Speed-Time Graphs

GCSELevel 4-5Edexcel iGCSE

Speed-Time Graphs Revision

Speed-Time Graphs

Speed-time graphs (or velocity-time graphs) graph an object’s movement over time.

Time is plotted along the $x$ -axis and is measured in a unit of time, usually seconds.

Speed is plotted along the $y$ -axis and is usually measured in meters per second ( $\text{m} / \text{s}$ or $\text{ms}^{-1}$ )

A unique feature about speed-time graphs is how much information we can attain from them

Speed-Time Graph Basics

We may discern from a speed time graph a lot of information about the movement of the object.

The gradient of the graph is the objects acceleration.

A negative gradient means an object is decelerating.

A flat line means the object is moving with a constant speed.

The area under the curve is the distance the object has travelled.

Describing Speed-Time Graphs

We need to be able to describe the speed-time graph. One way to fully describe a speed-time graph is to break up the journey into parts.

Here the graph has been broken up into four sections. $A$ , $B$ , $C$ and $D$ . Describing them individually is easier than describing the graph in one go. In this example

$A$ : The object accelerates from $0 \text{ ms}^{-1}$ to $15 \text{ ms}^{-1}$ over $15$ seconds.

$B$ : Then the object travels at a constant speed of $15\text{ ms}^{-1}$ for $10$ seconds.

$C$ : The object then accelerates again from $15\text{ ms}^{-1}$ to $25\text{ ms}^{-1}$ in $5$ seconds.

$D$ : The object then decelerates from $25\text{ ms}^{-1}$ to stationary in $20$ seconds.

Calculating Acceleration

$\text{Acceleration }$ $=$ $\text{ The Gradient of the Line}$

Using the previous example, we shall still divide the graph and find the gradient for each section.

Remember that

$\text{Gradient} = \dfrac{\text{Difference in } y}{\text{Difference in } x}$

given that we can find the acceleration, which is measured in meters per second $^{2}$ or $\text{ms}^{-2}$ .

$A$ : The gradient is then the difference in speed divided by the difference in time, which here gives us

$\text{Acceleration}= \dfrac{15-0}{15-0} = 1 \text{ ms}^{-2}$

$B$ : Here the graph is flat, so it has no acceleration.

$C$ : The acceleration here is

$\text{Acceleration}= \dfrac{25-15}{30-25} = 2\text{ ms}^{-2}$

$D$ : The acceleration here is

$\text{Acceleration}= \dfrac{0-25}{50-30} = -1.25 \text{ ms}^{-2}$

Negative acceleration is deceleration, so here we can say instead that it decelerates at a rate of 1.25 \text{ ms}^{-2}1.25 ms−2

Calculating Distance

$\text{Distance }$ $=$ $\text{ Area Under the Graph}$

Using the previous example, once again it makes sense to divide the graph into sections. Here the area between the line and the $x$ -axis forms a few easier to calculate shapes.

Section $A$ is a triangle, $B$ a rectangle, $C$ a trapezium and $D$ is another triangle. We know the height and with of all these sections, so the total distance travelled is just the sum of their areas.

$\textbf{Area A}= \dfrac{1}{2} \times 15 \times 15 = 112.5$

$\textbf{Area B}= 15 \times 10 = 150$

$\textbf{Area C}= \dfrac{15+25}{2} \times 5 = 100$

$\textbf{Area D}= \dfrac{1}{2} \times 25 \times 20 = 250$

So the total distance travelled is the sum of these areas

$\text{Distance Travelled} = 112.5+150+100+250$

$\text{Distance Travelled} = 612.5 \text{ m}$

Example: Drawing a Speed-Time Graph

A car journey begins with the car stationary. It then accelerates at a constant rate to $30 \text{mph}$ in $10$ seconds. It then travels at a constant speed for $30$ seconds, then speeds up linearly to $50\text{mph}$ in $5$ seconds only to decelerate for $20$ seconds until it comes to a stop.

Draw this speed-time graph.

[4 marks]

We shall break this journey up into four sections.

Firstly, the car starts at $0 \text{ mph}$ at $0$ seconds, then the car accelerates for $10$ seconds until it is at $30 \text{mph}$ , so there is straight line from $(0,0)$ to $(10, 30)$ .

Next, the car travels at a constant speed of $30\text{mph}$ for $30$ seconds, which gives us a flat line from $(10,30)$ to $(40, 30)$ .

Then the car accelerates for $5$ seconds until it is at $50 \text{mph}$ . For this we draw a straight line from $(40,30)$ to $(45,50)$ .

Finally the car decelerates for $20$ seconds until it is stationary. We then draw a line from $(45,50)$ to $(65, 0)$

This gives us a speed-time graph of

Speed-Time Graphs Example Questions

Question 1: For the following speed-time graph

calculate

a) The total distance travelled

[3 marks]

b) The acceleration between $0$ and $10$ seconds

[2 marks]

a) The total distance travelled is the area under the graph. It will then be easier to break the graph into sections with easy to work out sections.

Here, sections $A$ is a triangle, $B$ and $D$ are trapeziums and $C$ is a rectangle.

\text{Area }A = \dfrac{1}{2}\times 10 \times 5 = 25

$\text{Area }B = \dfrac{5+3}{2}\times 5 = 20$

$\text{Area }C = 15 \times 3 = 45$

$\text{Area }D = \dfrac{3+5}{2}\times 20 = 80$

So the total area is

$\text{Total Area} = 25+20+45+80=170$

So the distance travelled is $170 \text{ m}$

b) The acceleration is the gradient. Between $0$ and $10$ seconds it has a constant constant gradient as it is a straight line. So using the points $(0,0)$ and $(10,5)$ we get a gradient of

$\text{Gradient} = \dfrac{5-0}{10-0} = 0.5$

So the acceleration is $0.5 \text{ ms}^{-2}$

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Question 2: A metro train has a short journey. As it departs from the station and accelerates at a constant rate for $15$ seconds until is reaches a speed of $20 \text{ m/s}$ then it travels at a constant speed for $45$ seconds. It then decelerates at a rate of $1.25 \text{ ms}^{-2}$ until it comes to a halt.

Draw a speed-time graph for the train’s journey.

[4 marks]

We begin the journey at $0$ seconds and $0 \text{ m/s}$ . We can then draw a straight line from $(0,0)$ to $(15, 20)$ as the train accelerates. Then we may draw a flat line from $(15,20)$ to $(60, 20)$ as the train travels at a constant speed.

We need to work out how long the train takes to come to a stop. If it was increasing in speed at $1.25 \text{ ms}^{-2}$ to $20\text{ m/s}$ we would have

$1.25 t = 20$

$t = \dfrac{20}{1.25} = 16 \text{ s}$

Therefor it would take $16$ seconds for the train to decelerate from $20\text{ m/s}$ to $0\text{ m/s}$ at the same rate.

So we may draw a straight line from $(60, 20)$ to $(76, 0)$ , giving us a graph