclist. Work out the total distance travelled by the cyclist over the course of the journey.

We need to find the area underneath the graph. To do this, we will split it up into shapes we know how to calculate the area of, as seen below.

A is a triangle, B and C are trapeziums, and D is a rectangle. So, we get

\text{A}=\dfrac{1}{2}\times10\times15=75\text{m}

\text{B}=\dfrac{1}{2}\times(10+15)\times5=62.5\text{m}

\text{C}=\dfrac{1}{2}\times(10+20)\times5=75\text{m}

\text{D}=20\times25=500\text{m}

Therefore, the total distance travelled by the cyclist is

75+62.5+75+500=712.5\text{m}

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