# Speed-Time Graphs

## 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}

**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

**Question 3:** A journey is broken up into four sections as shown

a) Describe the acceleration for each section of the journey.

**[3 marks]**

b) Find the total distance travelled.

**[3 marks]**

a) For section A: The acceleration is the gradient. Given points (0,0) and (15, 3)

\text{Gradient} = \dfrac{3-0}{15-0} = 0.2

Acceleration is 0.2 \text{ ms}^2

For section B: The acceleration is the gradient. Given points (15,3) and (20, 5)

\text{Gradient} = \dfrac{5-3}{20-15} = 0.4

Acceleration is 0.4 \text{ ms}^2

For section C: As this is a flat line the acceleration is 0 \text{ ms}^2

For section D: The acceleration is the gradient. Given points (40,5) and (50, 0)

\text{Gradient} = \dfrac{0-5}{50-40} = -0.5

The deceleration is 0.5 \text{ ms}^2

b) The total distance travelled is the area under the graph.

\text{Area }A = \dfrac{1}{2}\times 15 \times 2 = 22.5

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

\text{Area }C = 20 \times 5 = 100

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

So the total area is

\text{Total Area} = 22.5+20+100+25= 167.5

So the distance travelled is 167.5 \text{ m}

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